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Effect of Slope on Bearing Capacity

By
Ubani Obinna
-

In some cases, the foundation of buildings may be founded on or very close to natural or man-made earth slopes. The bearing capacity of shallow foundations on slopes is reduced because the full formation of shear zones under ultimate loading conditions is not possible on the sides close to the slopes or edges. This is due to the fact that the zones of plastic flow in the soil on the slope side are smaller than those in a similar foundation on level ground, thereby resulting in a lower ultimate bearing capacity.

Meyerhof (1957) extended his bearing capacity theory to account for the effect of slope on the bearing capacity of foundations. Many other recent research works have been conducted on the effect of slope on bearing capacity using finite element modelling (Acharyya and Dey, 2017; Chakraboty and Kumar, 2013) and energy dissipation methods (Yang et al, 2007).

Effect of slope on bearing capacity
Figure 1: 4 Plastic zones and slip surfaces near rough strip foundation on top of slope (Meyerhof, 1957)

Figure 1 shows a section of a foundation with the failure surfaces under ultimate loading conditions. The distance b between the top edge of the slope and the foundation face determines the foundation’s stability. For a strip footing, the ultimate bearing capacity equation can be written as:

q = CNcq + 0.5γBNγq —– (1) (Meyerhof, 1957)

where γ = unit weight of soil, B = width of foundation, and Ncq and Nγq = resultant bearing capacity factors depending on the angle of inclination of the slope (β), angle of shearing resistance of the soil (ϕ), and the depth/width ratio D/B of the foundation.

The bearing capacity factors decrease with a greater inclination of the slope to a minimum for β = 90 degrees on purely cohesive material, and β = ϕ on cohesionless soil, when the slope becomes unstable. Because the bearing capacity of cohesionless soils is found to decrease approximately parabolically with an increase in slope angle, the decrease in bearing capacity is small in the case of clays but can be significant in the case of sands and gravels for inclinations of slopes used in practice (β < 30 degrees).

A solution of the slope stability has been obtained for a surcharge across the full horizontal top surface of a slope using dimensionless parameters called the stability number Ns, written as given in Equation (2);

Ns = c/γH —– (2)

Where c is the cohesion of the soil, γ is the unit weight of the soil, and H is the vertical height of the slope.

Eq. (3) represents the bearing capacity of a foundation on purely cohesive soil of great depth;

q = CNcq + γDf —— (3)

where the Ncq factor depends on b as well as β, and the stability number Ns. This bearing capacity factor decreases significantly with increasing height and to a lesser amount with increasing slope inclination, as shown in the lower regions of Figure 2. The bearing capacity factor increases with an increase in b for a given height and slope angle, and beyond a distance of about 2 to 4 times the height of the slope, the bearing capacity is independent of the slope angle.

These bearing capacity factors are given in Figures 2 and 3 for strip foundation in purely cohesive and cohesionless soils respectively.

S1
Figure 2: Bearing capacity factors for strip foundation on top of slope of purely cohesive material (Meyerhof, 1957)
s4
Figure 3: Bearing capacity factors for strip foundation on top of slope of cohesionless material (Meyerhof, 1957)

The bearing capacity factors increase with an increase in distance b. Beyond a distance of about 2 to 6 times the foundation width B, the bearing capacity is independent of the slope’s inclination and becomes the same as that of a foundation on a level surface.

References

[1] Acharyya R. and Dey A. (2017): Finite Element Investigation of the Bearing Capacity of Square Footings Resting on Sloping Ground. INAE Lett (2017) 2:97–105 DOI 10.1007/s41403-017-0028-6
[2] Debarghya Chakraborty & Jyant Kumar (2013) Bearing capacity of foundations on slopes, Geomechanics and Geoengineering, 8:4, 274-285, DOI: 10.1080/17486025.2013.770172
[3] Meyerhof G. G. (1957): The ultimate bearing capacity of foundations on slopes. The Proceedings of the Fourth International Conference on Soil Mechanics and Foundation Engineering, London, August 1957
[4] Yang, Xl., Wang, Zb., Zou, Jf. et al. (2007): Bearing capacity of foundation on slope determined by energy dissipation method and model experiments. J Cent. South Univ. Technol. 14, 125–128 (2007). https://doi.org/10.1007/s11771-007-0025-0

Cone Penetration Test (CPT) in Geotechnical Engineering

By
Ubani Obinna
-

The cone penetration test (CPT), often called the Dutch cone penetration test, is a versatile sounding method for determining the materials in a soil profile and estimating their engineering properties. This test is also known as the static penetration test, and it can be performed without the use of boreholes. In the original form of the test, a 60° cone with a 10 cm2 base area is pushed into the ground at a steady rate of around 20 mm/sec, and the resistance to penetration (called the point resistance) was measured.

The test’s results from cone penetration test can also be used to determine the soil’s bearing capacity at various depths below ground level. This test can also be used to determine the skin friction values that are used to establish the needed pile lengths in a given condition, in addition to bearing capacity values.

The current cone penetrometers used in the field of geotechnical engineering measure;

(a) the cone resistance to penetration, qc, which is equal to the vertical force applied to the cone divided by its horizontally projected area; and
(b) the frictional resistance, fc, which is measured by a sleeve located above the cone with the local soil surrounding it. The frictional resistance is calculated by dividing the vertical force applied to the sleeve by the surface area of the sleeve—in other words, the sum of friction and adhesion.
(c) Pore water pressure

To measure qc and fc, two types of penetrometers are commonly used:

(a) Mechanical friction-cone penetrometer
The penetrometer tip is attached to an inner set of rods in this case as shown in Figure 1. The tip is advanced into the soil to a depth of about 40 mm at first, providing cone resistance. The tip engages the friction sleeve with more thrusting. The rod force is the sum of the vertical forces on the cone and sleeve as the inner rod advances. The side resistance is calculated by subtracting the force on the cone.

mechanical cone penetrometer
Figure 1: Begemann friction-cone mechanical type penetrometer (Murthy, 2002)

(b) Electric friction-cone penetrometer
The tip is fastened to a string of steel rods in this case as shown in Figure 2. At a rate of 20 mm/sec, the tip is pushed into the soil. Wires from the transducers are threaded through the center of the rods and continuously give the cone and side resistances.

electrical cptu
Figure 2: Electric friction cone penetrometer

The static cone penetration test works best in soft or loose soils like silty sands, loose sands, layered deposits of sands, silts, and clays, as well as clayey deposits. Static cone penetration tests can be used to avoid the usage of test piles and loading tests in regions where some knowledge about the foundation strata is already available.

The results from a typical static cone penetration test are given in Figure 3;

Typical cone penetration test result
Figure 3: Typical Cone Penetration Test Result (Cavallaro et al, 2018)

For the cone resistance, qc, and the friction ratio, Fr, obtained from cone penetration tests, several correlations have been developed that are useful in evaluating the properties of soils encountered during an exploratory program. Fr stands for friction ratio and is defined as;

Fr = frictional resistance/cone resistance = fc/qc

In a more recent study on several soils in Greece, Anagnostopoulos et al. (2003) expressed Fr as;

Fr(%) = 1.45 – 1.36 logD50 (electric cone)
and
Fr(%) = 0.7811 – 1.611 logD50 (mechanical cone)

where D50 is size through which 50% of soil will pass through (mm). The D50 for soils on which the equations are based have been developed ranged from 0.001 mm to about 10 mm.

Soil classification based on friction ratio Fr was proposed by Sanglerat, (1972) and shown in the Table below;

Fr (%)Soil classification
0 – 0.5Loose gravel fill
0.5 – 2.0 Sands or gravels
2 – 5Clay sand mixtures and silts
> 5Clays, peats etc.

Correlation of Cone Penetration Test Result with Engineering Properties of Soil

The relative density of normally consolidated sand, Dr, and qc can be correlated according to the formula;
Dr (%) = 68[log(qc/√Poσ’o) – 1]

where;
Po = atmospheric pressure
σ’o = vertical effective stress

The variation of Dr, σ’o, qc for normally consolidated quartz sand is shown in Figure 4;

Variation
Figure 4: Relationship between relative density Dr and penetration resistance qc for uncemented quartz sands (Robertson and Campanella, 1983)

Correlation between qc and Drained Friction Angle (ϕ’) for Sand
On the basis of experimental results, Robertson and Campanella (1983) suggested the variation of qc, ϕ’, and σ’o for normally consolidated quartz sand. This relationship was expressed in (Kulhawy and Mayne, 1990);

ϕ’ = tan-1[0.1 + 0.38 log(qc/σ’o)]

Based on the cone penetration tests on the soils in the Venice Lagoon (Italy), Ricceri et al. (2002) proposed a similar relationship for soil with classifications of ML and SP-SM as;

ϕ’ = tan-1[0.38 + 0.27 log(qc/σ’o)]

In a more recent study, Lee et al. (2004) developed a correlation between ϕ’, qc, and the horizontal effective stress (σ’h) in the form;

ϕ’ = 15.575(qc/σ’h)0.1714

Correlation between qc and Undrained Shear Strength (Cu) for Clays

The undrained shear strength, cu, can be expressed as;

cu = (qc – σ’o)/NK

where
σ’o = total vertical stress
NK = bearing capacity factor

The bearing capacity factor, NK, may vary from 11 to 19 for normally consolidated clays and may approach 25 for overconsolidated clay. According to Mayne and Kemper (1988);

NK = 15 (for electric cone)
and
NK = 20 (for mechanical cone)

Based on tests in Greece, Anagnostopoulos et al. (2003) determined;

NK = 17.2 (for electric cone)
and
NK = 18.9 (for mechanical cone)

These field tests also showed that;
cu = fc/1.26 (for mechanical cones) and
cu = fc (for electrical cones)

Correlation between qc and Ultimate Bearing Capacity

To estimate the allowable bearing capacity from cone penetration tests, a set of empirical equations developed by Schmertmann (1978) and listed below are used;

For cohesionless soils:
Strip footing: qult = 28 – 0.0052 (300 – qc)1.5 (kg/cm2)
Square footing: qult = 48 – 0.009 (300 – qc)1.5 (kg/cm2)

For clay:
Strip footing: qult = 2 + 0.28qc (kg/cm2)
Square footing: qult = 5 + 0.34qc (kg/cm2)

where:
qult = ultimate bearing capacity
qc = cone friction averaged over the depth interval from about B/2 to 1.1B below the footing base with B is the foundation’s width.

References

[1] Anagnostopoulos, A., Koukis, G., Sabatakakis, N. et al. (2003): Empirical correlations of soil parameters based on Cone Penetration Tests (CPT) for Greek soils. Geotechnical and Geological Engineering 21, 377–387 (2003). https://doi.org/10.1023/B:GEGE.0000006064.47819.1a
[2] Cavallaro, A.; Capilleri, P.P.; Grasso, S. (2018): Site Characterization by Dynamic In Situ and Laboratory Tests for Liquefaction Potential Evaluation during Emilia Romagna Earthquake. Geosciences 2018, 8, 242. https://doi.org/10.3390/geosciences8070242
[3] Kulhawy, F. H., and Mayne, P. W. (1990). Manual on Estimating Soil Properties for Foundation Design, Electric Power Research Institute, Palo Alto, CA.
[4] Lee, J., Salgado, R., and Carraro, A. H. (2004). “Stiffness Degradation and Shear Strength of Silty Sand,” Canadian Geotechnical Journal, Vol. 41, No. 5, 831–843.
[5] Mayne, P. W., and Kemper J. B. (1988). “Profiling OCR in Stiff Clays by CPT and SPT,” Geotechnical Testing Journal, ASTM, Vol. 11, No. 2, 139–147.
[6] Murthy, V.N.S. (2002). Principles and Practices of Soil Mechanics and Foundation Engineering. CRC Press, Florida
[7] Ricceri, G., Simonin, P., and Cola, S. (2002). “Applicability of Piezocone and Dilatometer to Characterize the Soils of the Venice Lagoon” Geotechnical and Geological Engineering, Vol. 20, No. 2, 89–121.
[8] Robertson, P. K., and Campanella, R. G. (1983). “Interpretation of Cone Penetration Tests. Part I: Sand,” Canadian Geotechnical Journal, Vol. 20, No. 4, 718–733.
[9] Sanglerat, G. (1972). The Penetrometer and Soil Exploration. Elsevier Publishing Co., Amsterdam
[10] Schmertmann, J.H. (1978). Guidelines for Cone Penetration Test: Performance and Design. U.S. Dept. of Transportation, Washington, D.C.



Moving Wheel Loads from Overhead Electric Travelling Cranes

By
Ubani Obinna
-

When an overhead electric travelling crane is operating in a workshop or warehouse, a dynamic effect on the wheels is created as a result of a sudden drop of a full load, a slide of the sling, or a sudden braking action during the trip of a fully-loaded crane (where the load includes the crane’s self-weight). This consequently increases the static wheel load of the crane.

This effect is determined by multiplying the static wheel load by an impact factor to obtain the dynamic wheel load. Thus, maximum dynamic vertical wheel load = static wheel load × dynamic factor (ϕ).

The dynamic factor varies according to the crane’s duty class (loading class). The various loading classes are listed in Table 1. Table 2 lists the loads and dynamic parameters that should be addressed as part of a single crane motion. The values of the dynamic factors ϕi should be calculated using the formulae presented in Table 3.

Table 1: Recommendations for loading classes (based on Table B.1 in Eurocode 1, Part 3)a

ItemType of craneHoisting ClassS-Class
1Hand-operated cranesHC1S0, S1
2Assembly cranesHC1, HC2S0, S1
3Powerhouse cranesHC1S1, S2
4Storage cranes with intermittent operationHC2S4
5Storage cranes and spreader bar cranes, with continuous operationHC3, HC4S6, S7
6Workshop cranesH2, H3S3, S4
7Overhead travelling cranes and ram cranes, |with grab and magnet operationHC3, HC4S6, S7
8Casting cranesHC2, HC3S6, S7
9Soaking-pit cranesHC3, HC4S7, S8
10Stripper cranes, charging cranesHC4S8, S9
11Forging cranesHC4S6, S7

Table 2: Groups of loads and dynamic factors to be considered as one characteristic crane
action (based on Table 2.2 of Eurocode 1, Part 3)a

CRANE LOADING

Table 3: Dynamic factors ϕi for vertical loads (based on Table 2.4 of Eurocode 1, Part 3)

Dynamic factor ϕiValue of dynamic factor
ϕ10.9 < ϕ1 < 1.1 The two values 1.1 and 0.9 reflect the upper and lower values of vibrational pulses
ϕ2ϕ2 = ϕ2,min + β2vh, where vh = steady hoisting speed in m/s For ϕ2,min and β2, see Table 4
ϕ3ϕ3 = 1 − △m (1 + β3)/m, where △m = released or dropped part of the hoisting mass, m = total hoisting mass, β3 = 0.5 for cranes equipped with grabs or similar slow-release devices, and β3 = 1.0 for cranes equipped with magnets or similar rapid-release devices.
ϕ4ϕ4 = 1.0 provided that the tolerances for rail tracks as specified in EN 1993-6 are observed

Table 4: Values of β2 and ϕ2,min (based on Table 2.5 of Eurocode 1, Part 3)

Hoisting class of applianceβ2ϕ2,min
HC10.171.05
HC20.341.10
HC30.511.15
HC40.681.2

Example
The vertical dynamic factors can be evaluated as follows.
For hoisting classes HC1and HC2, for example, referring to Table 3, we have the dynamic factor ϕ1 for vertical loads: 0.9 < ϕ1 < 1.1. Assume ϕ1 = 0.9, the lower value for vibrational pulses. We also have;

ϕ2 = ϕ2,min + β2vh

where ϕ2,min = 1.05 and β2 = 0.17 for hoisting class HC1, and ϕ2,min = 1.1 and β2 = 0.34 for hoisting class HC2 (see Table 2.4).
In addition, Vb = steady hoisting speed = 1.3 m/s (assumed).

Therefore
for class HC1: dynamic factor = ϕ2 = 1.05 + 0.17 × 1.3 = 1.27;
for class HC2: dynamic factor = ϕ2 = 1.1 + 0.34 × 1.3 = 1.54.

Thus, referring to Table 2.2 and assuming the group of loads 1, we have the following:

for class HC1: ϕ = ϕ1ϕ2 = 0.9 × 1.27 = 1.14;
for class HC2: ϕ = ϕ1ϕ2 = 0.9 × 1.54 = 1.38.

The following vertical dynamic factors may be used as guidance for various hoisting classes:

  1. For hoisting class HC1, light-duty hand-operated cranes, assembly cranes, powerhouse cranes, and intermittently used storage cranes: dynamic factor ϕ = 1.1 (minimum) to 1.25.
  2. For hoisting class HC2, medium-duty cranes (normally in factories, workshops, and warehouses, and for casting and in scrapyards with continuous operation): dynamic factor ϕ = 1.25 to 1.4.
  3. For hoisting class HC3, heavy-duty cranes (in foundries and for intermittent grab and magnet work, forging, charging etc.): dynamic factor ϕ = 1.4 (minimum).

Generally, the crane manufacturer will provide the dynamic factor along with the crane wheel loads when details of the duty (class), the span of the crane, and the lifting capacity are given to the manufacturer.

crane lifting load

Transverse Horizontal Force (surge) on a Crane Girder

The following elements contribute to the transverse horizontal surge:

Thrust caused by the crab motor’s brakes being applied suddenly, causing the crab and cargo to come to a halt while traversing the crab girders. The frictional force created between the crab wheels and crab girders resists this thrust, which is then transferred to the crane’s crosshead girders and lastly transferred as point loads through the crane’s main wheels into the top flange of the crane girders.

Weights are frequently dragged across the manufacturing floor by a crane. This pulling motion creates a transverse horizontal component of force (a point load) on the crane girders through the crane wheels if the weight is particularly heavy. The crane girders receive the transverse horizontal force created by either of the above reasons or a combination of both through the double-flanged crane wheels on the end carriages, and cranes are intended to avoid derailment. It is difficult to quantify the value of this force because there are unknown components in addition to the preceding facts. The horizontal transverse force on each gantry girder must equal 10% of the load lifted, according to American specifications.

The British code of practice BS 2573-1: 1983 (British Standards Institution, 1983) specifies the following:

Value of total transverse horizontal force = 1/10 × weight of (lift load + crab). Eurocode 1, Part 3 stipulates the same value. Therefore value of total transverse horizontal force = 1/10 × weight of (lift load + crab).

This force should be shared equally between the two gantry girders.

Longitudinal horizontal force

The rapid application of brakes during crane travel causes frictional resistance to the locked wheels sliding along a rail attached to the gantry girder. This frictional resistance provides a horizontal force down the length of the gantry girder, which then transmits to the gantry girder’s supporting columns. Assume that steel sliding on steel has a coefficient of friction of 0.2. Consider the gantry girder’s maximum vertical wheel load, which happens when the weight lifted is at the closest allowed point to the girder.

So, maximum wheel load on the nearest gantry girder = maximum reaction from crane (load lifted + half the dead weight of crane) = W = R.

For example, if the load lifted is W1, the self-weight of the crane is W2, the distance of the load lifted from the nearest gantry girder is l and the crane span (centre to centre of crosshead) is L, then;

Maximum on-wheel load = W1(L − 1)/L + W2/2 = W = R.

Therefore longitudinal horizontal force developed = Rμ = 0.2R.

The American code of practice specifies that the longitudinal force is equal to 10% of the maximum wheel load. The British code of practice BS 2573 specifies that the longitudinal force is equal to 5% of the maximum wheel load, assumed to be acting on one gantry girder nearest to the load lifted. Eurocode 1 stipulates that the longitudinal force applied to the gantry girder should be calculated as follows (the equation numbers given in this chapter refer to Eurocode 1, Part 3):

HL,i = ϕ5Ki/nr

where
nr = number of gantry girders = 2
K = driving force (the value should be provided by the crane supplier),
ϕ5 = dynamic factor (see Table 5),
i = integer to identify the gantry girder (i = 1, 2).

Table 5: Dynamic factor ϕ5 (based on Table 2.6 of Eurocode 1, Part 3)

Value of the dynamic factor ϕ5Specific use
ϕ5 = 1For centrifugal forces
1.0 ≤ ϕ5 ≤ 1.5For systems where forces change smoothly
1.5 ≤ ϕ5 ≤ 2.0For cases where sudden changes can occur
ϕ5 = 3.0For drives with considerable backlash

Design of Steel Columns for Biaxial Bending | Eurocode 3

By
Ubani Obinna
-

When a column section is subjected to bending moment in the two axes in addition to a compressive axial force, the column is said to be biaxially loaded. The design of steel columns for biaxial bending involves the verification of the steel section’s capacity in bending, shear, compression, flexural buckling, and interaction of all these forces. Interaction formulas are available in EN 1993-1-1:2005 (Eurocode 3) for the design of members that are biaxially loaded.

Clause 6.2.9 of EN 1993-1-1:2005 describes the design of cross-sections subjected to combined bending and axial force (such as steel columns). Bending can occur along one or both major axes, with tensile or compressive axial forces (with no difference in treatment). Eurocode 3 provides several approaches for designing Class 1 and 2, Class 3, and Class 4 cross-sections in order to deal with the combined effects.

A basic linear interaction presented below and in equation (1) can be applied to all cross-sections (clause 6.2.1(7)). Although Class 4 cross-section resistances must be based on effective section properties. Furthermore, any additional moments arising from the resulting shift in neutral axis should be allowed for in class 4 sections. These extra moments necessitate the use of the expanded linear interaction expression.

NEd/NRd + My;Ed/My;Rd + Mz;Ed/Mz;Rd ≤ 1.0 ———— (1)

where NRd, My,Rd, and Mz,Rd are the design cross-sectional resistances, and any required reduction due to shear effects should be included (clause 6.2.8). The goal of equation (1) is to allow a designer to obtain a quick, approximate, and safe solution, possibly for initial member sizing, with the option to refine the calculations for final design.

Bi-axial bending with or without axial force (Class 1 and 2 sections)

EN 1993-1-1, like BS 5950: Part 1, treats bi-axial bending as a subset of the combined bending and axial force regulations. Clause 6.2.9.1 specifies the checks for Class 1 and 2 cross-sections subjected to bi-axial bending with or without axial forces (6). Although equation (1) shows a simple linear interaction expression, equation (2) represents a more sophisticated convex interaction expression that can result in large efficiency gains:

(My;Ed/MN;y;Rd)α + (Mz;Ed /MN;z;Rd)β ≤ 1.0 ———- (2)

in which α and β are constants, as defined below. Clause 6.2.9(6) allows α and β to be taken as unity, thus reverting to a conservative linear interaction.

For I- and H-sections:
α = 2 and β = 5n but β ≤ 1.0

For circular hollow sections:
α = 2 and β = 1

Rectangular hollow sections
α = β = 1.66/(1 – 1.13n2) but α = β ≤ 6.0

n = NEd / Nc,Rd

Worked Example

Verify the capacity of a 3500 mm tall column of UKC 254x254x89 in a commercial complex to withstand the following ultimate limit state actions;

Design of Steel Columns for Biaxial Bending

Axial load; NEd = 1500 kN; (Compression)
Major axis moment at end 1 – Bottom; My,Ed1 = 89.0 kNm
Major axis moment at end 2 – Top; My,Ed2 = 77.0 kNm
Minor axis moment at end 1 – Bottom; Mz,Ed1 = 7.9 kNm
Minor axis moment at end 2 – Top; Mz,Ed2 = 2.4 kNm
Major axis shear force; Vy,Ed = 56 kN
Minor axis shear force;Vz,Ed = 14 kN

Solution

Partial factors
Resistance of cross-sections; γM0 = 1
Resistance of members to instability; γM1 = 1
Resistance of cross-sections in tension to fracture;  γM2 = 1.1

Column details
Column section; UKC 254x254x89
Steel grade; S275
Yield strength; fy = 265 N/mm2
Ultimate strength;  fu = 410 N/mm2
Modulus of elasticity;  E = 210 kN/mm2
Poisson’s ratio;  υ = 0.3
Shear modulus; G = E / [2 × (1 + υ)] = 80.8 kN/mm2

Column geometry
System length for buckling – Major axis; Ly = 3500 mm
System length for buckling – Minor axis; Lz = 3500 mm

The column is part of a sway frame in the direction of the minor axis
The column is part of a sway frame in the direction of the major axis

Column loading
Axial load; NEd = 1500 kN; (Compression)
Major axis moment at end 1 – Bottom; My,Ed1 = 89.0 kNm
Major axis moment at end 2 – Top; My,Ed2 = 77.0 kNm

Minor axis moment at end 1 – Bottom; Mz,Ed1 = 7.9 kNm
Minor axis moment at end 2 – Top; Mz,Ed2 = 2.4 kNm
Major axis shear force; Vy,Ed = 56 kN
Minor axis shear force;Vz,Ed = 14 kN

Buckling length for flexural buckling – Major axis
End restraint factor; Ky = 1.000
Buckling length; Lcr_y = Ly × Ky = 3500 mm

Buckling length for flexural buckling – Minor axis
End restraint factor;  Kz = 1.000
Buckling length;  Lcr_z = Lz × Kz = 3500 mm

Web section classification (Table 5.2)
fy = 265 N/mm2
Coefficient depending on fy; ε = √(235/ fy) = 0.942
Depth between fillets; cw = h – 2 × (tf + r) = 200.3 mm
Ratio of c/t;  ratiow = cw / tw = 19.45
Length of web taken by axial load; lw = min(NEd / (fy × tw), cw) = 200.3 mm
For class 1 & 2 proportion in compression; α = (cw/2 + lw/2) / cw = 1.000

Limit for class 1 web;                                                     
Limit1w = (396 × e) / (13 × a – 1) = 31.08
The web is class 1

Flange section classification (Table 5.2)
Outstand length; cf = (b – tw) / 2 – r = 110.3; mm
Ratio of c/t; ratiof = cf / tf = 6.38

Conservatively assume uniform compression in flange

Limit for class 1 flange; Limit1f = 9 × e = 8.48
Limit for class 2 flange; Limit2f = 10 × e = 9.42
Limit for class 3 flange; Limit3f = 14 × e = 13.18

The section is class 1

Resistance of cross section (cl. 6.2)

Shear – Major axis (cl. 6.2.6)
Design shear force; Vy,Ed = 56.0 kN
Shear area; Avy = max((h – 2tf) × tw, A – 2 × b × tf + (tw + 2 × r) × tf) = 3081 mm2
fy = 265 N/mm2
Plastic shear resistance;  Vpl,y,Rd = Avy × (fy/√3)/ γM0 = 471.4 kN
Vy,Ed / Vpl,y,Rd = 0.119
PASS – Shear resistance exceeds the design shear force

Vy,Ed ≤ 0.5×Vpl,y,Rd – No reduction in fy required for bending/axial force

Shear – Minor axis (cl. 6.2.6)
Design shear force; Vz,Ed = 13.5 kN
Shear area; Avz = 2 × b × tf – (tw + 2 × r) × tf = 8250 mm2
Plastic shear resistance; Vpl,z,Rd = Avz × (fy /√3) / γM0 = 1262.3 kN
Vz,Ed / Vpl,z,Rd = 0.011
PASS – Shear resistance exceeds the design shear force
Vz,Ed ≤ 0.5×Vpl,z,Rd – No reduction in fy required for bending/axial force

Compression (cl. 6.2.4)
Design force; NEd = 1500 kN
Design resistance; Nc,Rd = Npl,Rd = A × fy / γM0 = 3003 kN
NEd / Nc,Rd = 0.5
PASS – The compression design resistance exceeds the design force

Bending – Major axis (cl. 6.2.5)
Design bending moment;  My,Ed = max(abs(My,Ed1), abs(My,Ed2)) = 89.0 kNm
Section modulus;  Wy = Wpl.y = 1223.9; cm3
Design resistance; Mc,y,Rd = Wy × fy / γM0 = 324.3 kNm
My,Ed / Mc,y,Rd = 0.274
PASS – The bending design resistance exceeds the design moment

Bending – Major axis(cl. 6.2.5)
Design bending moment; Mz,Ed = max(abs(Mz,Ed1), abs(Mz,Ed2)) = 7.9 kNm
Section modulus; Wz = Wpl.z = 575.3; cm3
Design resistance; Mc,z,Rd = Wz × fy / γM0 = 152.5 kNm
Mz,Ed / Mc,z,Rd = 0.052
PASS – The bending design resistance exceeds the design moment

Combined bending and axial force (cl. 6.2.9)
fy = 265 N/mm2;
Npl,Rd = A × fy / γM0 = 3003 kN
Ratio design axial to design plastic resistance; n = abs(NEd) / Npl,Rd = 0.500
Ratio web area to gross area; a = min(0.5, (A – 2 × b × tf) / A) = 0.217

Bending – Major axis (cl. 6.2.9.1)
Design bending moment; My,Ed = max(abs(My,Ed1), abs(My,Ed2)) = 89.0 kNm
Plastic design resistance; Mpl,y,Rd = Wpl.y × fy / γM0 = 324.3 kNm
Modified design resistance; MN,y,Rd = Mpl,y,Rd × min(1, (1 – n) / (1 – 0.5 × a)) = 182.1 kNm
My,Ed / MN,y,Rd = 0.489
PASS – Bending resistance in presence of axial load exceeds the design moment

Bending – Minor axis (cl. 6.2.9.1)
Design bending moment;Mz,Ed = max(abs(Mz,Ed1), abs(Mz,Ed2)) = 7.9 kNm
Plastic design resistance;  Mpl,z,Rd = Wpl.z × fy / γM0 = 152.5 kNm
Modified design resistance;MN,z,Rd = Mpl,z,Rd × [1 – ((n – a) / (1 – a))2] = 132.6; kNm
Mz,Ed / MN,z,Rd = 0.059
PASS – Bending resistance in presence of axial load exceeds the design moment

Biaxial bending
Exponent α; α = 2.00
Exponent β;  β = max(1, 5 × n) = 2.50

Section utilisation at end 1; URCS_1 = [abs(My,Ed1) / MN,y,Rd] α + [abs(Mz,Ed1) / MN,z,Rd] β = 0.240
Section utilisation at end 2; URCS_2 = [abs(My,Ed2) / MN,y,Rd] α + [abs(Mz,Ed2) / MN,z,Rd] β = 0.179
PASS – The cross-section resistance is adequate

Buckling resistance (cl. 6.3)
Yield strength for buckling resistance;  fy = 265 N/mm2

Flexural buckling – Major axis
Elastic critical buckling force; Ncr,y = π2 × E × Iy / Lcr_y2 = 24140 kN
Non-dimensional slenderness; λy = √(A × fy / Ncr,y) = 0.353
Buckling curve (Table 6.2);  b
Imperfection factor (Table 6.1); αy = 0.34
Parameter Φ;  Φy = 0.5 × [1 + αy × (λy – 0.2) + λy2] = 0.588
Reduction factor;  χy = min(1.0, 1 / [Φy + √(Φy 2 – λy2)]) = 0.944
Design buckling resistance; Nb,y,Rd = χy × A × fy  / γM1 = 2835.9 kN
NEd / Nb,y,Rd = 0.529
PASS – The flexural buckling resistance exceeds the design axial load

Flexural buckling – Minor axis
Elastic critical buckling force; Ncr,z = π2 × E × Iz / Lcr_z2 = 8219 kN
Non-dimensional slenderness; λz = √(A × fy / Ncr,z) = 0.604
Buckling curve (Table 6.2);  c
Imperfection factor (Table 6.1); αz = 0.49
Parameter Φ; Φz = 0.5 × [1 + αz × (λz – 0.2) + λz2] = 0.782
Reduction factor; χz = min(1.0, 1 / [Φz + √(Φz2 – λz2)]) = 0.783
Design buckling resistance; Nb,z,Rd = χz × A × fy  / γM1 = 2350.4 kN
NEd / Nb,z,Rd = 0.638
PASS – The flexural buckling resistance exceeds the design axial load

Torsional and torsional-flexural buckling (cl. 6.3.1.4)

Torsional buckling length factor; KT = 1.00
Effective buckling length; Lcr_T = KT × max(Ly, Lz) = 3500 mm
Distance from shear ctr to centroid along major axis;  y0 = 0.0 mm
z0 = 0 mm
Distance from shear ctr to centroid along minor axis; z0 = 0.0 mm

i0 = √(iy2 + iz2 + y02 + z02) = 129.9 mm
bT = 1 – (y0 / i0)2 = 1.000

Elastic critical torsional buckling force; Ncr,T = 1 / i02 × (G × It + π2 × E × Iw / Lcr_T2) = 12085 kN
Elastic critical torsional-flexural buckling force; Ncr,TF = Ncr,y/(2 × bT) × [1 + Ncr,T/Ncr,y – √[(1 – Ncr,T/Ncr,y)2 + 4 × (y0/i0)2 × Ncr,T/Ncr,y]]
Ncr,TF = 12085 kN

Non-dimensional slenderness; λT = √(A × fy / min(Ncr,T, Ncr,TF)) = 0.498
Buckling curve (Table 6.2); c
Imperfection factor (Table 6.1); αT = 0.49
Parameter Φ; ΦT = 0.5 × [1 + αT × (λT – 0.2) + λT2] = 0.697
Reduction factor; χT = min(1.0, 1 / [ΦT + √(ΦT 2 – λT2)]) = 0.844
Design buckling resistance; Nb,T,Rd = χT × A × fy  / γM1 = 2533.9 kN
NEd / Nb,T,Rd = 0.592
PASS – The torsional/torsional-flexural buckling resistance exceeds the design axial load

Minimum buckling resistance
Minimum buckling resistance; Nb,Rd = min(Nb,y,Rd, Nb,z,Rd, Nb,T,Rd) = 2350.4 kN
NEd / Nb,Rd = 0.638
PASS – The axial load buckling resistance exceeds the design axial load

Buckling resistance moment (cl.6.3.2.1)
Lateral torsional buckling length factor;  KLT = 1.00
Effective buckling length; Lcr_LT = KLT × Lz = 3500 mm
End moment factor; y = My,Ed2 / My,Ed1 = 0.865
Moment distribution correction factor (Table 6.6);  kc = 1 / (1.33 – 0.33 × y) = 0.957
C1 = 1 / kc2 = 1.091
Curvature factor;  g = √[1 – (Iz / Iy)] = 0.812
Poissons ratio;  υ = 0.3
Shear modulus;  G = E / [2 × (1 + υ)] = 80769 N/mm2

Elastic critical buckling moment; Mcr = C1 × π2 × E × Iz × √[Iw / Iz + Lcr_LT2 × G × It /(π2 × E × Iz)]/(Lcr_LT2 × g)
Mcr = 1739.3 kNm

Slenderness ratio for lateral torsional buckling; λLT = √[Wy × fy / Mcr] = 0.432
Limiting slenderness ratio; λLT,0 = 0.40
Correction factor for rolled sections; βr = 0.75
Buckling curve (Table 6.5); b
Imperfection factor (Table 6.1); αLT = 0.34

Parameter ΦLT;   ΦLT = 0.5 × [1 + αLT × (λLT – λLT,0) + βr × λLT2] = 0.575
Reduction factor; χLT = min(1.0, 1/λLT2, 1 / [ΦLT + √(ΦLT2 – βr × λLT2)]) = 0.988
Modification factor;   f = min(1 – 0.5 × (1 – kc) × [1 – 2 × (λLT – 0.8)2], 1) = 0.984
Modified LTB reduction factor – eq 6.58; χLT,mod = min(χLT / f, 1, 1/λLT2) = 1.000

Design buckling resistance moment; Mb,Rd = χLT,mod × Wy × fy / γM1 = 324.3 kNm
Design bending moment; My,Ed = max(abs(My,Ed1), abs(My,Ed2)) = 89.0 kNm
My,Ed / Mb,Rd = 0.274
PASS – The design buckling resistance moment exceeds the maximum design moment

Combined bending and axial compression (cl. 6.3.3)
Characteristic resistance to normal force;  NRk = A × fy = 3003 kN
Characteristic moment resistance – Major axis; My,Rk = Wpl.y × fy = 324.3 kNm
Characteristic moment resistance – Minor axis; Mz,Rk = Wpl.z × fy = 152.5 kNm

Moment factor – Major axis; Cmy = 0.9
Moment factor – Minor axis; Cmz = 0.9

Moment distribution factor for LTB;  yLT = My,Ed2 / My,Ed1 = 0.865
Moment factor for LTB;  CmLT = max(0.4, 0.6 + 0.4 × yLT) = 0.946

Interaction factor kyy; kyy = Cmy × [1 + min(0.8, λy – 0.2) × NEd / (χy × NRk / γM1)] = 0.973
Interaction factor kzy;  kzy = 1 – min(0.1, 0.1 × λz) × NEd / ((CmLT – 0.25) × (χz × NRk/γM1)) = 0.945
Interaction factor kzz; kzz = Cmz × [1 + min(1.4, 2 × λz – 0.6) × NEd / (χz × NRk / γM1)] = 1.250
Interaction factor kyz; kyz =  0.6 × kzz = 0.750

Section utilisation;                       
URB_1 = NEd / (χy × NRk / γM1) + kyy × My,Ed / (χLT × My,Rk / γM1) + kyz × Mz,Ed / (Mz,Rk / γM1)
URB_1 = 0.838

URB_2 = NEd / (χz × NRk / γM1) + kzy × My,Ed / (χLT × My,Rk / γM1) + kzz × Mz,Ed / (Mz,Rk / γM1)
URB_2 = 0.965

PASS – The buckling resistance is adequate

Alternatives to the Construction of Foundation of Duplexes on Good Soil

By
Ubani Obinna
-

Simple duplex buildings that are founded on soil with an allowable bearing capacity of 100 kN/m2 and above can be safely and economically supported on pad foundations. This article is aimed at proposing an alternative to the method of construction of foundations of duplexes on good soil (say, safe bearing capacity of 100 kN/m2 and above) in Nigeria.

In Nigeria, the most popular approach to the construction of the foundation of duplexes is the combination of pad foundation and strip foundation. Pad foundations are used to support the columns of the building, while the strip foundation is used to support the block wall (sandcrete masonry units).

step down of column base
Fig 1: Combination of a pad and strip foundation for a duplex

Typically, the size of a pad foundation depends on the expected service load coming from the column, the safe bearing capacity of the soil, and the allowable settlement (whichever governs the design). The thickness of the foundation and the reinforcements required is determined from ultimate limit state considerations such as bending, one-way shear, and punching shear.

On the other hand, the width of strip footings in the foundation of duplexes is usually kept between 675mm – 700 mm. The thickness of the concrete ranges from 150 mm to 225 mm if 9 inches hollow blocks are used. When the depth of the water table is very low below the surface, the depth of the foundation is usually kept between 900 mm and 1200 mm.

Since the strips are usually unreinforced, the recommended thickness of the strip foundation for 9 inches blocks (225 mm blocks) is about 225 mm, such that the pressure dispersal (at 45 degrees) will hit the edges of the foundation without introducing any punching shear (see Figure 2). This recommended option can be economically challenging during the construction of strips for duplexes and the advice of a structural engineer should be sought before a decision is taken.

Unreinforced strip footing construction
Fig 2: Schematics of a strip foundation for a duplex

During the structural design of residential duplexes, it is usually assumed that masonry units (block walls) do not carry any load (framed structure) even though sometimes construction methodology may suggest otherwise. Therefore at the foundation, we expect the strip footing to carry at most the self-weight of the ground floor wall and finishes, which is usually about 10.5 kN/m for a 3m high wall. The ground floor slab load may be transmitted to the strip foundation too.

Generically, the total load transmitted to the strip footing depends on the structural scheme and the construction methodology. For instance, some designers/builders prefer to chain foundations at the damp-proof course (DPC) level. In this case, the entire block wall load of the ground floor is not expected to be transmitted to the strip foundation, but to the columns through the plinth beams.

The Conventional Process of Constructing the Foundation of Duplexes in Nigeria

(a) Setting out
When it has been determined that pad foundations can be used to support a duplex in an area of low water table, the first step in the construction of the foundation is the ‘setting out’. For a regular duplex building, setting out can be achieved using wooden pegs and 2″ x 3″ softwood as the profile board (See Figure 3).

setting out
Fig 3: Typical setting out of a building using the 3-4-5 rule

Using the 3-4-5 method, builders’ square, and lines, the building can be set out accurately. Where available, total stations and laser setting out/levelling equipment can be used to make the job faster. The width of excavation, building lines, and wall centrelines are all established on the profile board using nails as shown in Figure 4.

foundation marker
Fig 4: Typical markings on a profile board

It is important to keep the profile board in place until the ground floor slab is cast. Furthermore, it can also be helpful to transfer reference levels and at least two coordinates (building lines/axes in both directions) to a permanent place peradventure you will need them later after the removal of the profile board.

(b) Excavation
After setting out the building, the next step is to commence excavation. The width of excavation is transferred from the profile board, while the depth of the foundation is determined from the design drawings. Using the reference level, the depth of the foundation is established by the site engineer. If the ground is sloping, the foundation can be stepped with the approval of the design engineer.

It may be possible to excavate the column bases first before the strip or vice versa. But in each case, some portion of the excavated soil must be moved away in order to accommodate the other. From experience, it may be handier to excavate the strips first before excavating the column bases.

In order to have a flat surface for laying blocks, the column bases may have to go deeper than the strips. For instance, if the column base is 300 mm thick, and the strip 150 mm thick, the column base may be stepped down by 200 mm (an additional 50 mm for the blinding) so that when it is cast, it will flush with the level of the strip as shown in Figure 5. For the foundation layout being considered, the excavation works should cost about ₦130,400 using manual labour.

Strip foundation and column base
Fig 5: Schematics of pad and strip foundation construction

(c) Leveling and Compaction of the Excavated Trenches
After the excavation of the trenches, the foundation base should be properly compacted and levelled to receive concrete. The levelling should ensure that the thickness of the concrete is the same wherever required. Wooden pegs or short reinforcement offcuts can be used to establish the required levels. Range and spirit level or laser can be used to transfer the levels from one point to another. For shallow foundations in lateritic or cohesive soils, the excavation should be able to stand on its own without caving in within the 900 – 1200 mm depth. If the foundation is founded on sand, side supports for the excavation will be required.

(d) Blinding
The column base should be blinded using concrete of the same strength as the foundation. The thickness of the blinding should be about 50 mm thick and should be properly consolidated.

(e) Reinforcement Works
The column base reinforcements and the starter bars should be prepared according to the structural drawing. Using the profile board and lines, the starter bars should be mounted in their correct position to avoid cranking of reinforcement later. The site engineer should ensure that the column starter bars are plumb and properly braced to prevent movement during concreting.

typical alignment of columns in duplex construction
Fig 6: Typical alignment of columns in duplex construction

The design engineer and the consulting architect should sign off the placement of the reinforcements after the iron benders are done. While the architect should check setting out and positioning, the structural engineer should check out the following;

(1) The blinding was properly done
(2) The correct bar size, yield strength, and spacing were used for the reinforcements. The Consulting Engineer should request the reinforcement tensile strength test result.
(3) Adequate concrete cover (50 mm) has been provided under the column base reinforcement, away from the blinding.
(4) The column starter bars are centrally positioned on the basket (or according to the drawings) to avoid eccentric loading on the foundation.
(5) The foundation has been pegged and excavated such that the thickness of the concrete when poured will meet the design specifications.

If the reinforcement placement is found satisfactory, approval should be given for concreting.

(f) Concreting
The strip foundation and the column bases should be poured to the specified thickness using the recommended concrete grade. The concrete should be properly vibrated to ensure that minimal voids exist in the concrete mass. An experienced mason should dress the surface and ensure that the established casting levels are properly followed.

(g) Blockwork
After the concreting, the blockwork should be done according to the drawings from the foundation up to the DPC level. The site engineer should ensure that the edges of the block adjacent to the column maintain a concrete cover of about 35 mm. Furthermore, the size of the column obtained should be consistent with the design specifications, since the blocks act as part of the permanent formwork for the column. The edges of the block close to the column should be perfectly aligned and plumb. Some engineers recommend filling the hollow sandcrete blocks with weak concrete.

block work in a foundation
Fig 7: Typical blockwork in the foundation of a duplex

(h) Casting of the column stubs
After the blockwork is complete, the column stubs should be cast. Before this is done, it is important to clean up all the soil that must have entered the base of the column preferably using a high-pressure water jet. This must be done before the carpenter places the formwork. The formwork for the column stubs must be properly braced to avoid bursting during the concrete pouring. The recommended concrete mix should be used, and be properly vibrated to avoid honeycombs.

(i) Backfilling and Compaction
The foundation is usually backfilled using the material excavated from the foundation. To make up level up to the DPC, high-quality lateritic materials or sand should be imported, placed, and compacted up to the required level. The earth material to be used should be non-expansive or problematic. After this process, the damp proof membrane should be installed, and hardcore/BRC wire mesh installed as recommended in the design specification.

(j) Casting of the ground-floor slab
Carpenters should prepare the edge formwork of the building and ensure that it is perfectly level. Levels should also be established internally so that the final surface of the ground floor slab will be flat. After this, the ground floor concrete is poured and finished as appropriate.

The casting of the ground-floor slab process completes the substructure works.

We can see that the process can be quite lengthy, and normally takes about one week to complete if there are no unnecessary delays and if an adequate workforce is used. A schematic representation of the final output is shown in Figure 8.

Conventional Duplex Foundation Construction Schematics
Fig 8: Schematics of complete substructure works in a building using strip foundation

Alternative Approach to Duplex Foundation Construction

If we theoretically assume that block walls do not carry any load, why do we go through the lengthy process of excavating the strips, casting the strip, and forming block walls from the foundation up to the DPC level? The new approach to be recommended in this section intends to boycott this process and probably save cost, and make the construction process faster. This approach may already be in practice somewhere else. The new approach is outlined below;

(1) Setting out
The setting out process is the same as outlined above, but the excavation lines for the width of the strip may be omitted.

(2) Excavation and blinding
At this stage, only the column bases will be excavated to the required depth, and the concrete for blinding poured as technically specified.

(3) Reinforcement Works
The reinforcement works for the column base and column starter bars should be prepared and installed as described in the approach above.

(4) Casting of the column base and stubs
The column base concrete should be poured first and the following day, 75mm thick kickers that depicts the dimensions and orientation of the columns should be formed. The column stub formwork should be installed and properly braced before the concreting of the column stub is done. The height of the column should be up to the grade level or as technically recommended.

(5) Casting of plinth beams
Reinforced concrete plinth beams should be cast near the ground level surface to receive the block walls that will go up to the required ground floor level. The ground to receive the plinth beams should be well prepared, levelled, and made firm to receive the blinding before the beam is cast. The plinth beams should run from column to column and to areas where blockwork is expected. The width of the beam should be the same width as the block to be used or larger. The depth can be determined by the structural engineer but is not expected to exceed 300 mm.

(6) Blockwork
Blockwork is laid from the plinth beam to the required DPC level. After which the filling, compaction, and casting of the ground floor are done. A schematic representation of the final output is shown in Figure 8.

Alternative to duplex construction
Fig 9: Schematics of complete substructure works in a building using plinth beams

Cost Comparison of both Alternatives

Let us now compare the cost of adopting both approaches in the construction of a duplex with the foundation layout shown in Figure 10.

foundation layout
Fig 10: Foundation layout of a typical duplex building

Constructing a plinth beam and ignoring strip foundation for the substructure layout shown in Figure 10 will incur the following costs;

(1) Cost of minor excavation and leveling of ground (say) = ₦20,000
(2) Concrete required = 6m3 × ₦42,500 = ₦255,000
(3) Y12 mm reinforcement required = 405 kg × ₦450 = ₦182,250
(4) Y8 mm required as links = 162 kg × ₦450 = ₦72,900
(5) Binding wire (say) = ₦13,000
(6) Formwork required 58 m2 of 1″ x 12″ planks = 58 m2 × ₦1905 = ₦110,490
(7) Bracing required 20 pcs pf 2″ x 3″ wood = 20 × ₦400 = ₦8000
(8) Total Labour and supervision cost = ₦120,000
Total cost of construction = ₦781,640

Alternatively, when the strip foundation is used, the likely costs are as follows;
(1) Excavation of strip foundation = ₦100,000
(2) Concrete required for the strip = 13.5 m3 × ₦42,500 = ₦573,750
(3) Blockwork to the ground level = 103 m2 × ₦3,700 = ₦381,100
(4) Backfilling of strip foundation (say) = ₦10,000
(5) Total Labour cost (concrete and block work) = ₦149,900
(6) Supervision cost (say) = ₦50,000
Total cost of construction = ₦1,264,750

Therefore using grade supported plinth beams to support blockwork instead of strip foundation in the substructure of a duplex can lead to savings in the cost of the substructure by about 38%.


Deflection of Beams

By
Ubani Obinna
-

When loads are applied on beams, they deflect. Deflection of beams is the downward movement a beam makes from its initial unloaded position to another deformed position when a load is applied to it. Since beams are usually treated as two-dimensional elements, the neutral axis of a beam is usually taken as the reference point for measuring the deflection of beams. The sag or curve that the deflected beam makes with the original neutral axis is called the elastic curve of the beam, while the angle of the elastic curve (in radians) is called the slope.

Calculation of deflection is very important in beams since it is a very important serviceability check in the design of structures. If deflection is excessive or beyond permissible limits, it may lead to cracks, damage to finishes, and misalignment of building services/fittings. Furthermore, the knowledge of deflections is the key in the analysis of statically indeterminate structures, since the equations of equilibrium are not sufficient for resolving such structures.

Relationship between Slope, Deflection, and Radius of Curvature

deflection of beams

The figure above shows a small portion of a beam bent into an arc. Let us consider a small portion of the beam PQ

Let
δs = Length of the beam PQ
O = Centre of the arc unto which the beam has been bent
θ = The angle which the tangent at P makes with the x-axis
θ + δθ = The angle which the tangent at Q makes with the x-axis

From geometry;
δs = Rδθ
Therefore R = δs/δθ = dx/δθ (assuming δs = dx)
1/R = dθ/dx —- (1)

If the coordinates of points P and Q are x and y;
tan θ = dy/dx or θ = dy/dx (taking tan θ = θ since θ is very small).

Differentiatign the above equation with respect to x, we get;

dθ/dx = d2y/dx2
1/R = d2y/dx2 —- (2)

From the bending equation of beams, we know that;
M/I = E/R —– (3)

Substituing equation (2) into (3), we obtain;

M = – EI d2y/dx2 —- (4)

Equation (4) is known as the elastic curve equation and represents to the relationship between the bending moment and the displacements of the structure without considering shear deformation.

Methods of Assessing Deflection of Beams

The slope and deflection of beams can be calculated using the following methods;

(1) Double integration method
(2) Macaulays’ method
(3) Moment Area method
(4) Castigliano’s theorem
(5) Virtual work method (unit load method)
(6) Vereschagin’s rule (graphical method)

Solved Examples on Deflection of Beams

The deflection of cantilever and simply supported beams can be easily calculated using the double integration method or Vereschagin’s rule. Let us consider the above-named methods with the examples below;

Cantilever with a concentrated load at the free end

Deflection of a cantilever beam with a point load at the free end

Let us consider the bending moment equation of a section x-x at a distance x from the fixed end A

Mx = -P(L – x)
From the elastic curve equation;

EI d2y/dx2 = P(L – x)

On integrating, we obtain;
EI dy/dx = P(Lx – x2/2) + C1
At A (fixed end), x = 0 and dy/dx = 0
Therefore C1 = 0

Hence;
EI dy/dx = P(Lx – x2/2) —- (a) is the equation for the slope

At x = L (free end, point B);
θB = EI dy/dx = P(L × L – L2/2) = P(L2 – L2/2) = PL2/2
Therefore;
θB = PL2/2EI

To get the deflection, let us integrate equation (a);
EIy = P(L·x2/2 – x3/6) + C2

At A (fixed end), x = 0 and y = 0;
Therefore, C2 = 0

EIy = P(L·x2/2 – x3/6) —- (b) is the equation for the deflection

At x = L (free end, point B);
yB = EI y = P(L·L2/2 – L3/6) = PL3/3
Therefore;
yB = PL3/3EI

This same problem can be solved by combining two different bending moment diagrams according to Vereschagin’s rule. The first bending moment diagram is produced by the externally applied load, while the second bending moment diagram is produced by a unit load applied at the point where the deflection is sought. In order to obtain the deflection at that point, the area of the first bending moment diagram is multiplied by the ordinate that its centroid makes with the second bending moment diagram.

This is based on the famous Mohr’s integral which is given by;

δ = 1/EI∫MṀ ds

For the problem above, the bending moment due to the externally applied load is given below;

Bending moment due to externally applied load

In order to obtain the deflection at point B, the external load P is replaced by a unit vertical load, and the bending moment diagram is plotted as shown below;

bending moment due to unit load

The deflection at point B is now obtained by combining the two bending moment diagrams according to Vereschagin’s rule.

diagram combination

As we all know, the area of the main diagram is given by A1 = bh/2 = (PL × L)/2 = PL2/2. Since the shape is a triangle, the centroid will occur at L/3 from point A. If the location of the centroid is traced down to the second diagram, the ordinate it makes (yg) can easily be calculated using similar triangles as 2L/3.

Note: When forming the flexibility matrix of structures using force method of structural analysis, the value obtained from the multiplication of A1 and yg is known as the ‘influence coefficient‘.

Therefore;
A1 = PL2/2
yg = 2L/3

δB = A1yg/EI = (PL2/2) × (2L/3) = PL3/3EI (This answer is the same as the one obtained using the double integration method).

To calculate the slope at point B, the vertical unit force at point B is replaced with a unit rotation (a unit moment), and the bending moment diagram obtained is combined accordingly with the bending moment diagram due to the externally applied load.

However, you may not need to bother so much about calculating the area, centroid, and ordinates of every shape since standard tables are available for different kinds of bending moment diagram combinations. You can download a copy HERE. An excerpt from the publication is given in the Table below.

diagram combination table

For instance, when a triangle is combined with a triangle, the influence coefficient is given by (1/3 × MA × ṀA × L).

Cantilever with a uniformly distributed load

Deflection of a cantilever beam with a uniformly distributed load


Let us consider the bending moment equation of a section X-X at a distance x from support A. The bending moment is given by;

Mx = -q(L – x)2/2
EI d2y/dx2 = q(L – x)2/2

On integrating;
EI dy/dx = -q(L – x)3/6 + C1

At x = 0, dy/dx = 0
Therefore, C1 = qL3/6

The general equation for slope is therefore given by;
EI dy/dx = -q(L – x)3/6 + qL3/6 —– (c)

At x = L (point B);
θB = qL3/6EI

To get the equation for deflection, we integrate equation (c);
EI y = q(L – x)4/24 + (qL3/6)x + C2

At x = 0, y = 0
Therefore, C2 = -qL4/24

The general deflection equation is therefore given by;
EI y = q(L – x)4/24 + (qL3/6)x – qL4/24 —– (d)

At x = L (point B);
yB = qL4/8EI

Using Vereschagin’s rule;

cantilever with UDL

The external load q (kN/m) is replaced by a unit vertical load placed at point B, and the bending moment diagram is plotted as shown below;

bending moment due to unit load

The deflection at point B is now obtained by combining the two bending moment diagrams according to Vereschagin’s rule.

diagram combination for UDL

δB = (1/4 × MA × ṀA × L) = (1/4 × qL2/2 × L × L) = qL4/8EI

Simply supported beam with a concentrated load at the mid-span

Deflection of a simply supported beam with a point load at the centre

At a section X-X from support A, the bending moment equation is given by;

Mx = Px/2
EI d2y/dx2 = -Px/2

Integrating;
EI dy/dx = -Px2/4 + C1

At x = L/2, dy/dx = 0 (slope is zero at the point of maximum deflection);
Therefore C1 = PL2/16

The general equation for slope is therefore given by;
EI dy/dx = -Px2/4 + PL2/16 ——- (e)

At x = 0;
θA = PL2/16EI (radians)

Integrating equation (e), we obtain;
EI y = -Px3/12 + (PL2/16)x + C2

At x= 0, y = 0
Therefore, C2 = 0

The general equation for deflection is therefore given by;
EI y = -Px3/12 + (PL2/16)x ——- (f)

At x = L/2;
ymax = PL3/48EI

Using Vereschagin’s rule;

SIMPLY SUPPORTED BEAM WITH POINT LOAD BMD

Let us replace the load P with a virtual unit load and plot the bending moment diagram;

UNIT LOAD ON SIMPLY SUPPORTED BEAM


On the combination of the two diagrams;

D COMB

δmax = 2(1/3 × PL/4 × L/4 × L/2) = 2(PL3/96) = PL3/48EI

Simply supported beam with a uniformly distributed load

Deflection of a simply supported beam with a UDL

At a section X-X from support A, the bending moment equation is given by;

Mx = (qL/2)x – qx2/2
EI d2y/dx2 = -(qL/2)x + qx2/2

On integrating;
EI dy/dx= -qLx2/4 + qx3/6 + C1

At x = L/2; dy/dx = 0;
C1 = qL3/24

Therefore, the general slope equation is given by;
EI dy/dx= -qLx2/4 + qx3/6 + qL3/24 —– (g)

At x = 0;
θA = PL3/24EI (radians)

On integrating the equation for slope;
EI y= -qLx3/12 + qx4/24 + (qL3/24)x + C2

At x = 0, y= 0;
C2 = 0

Therefore the general deflection equation;
EI y= -qLx3/12 + qx4/24 + (qL3/24)x —– (h)

At x = L/2;
ymax = 5qL4/384EI

Using Vereschagin’s rule;

udl

Let us replace the UDL with a virtual unit load and plot the bending moment diagram;

UNIT LOAD ON SIMPLY SUPPORTED BEAM

The diagram combination is now given by;

combination ty

δmax = 2(5/12 × qL2/8 × L/4 × L/2) = 2(5qL4/768) = 5qL4/384EI




Design of Deep Beams

By
Ubani Obinna
-

A deep beam is a beam with a depth that is comparable to the span length. As a result, deep beam theory is required for the design of such elements due to the shear warping of the cross-section and a combination of diagonal and flexural tension strains in the body of a deep beam.

In BS 8110, the designer is directed to specialist literature for the design of beams with a clear span less than twice the effective depth. A deep beam is defined in Eurocode 2 (EN 1992-1-1:2004) as one whose effective span is less than three times its entire depth.

The fundamental principle underlying the design of slender beams is that stress distribution across the section is proportional to the distance from the bending’s neutral axis, i.e. plane sections of the cross-section of a beam perpendicular to its axis remain plane after the beam is bent. However, this notion only applies to deep beams to a limited extent, resulting in designs that are often not conservative.

For beams with short span-to-depth ratios, section warpage is important, and the smaller the span-to-depth ratio, the more apparent the deviation from the linear stress theory. For beams with span-to-depth ratios less than 2.5, the divergence from the fundamental linear stress hypothesis should be considered. The load is carried to the supports via a compression force that combines the load and the reaction, according to deep beam theory.

Strut and tie model for simply supported deep beams
Fig 1: Strut and tie model for deep beam (Hanoon et al, 2016)

By implication, when compared to pure bending, the strain distribution in deep beams is no longer regarded linear, and shear deformations become prominent. Many countries have included design rules for these aspects in their design regulations as a result of the knowledge that a deep beam behaves differently from a slender beam.

The structural behaviour of a single-span deep beam after the concrete in tension has cracked is similar to that of a tied arch (Reynolds et al, 2008). The compression force at the centre of the arch rises from the support to a height at the crown equal to roughly half the beam’s span. Because the bending moment and the lever arm vary similarly along the length of the beam, the tension force in the tie is essentially constant along its length. The structural behaviour of a continuous deep beam is similar to that of a separate tied arch system for each span, paired with a suspension system centred over each internal support.

For deep beam reinforcement and detailing, Eurocode 2 recommends using one of two design techniques. The first methodology is connected to the strut-and-tie model application and the second methodology is related to a linear elastic analysis using the finite element method.

Strut and Tie Method (STM) for Deep Beams

Strut and Tie Modelling is an effective way of representing complex stress patterns as triangulated models. STM is based on the truss analogy concept and can be applied to numerous concrete structural parts. Essentially, strut-and-tie models (STM) are trusses consisting of struts, ties and nodes. It is commonly used to design non-standard concrete construction elements or parts of elements like pile caps, corbels, deep beams (depth > span/3), beams with holes, connections, and so on, where typical beam theory doesn’t always apply.

Strut and Tie Model for a Simple deep beam
Fig 2: Strut and Tie Model for a Simple deep beam (Source: Goodchild et al, 2014)

STM is a lower bound plastic theory, which implies it is safe as long as the following conditions are met:

  • The equilibrium has been satisfied
  • The structure is sufficiently ductile to allow the struts and connections to form.
  • Struts and ties are proportioned to withstand the forces they are designed to withstand.

The strut-and-tie model design method can be broken down into four stages:

  1. Define and isolate the B- and D-regions
  2. Develop an STM – a truss system to represent the stress flow through the D-region and calculate the member forces in the truss
  3. Dimension and design the STM members, as well as the truss members, to resist the design forces.
  4. Iterate as needed to optimise the STM to reduce strain energy.

Note: B (or beam or Bernoulli) regions in which plane sections remain plane and design is based on ‘normal’ beam theory. D (or discontinuity or disturbed) regions in which plane sections do not remain plane; so ‘normal’ beam theory may be considered inappropriate. D-regions arise as a result of discontinuities in loading or geometry and can be designed using STMs. Typical examples of D-regions include connections between beams and columns, corbels, openings in beams, deep beams and pile caps, etc.

For more information on strut and tie models, see ‘How to design concrete members using strut-and-tie models in accordance with Eurocode 2‘.

Worked Example on Deep Beam Design

A 6450 x 3000 beam, 225 mm thick is supported on 450 x 225 columns. As shown below, it spans 6.0 m and supports actions of gk = 55 kN/m and qk = 25 kN/m at the top and bottom of the beam. Assume C25/30 concrete, fyk = 500 MPa and cnom = 30 mm.

deep beam design

Solution

Bottle-shaped Strut and tie model
At ULS, a strut and tie model may be constructed to determine strut-and-tie forces: see Figure below. Here the UDLs top and bottom are resolved into two-point loads applied at ¼ spans at the top of the wall.

Total applied load = [2 × ( 55 × 1.35 + 25 × 1.5 )] = 223.5 kN/m × 6.45m = 1441.575 kN
Self weight = [3.0 × 0.225 × 25 × 1.35] × 6.45 = 146.939 kN
Total = 1588.51 kN (say 2 × 795 kN)

STRUT AND TIE MODEL FOR DEEP BEAM 1

MC90 gives z = 0.6-0.7 × minimum (h, L) = 0.67 × 3000 = 2000 mm
MC90 gives u ≈ 0.12 × minimum (h, L) = 0.12 × 3000 = 360 mm (i.e 180 mm to centreline)

Check θ
tan θ = 2000/1500 = 1.33 i.e. < 2/1 ∴ OK
θ = 53.13o

Forces
C12 = 795 kN
Length of C23 = (20002 + 15002)0.5 = 2500 mm

By trigonometry:
C23 = (2500/2000) × 795 = 993.75 kN
T35 = (1500/2500) × 993.75 = 596.25 kN

Choice
A fan-shaped stress field is appropriate for the ULS but not necessarily for the SLS where the lever arm can be determined from elastic analysis or alternatively in accordance with the recommendations of MC90. Designed reinforcement will not be required if the design bearing stress is less than σRdmax = 0.85υ’fcd: in that case the design loads will be safely transmitted to the supports through the fan-shaped stress field. Suspension reinforcement is required to transmit the bottom loading to the top of the beam. In addition, minimal horizontal reinforcement is required for crack control.

Check (fan) strut at node 3
Strut in bearing, C32
For CCT Node (and fan-shaped strut)
σRdmax = 0.85υ’fcd
where
υ’ = 1 – fck/250 = 1 – (25/250) = 0.90
fcd = αccfck/γm = 0.85 × 25 / 1.5 = 14.2
σRdmax = 10.8 MPa

σEd32 = Fc / ab

where
Fc = 993.75 kN
a = width of strut = (acol – cnom – 2so) sin 53.13 + u cos 53.13
= (450 – 30 + 2 × (30 + say 12 + 30/2)) sin 53.13 + 360 cos 53.13
= (450 – 144) sin 53.13 + 360 cos 53.13 = 532.799 mm

b = thickness = 225 mm
σEd32 = 993.75 × 103 / (532.799 × 225) = 8.289 MPa
i.e. < 10.8 MPa ∴ OK
NB: As σEd32 < σRdmax no further checks on strut 2-3 are necessary since the stress field is fan-shaped at the ULS.

Ties
a) Main tie
As,req = Ft / fyd = 596.25 × 103 / (500 / 1.15) = 1371 mm2
Try 8H16 (1608 mm2)

Check anchorage:
Assuming straight bar
lbd = αlbrqd = α(ϕ/4) (σsd/fbd)

where
α = 1.0 (assumed)
ϕ = diameter of bar = 16 mm
σsd= 500 / 1.15 = 435 MPa
fbd = 2.25η1fctk / γm = 2.25 × 1.0 × 1.0 × (1.8 / 1.5) = 2.7 MPa

lbd = 1.0 x (16/4) × (435/2.7) = 644 mm

Average length available = 450 – 30 + cot 53.13° × (360 / 2) = 555 mm – no good

lbd = 644 x 1371/1608 = 549 mm: OK
∴ Provide 8H16 (1608 mm2)

Vertical tie steel
Vertical tie steel is required to take loads from bottom level to top level.
As,req = (55 x 1.35 + 25 × 1.5) / (500/1.15) = 111.75 × 103 / ( 434.783) = 257 mm2/m


Minimum areas of reinforcement
Consider as a wall
Asvmin = 0.002Ac = 0.002 × 1000 × 225 = 450 mm2/m

Vertically, say minimum area and tie steel additive. Therefore provide
257 + 450 mm2/m = 707 mm2/m

Consider as a deep beam
Asdb,min = 0.2% Ac each surface: i.e. require 675 mm2/m
∴ Use H12@150 between each face (753 mm2/m each way each side)

Detailing Sketches

detailing sketches of deep beam

References
[1] Goodchild, C. H., Morrison, J., and Vollum, R. L. (2014): Strut-and-tie models: How to design concrete members using strut-and-tie models in accordance with Eurocode 2. The Concrete Centre, UK
[2] Hanoon, A. N., Jaafar, M. S., Hejazi F., and Abd Aziz F.N.A. (2016): Strut effectiveness factor for reinforced concrete deep beams under dynamic loading conditions, Case Studies in Structural Engineering, (6): 84-102, https://doi.org/10.1016/j.csse.2016.08.001.
[3] Reynolds C. E., Steedman J. C., and Threlfall A. J. (2008): Reynolds’s Reinforced Concrete Designer’s Handbook (11th Edition). Taylor and Francis

Sequence and Method of Erection of Steel Structures

By
Ubani Obinna
-

The sequence and methods of erection of steel structures are generally dependent on the layout and arrangement of the structural components. Erection of steel structures involves the on-site installation of steel members into a frame. Lifting and positioning the individual components into position, then joining them together, are all part of the process. Bolting is the most common method, however, site welding is also employed.

Generically, the erection of steel structures consists primarily of four tasks:

  1. Ensure that the foundations are suitable and safe before starting the erection process.
  2. Generally, cranes are used to lift and place components into place, however, jacking is also used. Bolted connections can be used to secure components in place, but they will not be fully tightened. Similarly, bracings may not be completely secure.
  3. Aligning the building, primarily by making sure that the column bases are lined and level, and that the steel columns/stanchions are plumb. To allow for column plumb adjustment, the packing in beam-to-column connections may need to be modified.
  4. Bolting-up the frame by ensuring that all bolted connections are fully installed in order to secure and make the frame rigid.
steel structure construction
Steel structure construction

Normally, the following sequence and method of erection are followed;

Sequence for Erection of Steel Structures

Stage 1:
Before the steelwork erection commences, the sizes and exact locations of the holding-down bolts on the foundation and base plates are double-checked. Sometimes, discrepancies can occur due to errors from setting out thereby causing the erection timetable to be delayed. Following these checks, the following procedures are carried out:

  • Install the columns on the base plates.
  • Align the columns properly to be straight and plumb.
  • Under the base plate, adjust the holding-down bolts with adjustable screws to maintain the required grout gap between the bottom of the base plate and the foundation.
  • Use temporary bracings to keep the columns true to their vertical position and prevent them from swaying in any direction.
  • The columns or stanchions are erected in parts and fastened together on site when they are quite long.

Stage 2:
Install the central portion of each rafter. When trusses are to be used, the whole section may be erected in one piece after fabrication on the ground.
Connect the truss to the column ends with bolts to form the full structural frame.

Stage 3:
After final alignments and adjustments of the frame placements, install the vertical column bracing and roof bracings to make the entire structure stable.

Stage 4:
Use bolted connections to secure all roof purlins and sheeting rails to the framework.

Stage 5:
If necessary, erect gantry crane girders (for industrial buildings and warehouses).

Stage 6:
Install the overhead cranes on the crane girders where required.

Stage 7:
Install the roof and side panels.

Stage 8:
After the erection is complete, fill the undersides of the base plates with non-shrink grout.

Normal construction follows the erection sequence outlined above. After preparing an erection program, a special sequence should be followed in the case of special constructions.

Lifting of Steel Members

Cranes and Mobile Elevating Work Platforms (MEWPs) are commonly used for the erection of steel structures, even though other methods may be utilized for steel bridge construction. Cranes are generally categorized into two categories: mobile and non-mobile. Truck-mounted cranes, crawler cranes, and all-terrain cranes fall into the first group, whereas tower cranes fall into the second.

crane lifting
Crane lifting a heavy steel member

The number of crane lifts required determines typical erection speeds and, as a result, the site programme of works. Pre-assembled units should be used to the maximum extent possible to reduce the number of crane lifts. If crane availability is an issue, steel decking, which can be installed by hand, is a better option than precast concrete units, which require a crane for individual placement. The designer can utilize a ‘piece count’ to estimate the number of lifts required and thus the erection time.

Alignment and Plumbing

Lining, leveling, and plumbing involve collaboration between the site engineer, who uses a survey instrument, and the erection crew, who tightens and shimmies the last bolts. The erection gang persuades the frame to move to a position acceptable to the checking engineer by using wedges, jacks, pull-lifts, and proprietary pulling devices such as Tirfors, and then bolts it up firmly. Some misalignment is overcome, and some are produced, as a result of this process. If the latter is unfavorable, local adjustments are made.

Connections

Bolted Connections
Bolted connections are preferable to site welding because it is faster, less sensitive to bad weather, and have less access restrictions and inspection requirements.

Property class 4.6 and 8.8 non-preloaded bolts to BS EN 15048 are commonly used in 2 mm clearance holes in structural bolting operations (for buildings) in the UK. M20 8.8 completely threaded bolts, which are the recommended solution, are widely accessible. Bolts with property class 4.6 are typically used only for fixing lighter components like purlins or sheeting rails, when 12 mm or 16 mm bolts can be used instead.

Bolting during erection of steel structures
Bolting of steel beams on site

Fully threaded bolts are commonly specified, implying that one bolt size can be used for a wide range of connections. It is advised that M20, 8.8 completely threaded bolts 60 mm long be used, as they can be used to make roughly 90% of basic connections. Preloaded bolts should only be utilized in situations where relative movement of linked parts (slip) is objectionable or if dynamic loading is a possibility.

It is best to avoid using different grade bolts of the same diameter on the same project. With non-preloaded bolts in typical clearance holes, washers are not required for strength. Bolts, nuts, and washers should be supplied with a corrosion-resistant coating that does not require further protection on-site when possible.

Welded Connections
If bolted connections can be made, site welding is usually avoided. When site welding is used, it is necessary to provide cover from bad weather as well as good access for both welding and inspection. Apart from the extra cost implications, providing such protection and access may have program consequences.

Erection Handover

The final goal of the erection process is to hand over the frame in good working order to the subsequent trades. The crucial criterion here is the erected frame’s positional precision, which is dependent on an understanding of how a steel frame’s erected position is regulated.

A steel-framed structure is a massive structure made up of a huge number of relatively thin and flexible components. Plumb and line accuracies of about 1 part in 1000 are targeted for the completed construction, employing components that can be fabricated with greater variability than 1 part in 1000 separately. Deformations such as the flexure of the structure under the self-weight of steel also have an impact on its real position.

Tests performed upon handover of a built steel structure could be deemed final under an inspection and test plan. All tests must have the following information in order to be meaningful:

  • The testing procedure
  • The test’s location and frequency
  • Criteria for Acceptance
  • Actions to be taken if compliance criteria is not obtained

How to Apply Wind Load on Roofs of Buildings

By
Ubani Obinna
-

Wind load is one of the significant actions on roofs. While other loads such as the self-weight of materials, imposed loads, service loads, and snow loads are pointed downwards, wind load on roofs tends to pull the roof upwards. As a result, when wind load is acting on roofs, there is a possibility of the reversal of internal forces in the members of the roofs trusses or rafters from tensile to compressive or vice versa. This is why the application of wind load on roofs is so important.

The method for the application of wind load on roofs is given in EN 1991-1-4:2005 (Eurocode 1 Part 4). The effect of wind on any structure (i.e. the response of the structure), depends on the size, shape, and dynamic properties of the structure. Wind action is represented by a simplified set of pressures or forces whose effects are equivalent to the extreme effects of the turbulent wind. It is important to note that the wind actions calculated using EN 1991-1-4 are characteristic values and are determined from the basic values of wind velocity or velocity pressure. Unless otherwise specified, wind actions are classified as variable fixed actions.

Simplified representation of wind pressures acting on building
Simplified representation of wind load on buildings

Procedure for the calculation of Wind Load on Roofs

(1) Determine the peak velocity pressure qp by calculating/obtaining the following values;

  • basic wind velocity Vb
  • reference height Ze
  • terrain category
  • characteristic peak velocity pressure qp
  • turbulence intensity Iv
  • mean wind velocity Vm
  • orography coefficient co(z)
  • roughness coefficient cr(z)

(2) Calculate the wind pressures using the pressure coefficients;

  • external pressure coefficient cpe
  • internal pressure coefficient Cpi
  • net pressure coefficient cp,net
  • external wind pressure: we = qp cpe
  • internal wind pressure: Wi = qp Cp

(3) Calculate the wind force

  • determine the structural factor: CsCd
  • wind force Fw calculated from force coefficients
  • wind force Fw calculated from pressure coefficients
wind pressure on roofs of buildings
Typical wind pressure on roofs

A pioneering wind tunnel experiment (Stanton, 1908) showed that when the roof slope is greater than 70 degrees from horizontal on the windward side, the roof surface can be treated as a vertical surface, with the external pressure coefficient cpe equal to +0.5 (positive). The positive normal wind pressure reduces as the roof slope lowers. The pressure drops to zero when the roof slope approaches 30 degrees. A negative normal pressure (suction) acts upwardly normal to the slope when the roof slope falls below 30°.

As the slope declines, the suction pressure increases until it reaches its maximum value when the slope is zero (i.e. a flat roof). If the roof slope is 45 degrees, for example, cpe = (45/100 – 0.2) = +0.25. If the roof slope is 30°, cpe = (30/60 – 0.5) = 0.0. If the roof slope is 10° cpe = (10/30 – 1.0) = -0.67 (upwards suction). If the roof slope is 0°, cpe = (0/30 – 1) = 1.0 (upwards suction).

Based on the same experiment results, cpe = -1.0 (upwards suction) for all roof slopes on the leeward roof slope. Thus, cpe = 1.0 (upwards suction) for the windward part of a flat roof and cpe = -0.5 (upwards suction) for the leeward side.

A building is subjected to internal pressures due to apertures in the walls, in addition to external wind pressures. Therefore, we have to also consider the internal pressure coefficients cpi. The result of wind blowing into a building through an opening facing the opposite direction as the wind blowing onto the building is the formation of internal pressure within the structure.

Positive internal pressure: When wind blows into an open-sided building or into a workshop through a large open door, the internal pressure seeks to force the roof and side coverings outwards, resulting in positive internal pressure.

Negative internal pressure (suction): This is formed within a building when the wind blows in the opposite direction, tending to pull the roof and side coverings inwards.

The coefficient of internal suction cpi = ±0.2 in normal permeability shops (covered with corrugated sheets), and ±0.5 in buildings with extensive apertures (in the case of industrial buildings). Internal suction, or pressure away from the interior surfaces, is shown by a negative number, whereas internal pressure is indicated by a positive value.

Worked Example

Calculate the wind action on the walls and roof of a building with the data given below. Consider when the wind is coming perpendicular (0°) to the length of the building, and normal to it (90°).

building details

Building data
Type of roof; Duopitch
Length of building;  L = 30000 mm
Width of building;  W = 15000 mm
Height to eaves; H = 6000 mm
Pitch of roof;  α0 = 15.0°
Total height;  h = 8010 mm

Tekla Tedds will be used for executing the wind load analysis on the building. The key to the pressure coefficients for a building with a duopitch roof is given below (Figure 7.8 EN 1991-1-1:2005). Note that e = B or 2h whichever is smaller, where b is the crosswind dimension.

duopitch roof
Pressure coefficients for wind load on roofs
(a) Key for pressure coefficients – wind direction θ = 0°
90 d
(a) Key for pressure coefficients – wind direction θ = 90°

Location of building; Onitsha, Anambra State, Nigeria
Wind speed velocity; vb,map = 40.0 m/s
Distance to shore; Lshore = 50.00 km
Altitude above sea level; Aalt = 50.0m

Altitude factor;calt = Aalt/1m × (0.001 + 1) = 1.050
Fundamental basic wind velocity; vb,0 = vb,map × calt = 42.0 m/s
Direction factor; cdir = 1.00
Season factor; cseason = 1.00
Shape parameter K; K = 0.2
Exponent n; n = 0.5
Air density; ρ = 1.226 kg/m3

Probability factor;  cprob = [(1 – K × ln(-ln(1-p)))/(1 – K × ln(-ln(0.98)))]n = 1.00
Basic wind velocity (Exp. 4.1); vb = cdir × cseason × vb,0 × cprob = 42.0 m/s
Reference mean velocity pressure;  qb = 0.5 × ρ × vb2 = 1.081 kN/m2

Orography
Orography factor not significant; co = 1.0
Terrain category; Country
Displacement height (sheltering effect excluded);   hdis = 0 mm

The velocity pressure for the windward face of the building with a 0 degree wind is to be considered as 1 part as the height h is less than b (cl.7.2.2). The velocity pressure for the windward face of the building with a 90 degree wind is to be considered as 1 part as the height h is less than b (cl.7.2.2)

Peak velocity pressure  – windward wall – Wind 0 deg and roof

Reference height (at which q is sought); z = 6000 mm
Displacement height (sheltering effects excluded); hdis = 0 mm
Exposure factor (Figure NA.7); ce = 2.05
Peak velocity pressure; qp = ce × qb = 2.22 kN/m2

Structural factor
Structural damping;  δs = 0.100
Height of element; hpart = 6000 mm
Size factor (Table NA.3);  cs = 0.884
Dynamic factor (Figure NA.9); cd = 1.003
Structural factor;  csCd = cs × cd = 0.887

Peak velocity pressure  – windward wall – Wind 90 deg and roof
Reference height (at which q is sought);  z = 8010 mm
Displacement height (sheltering effects excluded); hdis = 0 mm
Exposure factor (Figure NA.7);  ce = 2.23
Peak velocity pressure; qp = ce × qb = 2.41 kN/m2

Structural factor
Structural damping; δs = 0.100
Height of element;  hpart = 8010 mm
Size factor (Table NA.3);  cs = 0.911
Dynamic factor (Figure NA.9); cd = 1.016
Structural factor; csCd = cs × cd = 0.925

Structural factor – roof 0 deg

Structural damping; δs = 0.100
Height of element; hpart = 8010 mm
Size factor (Table NA.3); cs = 0.888
Dynamic factor (Figure NA.9); cd = 1.003
Structural factor; csCd = cs × cd = 0.891

Peak velocity pressure for internal pressure
Peak velocity pressure – internal (as roof pressure); qp,i = 2.41 kN/m2

Pressures and forces
Net pressure; p = csCd × qp × cpe – qp,i × cpi
Net force; Fw = pw × Aref

plan view 1
faces for wind action

Roof load case 1 – Wind 0, cpi 0.20, -cpe

ZoneExt pressure coefficient
cpe		Peak velocity pressure
qp, (kN/m2)		Net pressure
p (kN/m2)		Area
Aref (m2)		Net force
Fw (kN)
F (-ve)-1.102.41-2.8413.28-37.78
G (-ve)-0.802.41-2.2036.47-80.23
H (-ve)-0.402.41-1.34183.18-245.64
I (-ve)-0.502.41-1.56183.18-284.97
J (-ve)-1.302.41-3.2749.75-162.87

Total vertical net force; Fw,v = -783.83 kN
Total horizontal net force; Fw,h = 21.79 kN

Walls load case 1 – Wind 0, cpi 0.20, -cpe

ZoneExt pressure coefficient
cpe		Peak velocity pressure
qp, (kN/m2)		Net pressure
p (kN/m2)		Area
Aref (m2)		Net force
Fw (kN)
A-1.202.41-3.0520.60-62.77
B-0.802.41-2.1984.47-185.17
D0.742.220.97180.00174.37
E-0.382.22-1.22180.00-219.73

Overall loading

Equivalent leeward net force for overall section; Fl = Fw,wE = -219.7 kN
Net windward force for overall section; Fw = Fw,wD = 174.4 kN
Lack of correlation (cl.7.2.2(3) – Note);  fcorr = 0.85; as h/W is 0.534
Overall loading overall section; Fw,D = fcorr × (Fw – Fl + Fw,h) = 353.5 kN

Roof load case 2 – Wind 0, cpi -0.3, +cpe

ZoneExt pressure coefficient
cpe		Peak velocity pressure
qp, (kN/m2)		Net pressure
p (kN/m2)		Area
Aref (m2)		Net force
Fw (kN)
F (+ve)0.202.411.1513.2815.31
G (+ve)0.202.411.1536.4742.03
H (+ve)0.202.411.15183.18211.11
I (+ve)-0.502.41-0.35183.18-64.23
J (+ve)-1.302.41-2.0749.75-102.91

Total vertical net force; Fw,v = 97.86 kN
Total horizontal net force;  Fw,h = 112.74 kN

Walls load case 2 – Wind 0, cpi -0.3, +cpe

ZoneExt pressure coefficient
cpe		Peak velocity pressure
qp, (kN/m2)		Net pressure
p (kN/m2)		Area
Aref (m2)		Net force
Fw (kN)
A-1.202.41-1.8420.60-37.94
B-0.802.41-0.9984.47-83.38
D0.742.222.17180.00391.28
E-0.382.22-0.02180.00-2.83

Overall loading
Equivalent leeward net force for overall section; Fl = Fw,wE = -2.8 kN
Net windward force for overall section; Fw = Fw,wD = 391.3 kN
Lack of correlation (cl.7.2.2(3) – Note); fcorr = 0.85; as h/W is 0.534
Overall loading overall section; Fw,D = fcorr × (Fw – Fl + Fw,h) = 430.8 kN

90 degrees
faces 90 degrees

Roof load case 3 – Wind 90, cpi 0.20, -cpe

ZoneExt pressure coefficient
cpe		Peak velocity pressure
qp, (kN/m2)		Net pressure
p (kN/m2)		Area
Aref (m2)		Net force
Fw (kN)
F (-ve)-1.602.41-4.0511.65-47.17
G (-ve)-1.502.41-3.8311.65-44.58
H (-ve)-0.602.41-1.8293.17-169.59
I (-ve)-0.402.41-1.37349.41-480.12

Total vertical net force; Fw,v = -716.21 kN
Total horizontal net force;  Fw,h = 0.00 kN

Walls load case 3 – Wind 90, cpi 0.20, -cpe

ZoneExt pressure coefficient
cpe		Peak velocity pressure
qp, (kN/m2)		Net pressure
p (kN/m2)		Area
Aref (m2)		Net force
Fw (kN)
A-1.202.22-2.9418.00-52.99
B-0.802.22-2.1272.00-152.86
C-0.502.22-1.5190.00-135.69
D0.702.411.08105.07113.92
E-0.302.41-1.16105.07-122.01

Overall loading
Equiv leeward net force for overall section; Fl = Fw,wE = -122.0 kN
Net windward force for overall section; Fw = Fw,wD = 113.9 kN
Lack of correlation (cl.7.2.2(3) – Note); fcorr = 0.85; as h/L is 0.267
Overall loading overall section; Fw,D = fcorr × (Fw – Fl + Fw,h) = 200.5 kN

Roof load case 4 – Wind 90, cpi -0.3, +cpe

ZoneExt pressure coefficient
cpe		Peak velocity pressure
qp, (kN/m2)		Net pressure
p (kN/m2)		Area
Aref (m2)		Net force
Fw (kN)
F (+ve)0.202.411.1711.6513.62
G (+ve)0.202.411.1711.6513.62
H (+ve)0.202.411.1793.17108.93
I (+ve)0.202.411.17349.41408.48

Total vertical net force;  Fw,v = 526.08 kN
Total horizontal net force; Fw,h = 0.00 kN

Walls load case 4 – Wind 90, cpi -0.3, +cpe

ZoneExt pressure coefficient
cpe		Peak velocity pressure
qp, (kN/m2)		Net pressure
p (kN/m2)		Area
Aref (m2)		Net force
Fw (kN)
A-1.202.22-1.7418.00-31.30
B-0.802.22-0.9272.00-66.10
C-0.502.22-0.3090.00-27.24
D0.702.412.29105.07240.54
E-0.302.410.04105.074.60

Overall loading

Equiv leeward net force for overall section; Fl = Fw,wE = 4.6 kN
Net windward force for overall section;  Fw = Fw,wD = 240.5 kN
Lack of correlation (cl.7.2.2(3) – Note);  fcorr = 0.85; as h/L is 0.267
Overall loading overall section; Fw,D = fcorr × (Fw – Fl + Fw,h) = 200.5 kN

References
(1) BS EN 1991-1-4: 2005, Actions on structures. General actions. Wind actions
(2) Stanton, T. E., 1908. Experiments on wind pressure. Minutes of the Proceedings of the Institution of Civil Engineers, 171, 175–200.

On the Design of RC Tension Columns

By
Ubani Obinna
-

It is more common for reinforced concrete columns in a building to be in compression. This makes sense because, in a conventional load path of a building, gravity loads are transmitted through the columns to the foundation, which exerts an equal and opposite reaction on the column, thereby placing it in compression. For tension columns, the load is probably going somewhere else before being transmitted to the foundation. This is usually deliberate!

The most significant internal force associated with columns is the axial force, even though more often than not, columns are also subjected to bending moment and shear forces. Design equations and charts exist in almost all codes of practice for reinforced concrete columns subjected to compressive axial force, bending moment, and shear. However, when a concrete column is in tension, the approach is usually very cautious. It is advisable to avoid placing concrete columns in tension unless you cannot help it.

Concrete is very good in compression but weak in tension. For instance, for a block of concrete with a 28 days cylinder compressive strength of 25 N/mm2, we will expect the tensile strength to be about 2.6 N/mm2 (See Table 3.1 EN 1992-1-1:2004). In reinforced concrete design, we normally assume the tensile resistance of concrete to be zero unless we are dealing with serviceability issues such as cracking.

If a reinforced concrete column is to be in axial tension, the entire axial stress will be carried by the steel reinforcements, unlike when it is compression. The area of steel required to resist the tensile axial force will be given by;

Ast,req = Nt,Ed/0.87fyk

Where;
Ast,req = Area of steel required
Nt,Ed = Ultimate axial compressive force
fyk = characteristic yield strength of the reinforcement

However, this is not as simple and straightforward as it looks. For a concrete section subjected to a significant axial tensile force, cracking is going to be a major problem. As a result, additional reinforcements will be required to control the cracking, aside from the main reinforcements resisting the axial tension. The additional reinforcements will have to be caged as in the case of a wall reinforcement using smaller diameter bars. The column might have two layers of reinforcements; the inner layer for resisting the tensile force, and the outer layer for controlling the cracking (skin reinforcement).

The maximum bar diameter for the skin reinforcement can be estimated from equation 7.7N of EN 1992-1-1:2004 for members subjected to uniform axial tension;

φs = φ∗s(fct,eff/2.9)hcr/(8(h-d))

where:
φs is the adjusted maximum bar diameter
φ∗s is the maximum bar size given in Table 7.2N of EN 1992-1-1:2004
h is the overall depth of the section
hcr is the depth of the tensile zone immediately prior to cracking, considering the characteristic values of prestress and axial forces under the quasi-permanent combination of actions
d is the effective depth to the centroid of the outer layer of reinforcement

The discussion above has been based on the assumption that the tension column is subjected to pure axial force. However, what happens when there is bending and shear in the section? In that case, the interaction of axial force and bending will have to be considered, and this is expected to have an effect on the size of the tension column and the quantity of the reinforcements.

From the ongoing, it can be seen that structural steel or composite sections are the best for tension columns, provided the connection details are well designed by the structural engineer. The problem of cracking due to low tensile strength is the major challenge of constructing tension columns using reinforced concrete. However, if this can be overcome with the use of additional reinforcements, the problem can be solved at the expense of additional costs.

Furthermore, the reinforcements in tension columns must not be lapped and must be detailed in such a way that the reinforcement hooks and carries the load below it. For reinforcement continuity, mechanical couplers are expected to perform better than lapping.

Practical Applications of Tension Columns

Tension columns are usually floating columns, and transmit the load they are supporting to the member above them. This depends on the structural scheme adopted by the structural engineer. However, when tension columns are to be supported on the ground, uplift forces will have to be properly checked.

An example of the application of tension columns is in the United States Court House in Downtown Los Angeles. The building’s structural system is depicted in the diagram below. All of the columns are set around the perimeter of the structure, but they do not convey the load to the ground; instead, they take the load up. As a result, there is no compressive force applied to the columns.

couthouse building
3D Model of United States Court House (Source; Faizal Manzoor, 2020)
Building with tension column
Structural scheme and load path of the United States Court House (Source; Faizal Manzoor, 2020)

As can be seen above, the slab is not supporting the column’s weight; rather, it is hanging on to it. Therefore, the columns are subjected to axial tension due to the pull from the downward weight of the slab. The load is transferred from the slab to the column, which then passes through the roof truss, which is supported by shear walls. It is however important to note that steel columns were utilised in the construction described above.

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