Design of Raft Foundation

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January 29, 2018

Raft foundation is a type of shallow foundation that is mostly used on soils of low bearing capacity, where the foundation pressures need to be spread over a large area. They are also used in areas where the foundation soils are of varying compressibility and the foundation has to bridge over them. The geotechnical design of raft foundation ensures that there is no bearing capacity failure and that the settlement is kept to a minimum, while the structural design ensures that adequate thickness and reinforcements are provided to avoid structural failure of the raft plate.

The types of raft foundation commonly encountered in practice are;

Flat slab rafts for framed buildings
Beam and slab rafts for framed buildings
Cellular rafts
Piled rafts
Annular rafts
Strip rafts forming grid rafts (Grid foundations)

Factors Affecting Choice of Raft Foundation

A flat raft slab can be adopted and analysed using the ACI empirical method called the direct design method (DDM) when there is a symmetrical arrangement of spacing of columns and loadings that leads to uniform ground pressure. The DDM method involves analysing the flat raft foundation as flat slab (similar to the analysis and design of an inverted roof flat slab).
When there is uniform ground pressure but with unequal spans which do not satisfy the conditions for DDM, a flat slab can be adopted, but it has to be analyzed by the equivalent frame method (EFM). However, beam and slab rafts are generally preferred in such cases.
A beam and slab raft construction should be adopted when there is a symmetrical arrangement of unequally loaded columns but with uniform ground pressure.
On soils of very low bearing capacity, the loads on the raft can be compensated by excavation of soil, and the raft designed as a cellular raft.
When there is non-symmetrical loading with the centre of gravity of loads not coinciding with the centre of gravity of the area, the ground pressures will vary from place to place. In such cases, it is advisable to adopt a beam and slab construction for such cases. To some extent, the space around the loading area should be adjusted to give the least eccentricity. The slab should then be designed for the maximum pressures. The beam loading must be for the maximum values of the varying pressure in its location.
When the bearing capacity is satisfactory but there is excessive settlement, some parts of the load on the raft can be relieved by installing a few piles so that raft settlement can be reduced. This is known as piled raft foundation.

Raft foundations can be analysed using the rigid approach or flexible approach.

The conventional rigid combined footing approach is a method of analysing raft foundation using simple statics without any consideration of the elastic properties of the raft and the soil and their interaction. Here, the raft is analysed as a large beam member independently in both directions. The row of column loads perpendicular to the length of the beam is coupled together in single column load. Then for these column loads acting on the beam, the upward soil pressure is calculated and the moments and the shears at any section is determined by simple statics. Hence, the moment per unit width of the raft is determined by dividing the moment values by the corresponding width of the section (Gupta, 1997).

In order to obtain the upper bound values of the stresses, the raft is divided into strips bounded on the centre line of the column bays in each direction. Each of these strips is then analysed as independent combined footing by simple statics. Using the column loads on each strip the soil pressure under each strip is determined without reference to the planar distribution determined for the raft as a whole.

The eccentricity of the load and the pressure distribution below the raft which is considered to be linearly varying are taken into account in this analysis.

Approximate Rigid Analysis of Flat Raft Foundation with Eccentricity

In theory, the flat slab analysis is restricted to uniformly distributed load. However, if the eccentricity is small, we may proceed as follows. We adopt the following steps to analyze such a flat slab:

Step 1: Check eccentricity of the resultant load and find e_x and e_y with service loads.

Step 2: Find M_x and M_y with factored loads of all loads about XX- and YY-axes to calculate the varying base pressures.

Step 3: Find the moment of inertia of the raft slab in the x-x direction (I_xx) and y-y direction (I_yy).

Step 4: A flat slab has to be analyzed in frames formed by cutting the slab along mid-spans in the XX- and YY-directions. Find the ground pressures due to the eccentric loading at the end and mid points of the slab at the top and along the bottom.

Step 5: Check whether the maximum pressure exceeds the safe bearing capacity or not.

Step 6: Take each strip along YY. As the pressure in the strip varies, find the average pressures over the top edge and also at the bottom edge. (This average pressure for the interior frame will be pressure along the column line.) Draw the load diagram with the column loads P₁, P₂, etc. and base reaction in the strip along its length.

Step 7: Check for balance of downward and upward forces. If they do not match, modify the ground pressures for equilibrium.

Step 8: Draw the shear force and bending moment for the strip for design. As the loads from the top and the base pressure from below are known, we may use any of the following:

(a) Simple statics or neglecting colulmn loads used
(b) Direct design Methd (DDM), or
(c) Equivalent Frame Method (EFM) (we may use the transverse distribution percentage in detailing of steel).

(As a quick and short procedure, we first find the pressures along the boundary lines of the spans take and design each span for the average pressure in the span. We will also assume for simplicity that the load from the apron below the peripheral column is transmitted directly to the beams.)

Step 9: Distribute the moments using transverse distribution.

Design Example of a Flat Raft Foundation

It is desired to design the raft slab shown below to support the column loads for a building as given below. All columns are 230 x 450mm, the grade of concrete (f_ck) is 30 N/mm², and the yield strength of the reinforcement (f_yk) is 500 N/mm². The allowable bearing capacity for the supporting soil is 60 kN/m².

The service loads on the columns are given in the table below;

Total axial load = 18296 kN

Eccentricity along the x-direction
This is obtained by taking moment about grid 5
x = [22.4(770 + 1050 + 776 + 350) + 16.8(870 + 1450 + 860 + 660) + 11.2(1211 + 1850 + 1000 + 779) + 5.6(875 + 1400 + 865 + 660)] / 18296 = 11.258 m

e_x = x – (L/2) = 11.258 – (22.4/2) = 0.058 m

Eccentricity along the y-direction
This is obtained by taking moment about grid D
y = [15.1(770 + 870 + 1211 + 875 + 770) + 9.1(1050 + 1450 + 1850 + 1400 + 1050) + 3.1(776 + 860 + 1000 + 865 + 700) ] / 18296 = 7.8045 m

e_y = x – (B/2) = 7.8045 – (15.1/2) = 0.2545 m

Moment Due to Eccentricity

M_x = P.e_x = (18296 × 0.058) = 1061.168 kNm
M_y = P.e_y = (18296 × 0.2545) = 4656.332 kNm

Other geometrical properties

Moment of inertia of the raft slab about the x-direction;
I_x = (17.1 × 24.4³) / 12 = 20700.6672 m⁴

Moment of inertia of the raft slab about the y-direction;
I_y = (24.4 × 17.1³) / 12 = 10167.0957 m⁴

A = Area of raft slab = (17.1m × 24.4m) = 417.24 m²

The soil pressure at any point is given by the equation below;

P/A = (18296 kN / 417.24 m²) = 43.85 kN/m²
M_y/I_y = (4656.332 kNm / 10167.0957 m⁴) = 0.45798 kN/m²
M_x/I_x = (1061.168 kNm / 20700.6672 m⁴) = 0.05126 kN/m²

Substitutinh these values into equation (1) we can obtain the soil pressure at any pomt on the raft slab as follows;

σ = 43.85 ± 0.45798x ± 0.05126y

At  corner A1;
σ_A1 = 43.85 – (0.45798 × 12.2) + (0.05126 × 8.55) = 38.7 kN/m²

At  corner A5;
σ_A5 = 43.85 + (0.45798 × 12.2) + (0.05126 × 8.55) = 49.975 kN/m²

At  corner B1;
σ_B1 = 43.85 – (0.45798 × 12.2) + (0.05126 × 1.55) = 38.342 kN/m²

At  corner B5;
σ_B5 = 43.85 + (0.45798 × 12.2) + (0.05126 × 1.55) = 49.516 kN/m²

At  corner C1;
σ_C1 = 43.85 – (0.45798 × 12.2) – (0.05126 × 4.45) = 38.07 kN/m²

At  corner C5;
σ_C5 = 43.85 + (0.45798 × 12.2) – (0.05126× 4.45) = 49.209 kN/m²

At corner D1;
σ_D1 = 43.85 – (0.45798 × 12.2) – (0.05126 × 8.55) = 37.824 kN/m²

At corner D5;
σ_D5 = 43.85 + (0.45798 × 12.2) – (0.0512 × 8.55) = 48.999 kN/m²

At corner A2;
σ_A2 = 43.85 – (0.45798 × 5.6) + (0.05126 × 8.55) = 41.723 kN/m²

At corner A3;
σ_A3 = 43.85 + (0.05126 × 8.55) = 44.288 kN/m²

At corner A4;
σ_A4 = 43.85 + (0.45798 × 5.6) + (0.05126 × 8.55) = 46.852 kN/m²

At corner D2;
σ_D2 = 43.85 – (0.45798 × 5.6) – (0.05126 × 8.55) = 40.847 kN/m²

At corner D3;
σ_D3 = 43.85 – (0.05126× 8.55) = 43.411 KN/m²

At corner D4;
σ_D4 = 43.85 – (0.45798 × 5.6) – (0.05126× 8.55) = 40.847 KN/m²

A little consideration will show that the bearing capacity checks are satisfactory.

Limit State Calculations
A factor of 1.37 has been used to convert the load from service load to ultimate load.

For strip A – A bearing the most critical load;

Analysis of the raft strips

For strip A – A bearing the most critical load;

The analysis of the strip will be carried out using simple statics.

Check for balance of loads;

Total column load (summation of downward forces) = A1 + A2 + A3 + A4 + A5 = 770 + 870 + 1211 + 875 + 770 = 4496 kN

Width of strip = 4 m
Length of strip = 24.4 m
Taking the average load = (38.7 + 49.975) / 2 = 44.3375 kN/m²
Summation of upward force = (44.3375 × 4m × 24.4m) = 4327.34 kN

Check; 4496 kN – 4327.34 kN = 169 kN
Now the average ground pressure can be increased by [169 kN/(4m × 24.4m) = 1.73 kN/m²] which is 44.3375 + 1.73 = 46 kN/m². This has been however ignored in this analysis.

Multiplying by the width of the strip = 44.3375 kN/m²× 4m = 177.35 kN/m
On factoring at ultimate limit state = 1.37 × 177.35 = 242.9695 kN/m

Using simple statics and analysing as a continuous beam as shown below, the bending moment and shear forces along the strip can be determined.

This analysis is repeated using the same procedure for all the strips in the raft foundation.

Structural Design
Check for punching shear at column perimeter
We can determine the thickness of the raft slab by considering the punching shear at the column perimeter. At the column perimeter, the maximum punching shear stress should not be exceeded.

V_Ed < V_Rd,max

Where;
V_Ed = βV_Ed /u₀d

V_Rd,max = 0.5vf_cd

Considering Column B3 bearing the maximum axial load, V_Ed = 1.37 × 1850 kN = 2534.5 kN
β = 1.15 (an approximate value from clause 6.4.3(6))
u₀= column perimeter = 2(230) + 2(450) = 1360mm
v = 0.6[1 – f_ck/250] (strength reduction factor for concrete cracked in shear)
v = 0.6[1 – 30/250] = 0.528
f_cd = α_ccf_ck/γ_cf_cd = (1.0 × 30)/1.5 = 20 N/mm²
V_Rd,max = 0.5 × 0.528 × 20 = 5.28 N/mm²
V_Ed = (1.15 × 2534.5 × 1000) / (1360 mm × d)

Therefore;
2914675/1360d = 5.28
On solving; d_min = 405.89 mm

The basic control perimeter for punching shear check is normally taken at 2d, but when the concentrated force is resisted by high pressure as can be found in foundations, the punching control perimeter is taken at less than 2d.

Let us consider a trial footing depth of 700mm.

Effective depth (d) = 700 – 70 – (20/2) = 620 mm (concrete cover is taken as 70mm and assumed diameter of bar is 20mm)

Design of bottom reinforcement
M_Ed = 781.67 kNm

A_s1 = M_Ed/(0.87f_yk z)
A_s1= (781.67 × 10⁶)/(0.87 × 500 × 0.95 × 620) = 3050.836 mm²

Minimum area of reinforcement A_smin= 0.0013bd = 0.0013 × 1000 × 620 = 806 mm²/m

Provide 18H16 @ 225mm c/c BOT (A_Sprov = 893 mm²/m or 3216 mm²) along the strip.

Design of top reinforcement
M_Ed = 529.47 kNm

A_s1 = M_Ed/(0.87f_yk z)
A_s1= (529.47 × 10⁶)/(0.87 × 500 × 0.95 × 620) = 2066.506 mm²

Minimum area of reinforcement = 0.0013bd = 0.0013 × 1000 × 620 = 806 mm²/m (clause 9.2.1.1(1))
Also provide 18H16 @ 225mm c/c TOP (A_Sprov = 893 mm²/m or 3216 mm²) along the strip.

Beam shear
Check critical section d away from column face considering the highest force in the shear force diagram.
The point of contraflexure for shear between column A1 and A2 is 2.315 m.

The shear force at d from column A2 (towards the left) can therefore be calculated using similar triangle. The length d from column face is 0.15 m + 0.62 = 0.77 m (width of column is 230 mm).
V_Ed = 611 kN
v_Ed = V_Ed/bd = (611 × 1000) /(4000 × 620) = 0.246 N/mm²

v_{Rd, c} = C_{Rd, c} × k × (100 × ρ₁ × f_ck) ^0.3333
C_{Rd, c} = 0.12
k = 1 + √ (200/d) = 1 + √ (200/620) = 1.568
ρ = A_s/bd = 3216/(4000 × 620) = 0.00129
v_{Rd, c} = 0.12 × 1.568 × (100 × 0.001296 × 30)^0.333 = 0.295 N/mm²
V_min = 0.035k^(3/2) f_ck^0.5
V_min = 0.035 × (1.568)^1.5 × 30^0.5 = 0.376 N/mm²
=> v_Ed (0.246 N/mm²) < v_Rd,c (0.376 N/mm²) beam shear is ok

Punching Shear (Use column A3)
Punching shear: Basic control perimeter at 2d from face of column
v_Ed = βV_Ed/u_id < v_Rd,c

β = 1,
a = 2d = 2(620) = 1240 mm
Perimeter length, u = 2(c₁ + c₂ + π × a)
c₁ = 230 mm; c₂ = 450 mm
u_i = 2(230 + 450 + π × 1240) = 9151 mm

V_Ed = load minus net upward force within the area of the control perimeter)
Area inside the perimeter A = c₁ × c₂ + 2 × (c₁ + c₂) × a + π × a² = 0.23 × 0.45 + 2 × (0.23 + 0.45) × 1.24 + π × 1.24² = 6.62 m²
Average earth pressure at ultimate limit state = 1.37 x 44.3357 = 60.733 kN/m² (note that the SLS actual soil pressure under the column is44.288 kN/m²). Since this is meant to beneficial, the lowest value can be used.

V_Ed = 1.37 × 1211 kN = 1659.07 kN

V_Ed,_red = 1659.07 – (60.733 x 6.62) = 1257 kN
v_Ed = (1257 x 10³)/(9151 x 620) = 0.221 N/mm²

v_{Rd, c} = 0.12 × 1.568 × (100 × 0.001296 × 30)^0.333 = 0.295 N/mm²
V_min = 0.035 × (1.568)^1.5 × 30^0.5 = 0.376 N/mm²
Punching shear at 2d is therefore okay

This shows that shear is ok.
Using this approach, the entire reinforcement for the mat foundation can be obtained in the longitudinal and transverse directions.

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Design of Combined Footings

By

Ubani Obinna Uzodimma

-

January 25, 2018

A combined footing is a type of shallow foundation where a common base is provided for two closely spaced columns. When a common base in form of a strip is provided for three or more rows of columns, it is better described as a continuous footing. Combined footings are normally employed in cases where two or more columns are closely spaced such that their individual pad footings would overlap each other. They can also be used where property boundaries would not permit the design of separate bases.

The design of a combined footing involves the geotechnical and structural design, where the proper proportioning of the dimensions, thickness of the base, and proper reinforcement is provided. The foundation of a combined footing must not undergo excessive settlement or shear failure, and the footing itself must be strong enough to resist the bending moment and shear forces as a result of the superstructure load. The thickness of combined footings is usually governed by shear force considerations.

In the design of combined footings, the centre of gravity of the two column loads should as practically as possible coincide with the centre of area of the base. The base should preferably be rectangular in plan and symmetrically disposed about the line of loads. If it is not practicable to proportion the base as described above, the load will be eccentric, and the centre of pressure of the higher ground pressure will have the same eccentricity relative to the centre of the base.

If the base is thick enough, the pressure distribution diagram will be trapezoidal for eccentric loads, but uniform for concentric loads. If the base is relatively thin, the pressure distribution will be variable, with the maximum pressure occurring directly under each column.

The most common method of design of combined footings is the rigid approach, which makes the following assumptions.

The footing is infinitely rigid, and therefore, the deflection of the footing does not influence the pressure distribution.
The soil pressure is distributed in a straight line or a plane surface such that the centroid of the soil pressure coincides with the line of action of the resultant force of all the loads acting on the foundation.

However, flexible method can also be used for the design of combined footings, taking into account soil-structure interaction.

Solved example on the design of combined footing

Two (300 x 300)mm square columns spaced at a distance of 2.45 m c/c are loaded as shown below. The foundation is founded on soil of a bearing capacity of 125 kN/m². It is desired to design the footing to satisfy all requirements using the concrete grade of 30 N/mm² and steel of yield strength 500 N/mm². The concrete cover is 50mm.

STRUCTURAL%2BDESIGN%2BOF%2BCOMBINED%2BFOOTING — Combined footing

At serviceability limit state;
P_Ed = 1.0G_k + 1.0Q_k
Total service load on column P₁ = 665 + 122 = 787 kN
Total service load on column P₂ = 825 + 145 = 970 kN
Total load on both columns = 787 + 970 = 1757 kN

Assuming 10% of the service load to account for the self-weight of the footing;
S_w= 0.1 × 1757 = 175.7 kN

Area of footing required = Total service load/Allowable bearing capacity = (1757 + 175.7)/125 = 15.46 m²

Adopt a rectangular base = 6.5m x 2.5m (Area provided = 16.25 m²)

To locate the centroid of the footing, let us take moment about column P₁;
(970 × 2.45) – 1757x = 0
On solving, x = 1.352m from column P₁

Let the projection of the footing from the lighter column be L₁
Therefore;
L₁ + 1.352m = (Total length of footing)/2 = 6.5/2 = 3.25 m
Therefore, L₁ = 3.25 – 1.352 = 1.898 m (say 1.9 m)
Hence L₃ = 6.5 – (1.9 + 2.45) = 2.15 m

Alternatively, the footing can be proportioned using a straight-forward formula based on the calculations done above;

L₃ = 0.5L – P₁L₂/(P₁ + P₂) = (6.5/2) – (787 × 2.45)/(787 + 970) = 2.15 m

The final disposition of the footing is given below assuming a trial footing depth of 600 mm.

We can check the preliminary thickness of 600 mm by considering the punching shear at the column perimeter. At the column perimeter, the maximum punching shear stress should not be exceeded.

At ultimate limit state;
P₁ = 1.35G_k + 1.5Q_k= 1.35(665) + 1.5(122) = 1080.75 kN
P₂ = 1.35G_k + 1.5Q_k= 1.35(825) + 1.5(145) = 1331.25 kN
Total ultimate load = 1080.75 + 1331.25 = 2412 kN

V_Ed< V_Rd,max

Where;
V_Ed = βV_Ed /u₀d

V_Rd,max = 0.5vf_cd

Considering Column P₂ bearing the maximum axial load, V_Ed = 1331.25 kN

β = 1.5 (an approximate value from clause 6.4.3(6) of EN 1992-1-1:2004)
u₀= column perimeter = 2(300) + 2(300) = 1200 mm
v = 0.6[1 – f_ck/250] (strength reduction factor for concrete cracked in shear)
v = 0.6[1 – 30/250] = 0.528
f_cd = α_ccf_ck/γ_cf_cd = (1.0 × 30)/1.5 = 20 N/mm²V_Rd,max = 0.5 × 0.528 × 20 = 5.28 N/mm²
V_Ed = (1.5 × 1331.25 × 1000) / (1200mm × d)

Therefore;
1996874/1200d = 5.28

On solving;
The minimum thickness of footing d_min = 315.263 mm

Hence the depth provided exceeds the minimum required.

Earth pressure intensity at ultimate limit state = (1080.75 + 1331.25)/16.25 = 148.43 kN/m²

For a width of 2.5m, q = 148.43 × 2.5 = 371.075 kN/m

Bending moment at column P₁ = (371 × 1.9²)/2 = 669.65 kNm
Bending moment at column P₂ = (371 × 2.15²)/2 = 857.47 kNm

The maximum bending moment in the span will occur at the point of zero shear;
M_x = 371x²/2 – 1080.75(x – 1.9) = 185.5x² – 1080.75x + 2055.425
∂M_x/∂x = 371x – 1080.75 = 0
On solving; x = 1080.75/371 = 2.91 m

M_max= 185.5(2.91)² – (1080.75 × 2.91) + 2055.425 = 479.275 kNm

The shear force just to the left of column P₁ = (371 × 1.9) = 704.9 kN
The shear force just to the right of column P₁ = 704.9 – 1080.75 = -375.85 kN
The shear force just to the left of column P₂ = -375.85 + (371 × 2.45) = 533.1 kN
The shear force just to the right of column P₂ = 533.1 – 1331.25 = -798.15 kN

As an be seen, there is no hogging moment in the combined footing.

Bottom Reinforcement
We are supposed to take our maximum moment from the column face which can be easily determined as shown below.

M_Ed = 857.47 kNm
Effective depth (d) = 600 – 50 – 10 = 540 mm (assuming 20mm bars).
b = 2000mm

k = M_Ed/(f_ckbd² )
k = (857.47 × 10⁶)/(30 × 2500 × 540² ) = 0.039

Since k < 0.167 No compression reinforcement required
z = d[0.5 + √(0.25 – 0.882k)] = z = d[0.5 + √(0.25 – (0.882 × 0.039)] = 0.95d

A_s1 = M_Ed/(0.87f_yk z)
A_s1= (857.47 × 10⁶)/(0.87 × 500 × 0.95 × 540) = 3842 mm²
Provide 13H20mm @ 200mm c/c BOT (A_Sprov = 4082 mm²)

Top reinforcement
There is no hogging moment on the footing as shown in the bending moment diagram. However, we can provide minimum reinforcement at the top.

A_smin = 0.0013bd = (0.0013 × 1000 × 540) = 702 mm²
Provide H12mm @ 150mm c/c BOT (A_Sprov = 753 mm²/m)

Transverse Reinforcement
Cantilever arm = (2.5 – 0.3)/2 = 1.1 m
Designing per unit strip (per 1000 mm)
M_Ed = [148.43 × 1.1²]/2 = 89.8 kNm
Effective depth (d) = 600 – 50 – 10 = 540 mm (assuming H20mm bars).

k = M_Ed/(f_ckbd² )
k = (89.8 × 10⁶)/(30 × 1000 × 540² ) = 0.010

Since k < 0.167 No compression reinforcement required
z = d[0.5+ √(0.25 – 0.882k)] = z = d[0.5+ √(0.25 – (0.882 × 0.010)] = 0.95d

A_s1 = M_Ed/(0.87f_yk z)
A_s1= (89.8 × 10⁶)/(0.87 × 500 × 0.95 × 540) = 402 mm²
Provide H12mm @ 200mm c/c BOT (A_Sprov = 565 mm²/m)

Check for shear
Clause 6.4.2(2) of EN 1992-1-1:2004 states that control perimeters at a distance less than 2d should be considered where the concentrated force is opposed by a high pressure (e.g. soil pressure on a base), or by the effects of a load or reaction within a distance 2d of the periphery of the area of application of the force.

Beam shear
Check critical section d away from the most loaded column face
V_Ed = 148.43 x (2 – 0.54) x 2.5 = 541.7 kN
v_Ed = (541.7 x 10³)/(2500 x 540) = 0.401 N/mm²

v_{Rd, c} = C_{Rd, c} × k × (100 × ρ₁ × f_ck) ^0.3333
C_{Rd, c} = 0.12
k = 1 + √ (200/d) = 1 + √ (200/540) = 1.608
ρ = 4082/(540 × 2500) = 0.00302
v_{Rd, c} = 0.12 × 1.608 × (100 × 0.00302 × 30)^0.333 = 0.402 N/mm²
=> v_Ed (0.401 N/mm²) < v_Rd,c (0.402 N/mm²) Beam shear ok

Punching Shear
Punching shear: Basic control perimeter at 2d from face of column
v_Ed = βV_Ed/u_id < v_Rd,c

β = 1,
u_i = (300 x 4 + 540 x 2 x 2 x π) = 7986 mm

V_Ed = load minus net upward force within the area of the control perimeter)
V_Ed = 1331.25 – 148.43 x (0.30² + π x 1.080² + 1.080 x 0.30 x 4) = 581.6 kN
v_Ed = (581.6 x 10³)/(7986 x 540) = 0.1348 N/mm²

Punching shear resistance of the section
V_Rd,c = [C_Rd,c.k.(100ρ₁ f_ck)^(1/3) ] × 2d/a ≥ (V_min × 2d/a)
a = 540 + 540 = 1080 mm

C_Rd,c = 0.18/γ_c = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/540) = 1.606 > 2.0, therefore, k = 1.608

V_min = 0.035k^(3/2) f_ck^0.5
V_min = 0.035 × (1.608)^1.5 × 30^0.5 = 0.391 N/mm²
ρ_x (in the transverse direction) = A_s/bd = 565/(1000 × 524) = 0.001078
ρ_z (in the longitudinal direction) = A_s/bd = 4082/(2500 × 540) = 0.003023

ρ₁ = √(ρ_x× ρ_z) = 0.0018054 < 0.02

V_Rd,c = [0.12 × 1.608 (100 × 0.0018054 × 30 )^(1/3)] × (2 × 540)/1080 = 0.338 N/mm² (Take V_min = 0.391 N/mm²)

0.1348 N/mm² < 0.391 N/mm²

The section is ok for punching shear.

With this, the reinforcement detailing can be drawn.

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Analysis of Statically Indeterminate Truss Using Force Method: Solved Example

By

Ubani Obinna Uzodimma

-

January 24, 2018

Introduction
Indeterminate trusses are analysed usually analysed using force method or direct stiffness method. In this post, we are going to analyse step by step, the analysis of the truss loaded as shown below.

(EA = Constant)

Solution

Step 1: Determine the degree of static indeterminacy
For a truss to be determinate;
m + r = 2j
Where;
m = number of members = 13
r = number of support reactions = 4
j = number of joints = 8

Therefore;
13 + 4 – 2(8) = 1
Therefore, the truss is indeterminate to the 1st order at the supports.

Step 2: Select redundant and remove constraint
To solve for the single degree of indeterminacy, the structure has to be reduced to a statically determinate and stable structure. This can be achieved by removing a redundant support, and a little consideration will show that for the structure to be stable, only a horizontal redundant will have to be removed. In this case, let us remove the horizontal support at H. This will give us the basic system that is given below.

Step 3: Analyse the basic system completely and obtain the internal forces
We now have to obtain the support reactions and internal forces using the principles of statics.

Support Reactions

∑M_H = 0 (clockwise positive)

16Ay – (4 × 12) – (10 × 8) – (5 × 4) – (3 × 2) = 0

16A_y = 154

Ay = 9.63 kN

∑M_A = 0 (anticlockwise positive)

16Hy – (5 × 12) – (10 × 8) – (4 × 4) + (3 × 2) = 0

16Hy = 150

Hy = 9.37 kN

∑F_X = 0

Ay + 3 = 0

Ay = -3 kN

Read Also…..

Practical Analysis and Design of Roof Trusses to Eurocode 3

Analysis of Indeterminate Trusses Using Stiffness Method

Internal Forces

Joint A

Geometrical Properties
θ = tan^-1(2/4) = 26.57°

∑F_y = 0
-F_AC sinθ + 9.63 = 0
F_AC = 9.63/sin 26.57° = 21.53 kN (Tension)

∑F_x = 0
F_AB + F_AC cosθ – 3 = 0
F_AB = 3 – (21.53 cos 26.57) = -16.27 kN (Compression)

Joint B

∑F_y = 0
-F_BC – 4 = 0
F_BC = -4kN (Compression)

∑F_x = 0
-F_AB + F_BD= 0
F_AB = F_BD= -16.27 kN (Compression)

To simplify our analysis, we can obtain the forces in members CE and CD through the method of sections.

∑M_C = 0

(9.63 × 4) – (3 × 2) + (2 × F_BD) = 0

F_BD = -16.27 kN (Compression, verification of our answer above)

∑M_D = 0

(9.63 × 8) – (4 × 4) + (2 × F_CE) = 0

F_BD = 30.52 kN (Tension)

∑M_A = 0

(30.52 × 2) – (4 × 4) + (8 × F_CDsinθ) = 0

F_CD = -12.59 kN (Compression)

Joint H

∑F_y = 0
-F_GH sinθ + 9.37 = 0
F_GH = 9.37/sin 26.57° = 20.95 kN (Tension)

∑F_x = 0
-F_FH – F_GH cosθ = 0
F_FH = -(20.95 cos 26.57) = -18.74 kN (Compression)

Joint E

∑F_y = 0
F_ED + 0 = 0
F_ED = 0 (No force)

∑F_x = 0
-F_CE + F_EG= 0
F_CE = F_GE= 30.52 kN (Tension)

Joint D

∑F_y = 0
-F_DC sinθ – F_DG sinθ – 10 = 0
-(-12.59 sinθ) – F_DG sinθ – 10 = 0
F_DG = -4.37/sin 26.57° = -9.77 kN (Compression)

Step 4: Calculate the deformation at the redundant
Having obtained the internal forces at the redundant, we can now use virtual work method to calculate the horizontal translation at support H which corresponds to the removed redundant Hx. This is done by removing the external load on the basic system and applying a unit horizontal force in the direction of the removed force.

By looking at what is happening at the above structure, you will agree with me that the entire top chord is in a uniform compression of -1.0 kN.

Hence;
F_AB = F_BD= F_DF= F_FH= -1.0 (Compression)
All other members of the truss have zero forces.

Since the horizontal displacement at H in the original structure is equal to zero, it means that the horizontal reaction at support H must produce a deflection in the opposite direction that will counter the deflection.

Deformation of the basic system due to externally applied load can be obtained by using the relationship below;

Where;
m = number of members
n = Internal forces due to virtual load
N = Internal forces due to externally applied load
L = Length of member
A = Cross sectional area of member
E = Modulus of Elasticity of member

To analyse it, we normally present this in tabular form. Due to the fact that the force in the members in the virtual load state are zero in all members except the top chord, we will just focus on the top chord.

δ_1P = 2(65.08/AE) + 2(74.96/AE) = 280.08/AE

The deflection at point H due to the virtual load can be obtained using the same relationship, and this is shown in the table below;

δ₁₁ = 4(4/AE)) = 16/AE

The appropriate cannonical equation is therefore given by;

δ₁₁X₁+ δ_1P = 0

On substituting;
16X₁+ 280.08 = 0
X₁= -17.51 kN ←

Therefore, the horizontal reaction at H (Hx) is 17.51 kN←

Having obtained this, we can now analyse the truss again to obtain the final internal forces.

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ProtaStructures Conference/Live Event Coming Up in Nigeria

By

Ubani Obinna Uzodimma

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January 23, 2018

ProtaSoftware in conjuction with ApeleDesigns invites you to the first ever live event/conference in Nigeria. ProtaStructure is the all-in-one package for multi-material modelling with steel, concrete and composite members, 3-D finite element analysis and code-compliant design of building structures.

Theme:

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DATE: Saturday, March 03, 2018
TIME: 09:30am – 02:30pm

VENUE: NECA House Plot A2, Hakeem Balogun Street, Central Business District, Alausa, Ikeja, Lagos, NIGERIA.

Registration: FREE

Please register at www.protasoftware.com/events

(note: the venue has limited seats which will be allocated on a first come first serve basis)

In this year’s seminar series, we will explore innovations Prota is making in helping engineers take advantage of changes in the way projects are designed, documented, coordinated, and constructed. Attendees will learn how 3D modelling technology can be effectively used to design multi-material structures including steel and concrete elements, automatically produce and manage high quality drawing and calculation documentation, and share 3D BIM information with Architects, Owners and other project stakeholders.

Speakers:

Joseph Kubin [Msc] – CEO, Prota and Creator of CSC Orion

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Event Programme:

9:30 am Registration

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10.15 am Unveiling ProtaStructure 2018 – Integrated Multi-Material Design Solution for Building Systems

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11.45 am Morning Tea Break

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2:00 pm Question & Answer

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How to Apply Load Model 1 (LM 1) on Highway Bridges

By

Ubani Obinna Uzodimma

-

January 23, 2018

Table of Contents

Highway bridges are designed to resist various systems of loads (actions) and load combinations. One of the most prominent actions on highway bridges is vehicular traffic. Practically, traffic actions on bridges consist of different vehicular categories, such as standard automobiles, lorries (trucks), and specialized industrial transport vehicles.

However, for the purpose of bridge design, traffic actions on highway bridges can simplified using traffic Load Models specified in the Eurocodes. Other codes in the world such as AASHTO, IRC, etc have their specialised loadings for bridges. The resulting loads manifest as a complex interplay of vertical and horizontal forces, further categorized as static (constant) and dynamic (variable).

In the Eurocodes, traffic load models were calibrated for highway bridges having carriageway width less than 42m, and span length up to 200m. The load models aim to reproduce the real values of the effects induced in the bridges by real traffic. Therefore, these models are artificial models and not necessarily real vehicles.

Calibration of the traffic models was based on real traffic data recorded in experiments performed in Europe between 1980 and 1994.

European Bridge Load Models

– Load Model 1 (LM 1): This load model reproduces traffic loads to be used for local and global verifications. It is made up of concentrated load and UDL, which can be thought of in terms of HA load of BS 5900, even though they are very different.

– Load Model 2 (LM 2): This load model reproduces effects on short structural members and is composed of single axle load on specific rectangular tyre contact areas.

– Load Model 3 (LM 3): This is for special vehicles, and it is to be considered on request. It represents abnormal vehicles.

– Load Model 4 (LM 4): Crowd loading.

To ensure a robust and safe bridge design, it is important to employ load models that are specifically tailored to the location of the bridge. By accounting for the aforementioned variations in traffic patterns and vehicle types, location-specific load models provide a more accurate representation of the actual loading conditions the bridge will encounter throughout its service life.

Division of Carriageway into notional lanes

A carriageway is the part of the roadway sustained by a single structure (deck, pier etc). It includes all the physical lanes, the hard shoulders, the hard strips, etc. The carriageway is divided into notional lanes and the remaining area.

The subdivision of roads into notional lanes is summarised in the table below;

Carriageway width (w)	Number of notional lanes (n)	Width of the notional lane	Width of the remaining area
w < 5.4 m	1	3m	w – 3m
5.4 ≤ w < 6	2	0.52	0
6m ≤ w	int(w/3)	3m	w – (3 x n)

Load Model 1

Load Model 1 represents the effects of normal traffic and comprises tandem axles (TS) superimposed over a uniformly distributed load (UDL) whose intensity remains constant with the loaded length. The model is very different from the Type HA loading given in BD37. Type HA loading consists of a uniformly distributed load, the intensity which varies with the loaded length, and a constant Knife Edge Load (KEL) of 120 kN.

There are also lane factors for different lengths which account for the simultaneity of loading in adjacent lanes as a function of loaded length. Eurocode (EN 1991-2) load model also differs from BD37 in the way that the carriageway is divided into notional lanes. In EN 1991-2, the notional lane width is constant at 3.0m except for a small range of carriageway width between 5.4m and 6.0m when the lane width varies from 2.7m to 3.0m. (See Table above).

The load model 1 which generally represents normal vehicles consists of two sub-systems;

(1) A system of two concentrated axle loads representing a tandem system weighing 2.α_kQ_k

(2) A system of distributed load having a weight density per square metre of α_kq_k

The characteristic values of LM 1 forces;

The Tandem System

For the assessment of local effects, a tandem axle system should be placed at the location on the bridge deck that experiences the most unfavourable loading conditions. In scenarios where two tandem systems are considered on adjacent design lanes, these systems may be positioned closer together. However, the minimum centre-to-centre spacing between wheel axles within each tandem system must not be less than 0.50 meters.

In Load Model 1, each axle within a tandem system shall be modelled as possessing two identical wheels. Consequently, the load acting on each individual wheel is calculated as one-half the product of the factor α, the characteristic axle load (Q_k), and a distribution factor.

The contact surface of each wheel shall be idealized as a square with a side length of 0.40 meters. This simplifies the analysis by representing the wheel’s footprint as a uniformly distributed load over a defined area of 0.16 m².

The Uniformly Distributed Load

This load model represents a constant weight distributed per unit area across a designated lane width. The characteristic load intensity (q_k) is multiplied by an adjustment factor (α_ik) to obtain the characteristic value (α_ikq_k) for UDL application. These adjustment factors account for potential variations in traffic patterns and weight distributions.

Load Model 1, incorporating the UDL system just described, should be applied to each designated traffic lane (“notional lane”) as well as any remaining bridge areas not designated as traffic lanes. For each notional lane “i,” the characteristic load magnitudes are denoted as α_ikQ_ikand α_ikq_ik, respectively, referencing the values provided in Table 4.2 of EN-1991-2.

These symbols represent the adjusted axle load (α_ikQ_ik) and the adjusted UDL load (α_ikq_ik) specific to lane “i.” In the remaining bridge areas not designated for traffic, the design load magnitude is denoted as α_rkq_rk, where α_rk is the corresponding adjustment factor and q_rk is the characteristic load intensity for that area.

Selection of Adjustment Factors

The specific values assigned to the adjustment factors depend on the anticipated traffic patterns on the bridge and may also vary based on the classification of the route (e.g., highway versus local road). They are however usually specified in the national annex. If no specific information is available regarding these factors, they should be assumed to be equal to 1.0 for all cases.

The adjustment factors specified in Eurocode 1, National Application Document (NAD, UK), and UK National Annex are shown in Table below.

Rules for applying Load Model 1

In each notional lane, only one tandem system should be considered, situated in the most unfavourable position.
The tandem system should be considered travelling in the longitudinal axis of the bridge.
When present, the tandem system should be considered in full i.e with all four wheels
The UDL’s are applied longitudinally and transversally on the unfavourable parts of the influence surface.
The two load systems can insist on the same area.
The dynamic impact factor is included in the two load systems.
When static verification is governed by the combination of local and global effects, the same load arrangement should be considered for the calculation of local and global effects.

Solved Example

A simply supported bridge deck has a carriageway width of 8m and a span of 18m. Determine the maximum bending moment on the bridge deck due to Load Model 1.

Solution

Carriageway width w = 8 m
Number of notional lanes = 8/3 = 2
Width of the remaining area = 8m – (3 x 2) = 2m

Let us analyse for the worst load condition on Notional Lane 1;

Axle Load
Each axle load of TS = α_Q1Q_1k = 1.0 x 300 = 300 kN

Figure 2 illustrates the calculation of bending moment (M_y) and shear force (V_z) at a specific section “s” located at a distance “x” from the support of a simply supported beam. The beam is subjected to two point loads positioned 1.2 metres apart, representing the configuration of a Load Model 1 Tandem System (LM1 TS) oriented longitudinally. The calculations are performed using the influence line method.

The influence line for the bending moment is drawn with their positive values below the reference line. Load effects from that model are as follows:

M_y= φ₁ + φ₂

Where φ₁ and φ₂ are the corresponding influence line ordinates under the point loads.
φ₁ = x (L – x) /L
φ₂ = x (L – x – 1.2) /L

Notably, the maximum bending moment induced by these tandem wheel loads does not occur precisely at the midpoint of the beam span. The location of the section experiencing the greatest bending moment can be determined by solving the following equation:

dM/dx = (2L – 4x – 1.2)/L = 0

The section with the maximum bending moment from the Load Model 1 tandem system is at a distance from the support:
x = 0.5L – 0.3

Therefore, in a simply supported beam system, the maximum bending moment for the Load Model 1 tandem system occurs under the axle when the vehicle is positioned such that the midpoint of the beam span coincides with the midpoint of the distance between the centre of gravity (centroid) of the entire vehicle load and the axle closest to that support.

With an 18m span and the axles spaced 1.2m apart, the rear axle is placed at 0.3m from the mid-span.

Support Reactions

Let ∑M_B = 0
18V_A – (300 x 9.3) – (300 x 8.1) = 0
V_A = 290 kN

Let ∑M_A = 0
18V_B – (300 x 8.7) – (300 x 9.9) = 0
V_B = 310 kN

M₁ = 290 x 8.7 = 2523 kNm
M₂= 310 x 8.1 = 2511 kNm

Maximum bending moment on 1m width of the lane due to TS = 2523/3 = 841 kNm/m

Uniformly Distributed Load
Adjustment factor α_q1 = 0.61
Hence UDL system = 0.61 x 9 = 5.5 kN/m²

The maximum mid-span moment on 1m width of lane due to UDL = 5.5 x 18²/8 = 222.75 kNm/m

Total maximum moment due to LM1 = 841 + 222.75 = 1063.75 kNm per metre width of lane.

Sources and Citations
Pietro Croce et al (2010): Guidebook 2 Design of Bridges; Published by Czech Technical University in Prague, Klokner Institute. ISBN 978-80-01-04617-3

Design Example of Steel Beams According to Eurocode 3

By

Ubani Obinna Uzodimma

-

January 22, 2018

Introduction
This post deals with the design of simply supported I-beam section subjected to permanent and variable loads according to Eurocode 3. The design involves selecting the appropriate section that will satisfy limit state requirements.

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Problem Statement
It is desired to select an appropriate section to satisfy ultimate and serviceability limit state requirements for a laterally restrained simply supported beam that is subjected to the following loads;
Permanent Load Gk = 38 kN/m
Variable Load Qk = 12 kN/m

The length of the beam = 7.5m

Solution

At ultimate limit state,
P_Ed = 1.35Gk + 1.5Qk
P_Ed = 1.35(38) + 1.5(12) = 69.3 kN/m

An advanced UK beam S275 is to be used for this design.
F_y = 275 N/mm²
γ_m0= 1.0 (Clause 6.1(1) NA 2.15 BS EN 1993-1- 1:2005)

The required section is supposed to have a plastic modulus about the y-y axis that is greater than;
W_pl,y = M_y,Edγ_m0/F_y

W_pl,y = (487.265 × 10³ × 1.0)/275 = 1771.872 cm³

From steel tables, try section 457 x 191 x 82 W_pl,y = 1830 cm³

CHECK OUT ALSO….
Design of Steel Beams to BS 5950 – 1: 2000
Structural Analysis of Compound Arch-Frame Structure

Properties
h = 460.0mm; b = 191.3mm; d = 407.6mm; t_w = 9.9mm; t_f; r = 10.2mm; A = 104 cm⁴; I_y = 37100 cm⁴; I_z = 1870 cm⁴; W_el,y = 1610 cm³; W_el,y = 1830 cm³

h_w = h – 2t_f = 428.0mm

E (Modulus of elasticity) = 210000 N/mm²(Clause 3.2.6(1))

Classification of section
ε = √(235/F_y) = √(235/275) = 0.92 (Table 5.2 BS EN 1993-1- 1:2005)

Outstand flange: flange under uniform compression c = (b – t_w – 2r)/2 = [191.3 – 9.9 – 2(10.2)]/2 = 80.5mm

c/t_f = 80.5/16.0 = 5.03

The limiting value for class 1 is c/t_f ≤ 9ε = 9 × 0.92
5.03 < 8.28
Therefore, outstand flange in compression is class 1

Internal Compression Part (Web under pure bending)
c = d = 407.6mm
c/t_w = 407.9/9.9 = 41.17

The limiting value for class 1 is c/t_w ≤ 72ε = 72 × 0.92 = 66.24
41.17 < 66.24
Therefore, the web is plastic. Therefore, the entire section is class 1 plastic.

Member Resistance Verification
Moment Resistance
For the structure under consideration, the maximum bending moment occurs where the shear force is zeo. Therefore, the bending moment does not need to be reduced for the presence of shear force (clause 6.2.8(2))

M_Ed/M_c,Rd ≤ 1.0 (clause 6.2.5(1))

M_c,Rd = M_pl,Rd= (M_pl,y× F_y)/γ_m0

M_c,Rd = M_pl,Rd= [(1830 × 275)/1.0] × 10^-3 = 503 kNm

M_Ed/M_c,Rd = 487.265/503 = 0.9687 < 1.0 Ok

Shear Resistance (clause 6.6.2)
The basic design requirement is;

V_Ed/V_c,Rd ≤ 1.0

V_c,Rd = V_pl,Rd= A_v(F_y/ √3)/γ_m0(for class 1 sections)
For rolled I-section with shear parallel to the web, the shear area is;

A_v = A – 2bt_f + (t_w + 2r)t_f (for class 1 sections) but not less than ηh_wt_w

A_v = (104 × 10² – (2 × 191.3 × 16) + [9.9 + 2(10.2)] × 16 = 4763 mm²
η = 1.0 (conservative)
ηh_wt_w= (1.0 × 428 × 9.9) = 4237 mm²
4763 > 4237
Therefore, A_v = 4763 mm²

The shear resistance is therefore;
V_c,Rd = V_pl,Rd= [4763 × (275/ √3)/1.0] × 10^-3 = 756 kN

V_Ed/V_c,Rd = 259.875/756 = 0.343 < 1.0 Ok

Shear Buckling
Shear buckling of the unstiffnened web will not need to be considered if;

h_w/t_w≤ 72ε/η

h_w/t_w= 428.0/9.9 = 43
72ε/η = (72 × 0.92)/1.0 = 66

43 < 66 Therefore shear buckling need not be considered.

Serviceability limit state
Vertical deflections are computed based on variable loads. Permanent loads need not be considered.(BS EN 1993-1-1 NA 2.23)

Q_k = 12 kN/m

w = 5ql⁴/384EI

w = (5 × 12 × 7500⁴)/(384 × 210000 × 37100 × 10⁴) = 6.345mm
Span/360 = 7500/360 = 20.833mm (BS EN 1993-1-1 NA 2.23)

6.345mm < 20.833mm. Therefore, deflection is satisfactory

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Precise Moment Distribution Analysis of Non-sway Frames

By

Ubani Obinna Uzodimma

-

January 22, 2018

Precise moment distribution is a variation of moment distribution method which aims to shorten the time spent carrying out the normal moment distribution method. This is achieved by introducing continuity and distribution factors, which in turn means that only one distribution and carry over is required to achieve the final bending moment values. This post explores this method of structural analysis, and a solved example is used to drive the point home.

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Procedure for Carrying Out Precise Moment Distribution method (Reynolds and Steedman, 2005)

(1) Calculate fixed end moments
(2) Determine continuity factor for each span from the general expression;

Where;
ϕ_n+1 and k_n+1 are continuity and stiffness of span being considered
∑[k_n/(2 – ϕ_n)] is the sum of values of [k_n/(2- ϕ_n)] of all remaining members meeting at that joint
If far end of column is fully fixed, then [k_n/(2- ϕ_n)] = 2k_n/3, since ϕ_n= 0.5
If far end of column is pinned, then [k_n/(2- ϕ_n)] = k_n/2

You have to work from left to right, and from right to left

(3) Calculate distribution factors at joints from general expression;

Where ϕ_BCand ϕ_CBare continuity factors obtained from step 2.
Unlike in continuous beams, sum of distribution factors each side of support will not equal to unity due to action of columns. Distribution factor for each column is obtained by dividing total column distribution factor in proportion to stiffness of columns.

(4) To carry over moment at supports, multiply distributed balancing moment at left hand end of member by continuity factor obtained from working RIGHT to LEFT, and carry over this value to the right hand side. At this point balance carried over moment by dividing an equal amount of opposite sign between then remaining members meeting at that joint with respect to their values of [k_LR/(2- ϕ_RL)]

(5) Undertake one complete carry over operation working from left to right and then from right to left from each joint at which fixed end moment occurs and sum results to obtain final moment of the system.

Solved Example
For the non-sway frame loaded as shown below, obtained the bending moments on the frames using precised moment distribution method.

Solution
(1) Fixed End Moment
F_CD = qL₂/12 = (-15 × 8²)/12 = -80 kNm
F_DC = qL₂/12 = (15 × 8²)/12 = 80 kNm

Relative Stiffnesses
For columns K = 2EI/h = 2/3
For beam K = EI/L = 1/8

For columns with far ends fully fixed;
2k_n/3 = 2(2/3)/3 = 4/9

Continuity factors

Distribution Factors (DF)

DF_CD = DF_DC = (1 – 2ϕ_CD)/(1 – ϕ_CDϕ_DC)
DF_CD = DF_DC = (1 – 2 × 0.467) / (1 – 0.467 × 0.467) = 0.0844

The remaining distribution factor at each node = 1 – DF_CD = 1 – 0.0844 = 0.9156

This distribution factor is now shared among the columns according to their stiffnesses. Since the stiffnesses are equal = 0.9156/2 = 0.4578

Now the moments are distributed by multiplying the fixed end moments by the distribution factor in the opposite sign.

E.g For the left hand side of the beam; M = -80 × 0.084 = +6.72 (note the change in sign)
For the upper column (UC) ; M = -80 × 0.458 = +36.64 (note the change in sign)
Using this same approach, you can compute the remaining.

Carry over
The moment of +6.72 is carried over from DC to CD based on the continuity factor, and once again this is done with the reverse sign (check step 4 above).

For instance, +6.72 is multiplied by 0.467 which gives -3.138 (note the change in sign).
This moment of -3.138 is shared in opposite sign to the remaining members in the proportion of their [k_LR/(2 – ϕ_RL)].

For instance for the columns, [k_LR/(2 – ϕ_RL)] = [(2/3) /(2 – 0.467)] = 0.4348
For the beam, [k_LR/(2 – ϕ_RL)] = [(1/8) /(2 – 0.467)] = 0.08155

Now, the moment shared to each column (in opposite sign) is given by;
MUC = MLC = -3.138 × [0.4328/(0.4348 + 0.4348 + 0.08155] = 2.282 kNm

The final moments are now obtained by summing up the fixed end moment, the distributed moment, and the carried over moment.

These steps are normally prepared using the simple table below;

UC = Upper column; LC = Lower Column

To verify our result, kindly look at the result from Staad Pro for the same frame;

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How to Load Columns in RC Building Design

By

Ubani Obinna Uzodimma

-

January 13, 2018

When carrying out the manual design of reinforced concrete structures, column loads are usually assessed by considering the support reactions from beams they are supporting, or by the tributary area method. The latter is more popular due to its simplicity and speed, but usually fails to capture all the loads that are imposed on the columns (such as wall loads), while the former is more complex and time-consuming, it usually very representative of all the possible loads that are imposed on the column.

Steps on how to load a column from beam support reaction

The following steps can be adopted when using beam support reactions to obtain the axial load on columns.

(1) Load the floor slab adequately, and factor the loads at ultimate limit state, using the appropriate load combination.

(2) Load all the beams that are connected to the column.

(3) Transfer the loads from slab to the beam using the appropriate relationship. Based on yield lines, the loads are usually triangular or trapezoidal, but this is cumbersome to analyse manually. An equivalent UDL (reasonably accurate) can be used to transfer slab loads to beams by employing the formulas below.

One way slabs
Long span q = nl_x/2
Short slab q = nl_x/5

Two way slabs
Long span q = nl_x/2(1 – 1/3k²)
Short span q = nl_x/3

Where;
q = Load transferred from slab to the beam
n = load at ultimate limit state
1.4g_k + 1.6q_k(BS 8110)
1.35g_k + 1.5q_k (Eurocode 2)
k = L_y/L_x
L_y= Length of long span of slab
L_x = Length of short span of the slab

(4) Analyse the floor beams completely using any suitable method of your choice, while also considering any additional load that may be on the beam such as wall load and finishes.

(5) Obtain the support reactions of the beam, which represents the load that is transferred from the floor to the column.

Design Example

In this post, the floor plan shown below is for a shopping complex, and it desired to obtain the column axial loads at ultimate limit state.

Design Data
Size of all columns = 230 x 230mm
Size of all beams = 450 x 230mm
Thickness of slab = 200mm
Unit weight of concrete = 25 kN/m³
Unit weight of sandcrete block = 3.47 kN/m²
f_ck= 25 N/mm²
f_yk = 500 N/mm²

Load combination = 1.35g_k+ 1.5q_k
Design variable load (q_k) = 4 kN/m²

k = L_y/L_x = 6/5 = 1.2 (in all cases)
Only external beams are carrying block work load.

Load Analysis

Roof beam
Permanent load on roof beams g_k = 6 kN/m
Variable load on roof beam q_k = 1.5 kN/m
(These values are assumed)
At ultimate limit state, n = 1.35(6) + 1.5(1.5) = 10.35 kN/m

Slab Load Analysis
Concrete own weight = 25 kN/m³× 0.2m = 5.0 kN/m²
Screeding and Finishes (say) = 1.35 kN/m²
Partition allowance = 1.5 kN/m²
Total (g_k) = 7.85 kN/m²
Variable action (q_k) = 4 kN/m²

n = 1.35gk + 1.5qk = 1.35(7.85) + 1.5(4) = 16.6 KN/m²

Wall load on beam
Unit weight of sandcrete block = 3.47 kN/m²
Height of wall = 3.5m
Wall load on beam = 3.47 kN/m²× 3.5m = 12.145 kN/m

Load on Beams
External Longitudinal beams (Axis 1:A-D and Axis 3:A-D)
Self weight of beam (factored) = 1.35 × 0.23m × 0.25m × 25 = 1.94 kN/m
Load from slab = nl_x/2(1 – 1/3k²) = [(16.6 × 5)/2] × (1 – 1/(3 ×1.2²)) = 31.893 kN/m
Load from block work = 12.145 kN/m
Total load = 45.978 kN/m

External Transverse beams (Axis A:1-3 and Axis D:1-3)
Self weight of beam (factored) = 1.35 × 0.23m × 0.25m × 25 = 1.94 kN/m
Load from slab = nl_x/3 = (16.6 × 5)/3 = 27.667 kN/m
Load from block work = 12.145 kN/m
Total load = 41.752 kN/m

Internal Longitudinal Beam (Axis 2:A-D)
Self weight of beam (factored) = 1.35 × 0.23m × 0.25m × 25 = 1.94 kN/m
Load from slab = nl_x/2(1 – 1/3k²) = [(16.6 × 5)/2] × (1 – 1/(3 ×1.2²)) = 2 × 31.893 kN/m = 63.786 kN/m (we multiplied by two because this beam is receiving slab load from two directions.)
Load from block work = 0 (there are no block works inside the building)
Total load = 65.726 kN/m

Internal Transverse beams (Axis B:1-3 and Axis C:1-3)
Self weight of beam (factored) = 1.35 × 0.23m × 0.25m × 25 = 1.94 kN/m
Load from slab = nlx/3 = (16.6 × 5)/3 = 2 × 27.667 kN/m = 55.334 kN/m (slab load is coming from two directions)
Load from block work = 0
Total load = 57.274 kN/m

Self weight of Columns
Self weight of column (factored) = 1.35 × 0.23m × 0.23m × 3.75m × 25 kN/m³ = 6.695 kN

Structural Analysis
A little consideration will show that the support reactions for the floor beams (equal spans) can be obtained by considering the shear force coefficients given below;

**Fig 3:** Shear force coefficient (2 – span beam)

Fig 4: Shear force coefficient (3 span beam)

Load Analysis of Column A1
1st Floor
Load from longitudinal roof beam (1:A-D) = 0.4 × 10.35 kN/m × 6m = 24.84 kN
Load from transverse roof beam (A:1-3) = 0.375 × 10.35 kN/m × 5m = 19.41 kN
Self weight of column = 6.695 kN
Total = 50.945 kN

Ground Floor
Load from above = 50.945 KN
Load from longitudinal floor beam (1:A-D) = 0.4 × 45.978 kN/m × 6m = 110.347 kN
Load from transverse floor beam (A:1-3) = 0.375 × 41.752 kN/m × 5m = 78.285 kN
Self weight of column = 6.695 kN
Total = 246.272 kN

Load Analysis of Column A2
1st Floor
Load from longitudinal roof beam (2:A-D) = 0.4 × 10.35 kN/m × 6m = 24.84 kN
Load from transverse roof beam (A:1-3) = 2(0.625 × 10.35 kN/m × 5m) = 64.687 kN
Self weight of column = 6.695 kN
Total = 96.222 kN

Ground Floor
Load from above = 96.222 KN
Load from longitudinal floor beam (2:A-D) = 0.4 × 65.726 kN/m × 6m = 157.742 kN
Load from transverse floor beam (A:1-3) = 2(0.375 × 41.752 kN/m × 5m) = 156.57 kN
Self weight of column = 6.695 kN
Total = 417.229 kN

Load Analysis of Column B1
1st Floor
Load from longitudinal roof beam (1:A-D) = (0.5 × 10.35 kN/m × 6m) + (0.6 × 10.35 kN/m × 6m) = 68.31 kN
Load from transverse roof beam (B:1-3) = 0.375 × 10.35 kN/m × 5m = 19.41 kN
Self weight of column = 6.695 kN
Total = 94.415 kN

Ground Floor
Load from above = 94.415 KN
Load from longitudinal floor beam (1:A-D) = (0.5 × 45.978 kN/m × 6m) + (0.6 × 45.978 kN/m × 6m) = 303.455 kN
Load from transverse floor beam (B:1-3) = 0.375 × 57.274 kN/m × 5m = 107.388 kN
Self weight of column = 6.695 kN
Total = 511.953 kN

Load Analysis of Column B2
1st Floor
Load from longitudinal roof beam (2:A-D) = (0.5 × 10.35 kN/m × 6m) + (0.6 × 10.35 kN/m × 6m)= 68.31 kN
Load from transverse roof beam (A:1-3) = 2(0.625 × 10.35 kN/m × 5m) = 64.687 kN
Self weight of column = 6.695 kN
Total = 139.692 kN

Ground Floor
Load from above = 139.692 kN
Load from longitudinal floor beam (2:A-D) = (0.5 × 65.726 kN/m × 6m) + (0.6 × 65.726 kN/m × 6m) = 433.7916 kN
Load from transverse floor beam (A:1-3) = 2(0.625 × 57.274 kN/m × 5m) = 357.9625 kN
Self weight of column = 6.695 kN
Total = 938.141 kN

As you can see, that was very straightforward with the use of shear force coefficients, given the fact that the beams were of equal span. When the beams are not equal, you have to carry out the analysis, and transfer the shear forces accurately. Now, let us compare load analysis of column B2, by the tributary area method.

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Design of Pile Foundation Using Pile Load Test (Eurocode 7)

By

Ubani Obinna Uzodimma

-

January 11, 2018

Pile load test is the most reliable method of estimating the load carrying capacity of a pile, but it is rather expensive. Load tests are performed on-site on test piles to verify the design capacity of the piles. A pile load test consists of applying increments of static load to a test pile and measuring the settlement of the pile. The load is usually jacked onto the pile using either large dead weight or a beam connected to uplifts anchor piles to supply reaction for the jack. Load tests could be by constant rate of penetration test (CRP) or maintained load Test (ML).

In arriving at the ultimate load of a pile from pile load test, several definitions of failure have been given some of which are as follows;

(1) Terzaghi (1942) says that a failure load is that which causes a settlement equal to 10% of the pile diameter.
(2) The CP 2004 defines failure load as that load at which the rate of settlement continues undiminished without further increase increment of load unless this rate is so slow as to indicate that this settlement may be due to consolidation of the soil. This definition is a bit difficult to apply in practice especially in cohesive soils.
(3) German DIN 4026 defines failure load as that load which causes irreversible settlement at the pile head equal to 2.5% of the pile diameter.

Design Example
It is desired to estimate the number of piles in a group of piles to carry the following loads;

Permanent load Gk = 3600 KN
Variable Load Qk = 1740 KN

Pile load test has been carried out on 3 piles on the site and the result is as shown below. The piles are 750mm diameter and 15m long. Assume settlement of 10% of the pile diameter as failure criterion.

Solution

(1)The ultimate resistance of the pile is the load at settlement of 10% of the pile diameter.

Settlement = 750 × (10/100) = 75mm

(2) From the load settlement graph for each pile;
Pile 1 R_m = 4156.25 KN
Pile 2 Rm= 4318.325 KN
Pile 3 R_m = 4887.8 KN

(3) The mean and minimum measured pile resistances are;
R_m,mean = 4454.125 KN
R_m,min = 4156.25 KN

(4) The characteristic pile resistance is obtained by dividing the mean and minimum measured pile resistances by the correlation factors ξ₁and ξ₂and choosing the minimum value. The equation below is given by equation 7.2 of Eurocode 7.

For 3 number of test piles (Table A9, EC7);
ξ₁ = 1.20
ξ₁ = 1.05

R_c,k = min{4454.125/1.2 , 4156.25/1.05} = 3711.77 KN

Design Approach 1

Combinations of sets of partial factors
DA1.C1 —– A1 + M1 + R1
DA1.C2 ——- A2 + M1 or M2 + R4

Partial factors for actions;
A1 γ_G = 1.35 γ_Q = 1.5
A2 γ_G = 1.0 γ_Q = 1.30

Partial factors for materials
M1 and M2 not relevant (γϕ’ = 1.0, not used)

Partial Resistance factors
R1 γ_t = 1.15 (Total/combined compression)
R4 γ_t = 1.5

DA1.C1 F_c,d = 1.35Gk + 1.5Qk = (1.35 × 3600) + (1.5 × 1740) = 7470 KN
DA1.C2 F_c,d = 1.0Gk + 1.3Qk = (1.0 × 3600) + (1.3 × 1740) = 5862 KN

For a single pile;
DA1.C1 R_c,d = R_c,k/γ_t = 3711.77/1.15 = 3227.62 KN
DA1.C2 R_c,d = R_c,k/γ_t = 3711.77/1.5 = 2474.513 KN

Assuming no pile group effect, for n piles,
Resistance = n × R_c,d
Hence,
DA1.C1 n ≥ F_c,d/R_c,d= 7470/3227.62 = 2.31
DA1.C2 n ≥ F_c,d/R_c,d= 5862/2474.513 = 2.36

Therefore, DA1.C2 controls, and 3 number of piles will be required.

Therefore
R_c,d= 3 × 2474.513 = 7423.539
F_c,d = 5862 KN

F_c,d/R_c,d= 5862/7423.539 = 0.789 < 1.0 Ok

Design Approach 2

Combinations of sets of partial factors
DA2 —– A1 + M1 + R2

Partial factors for actions;
A1 γ_G = 1.35 γ_Q = 1.5

Partial factors for materials
M1 not relevant (γϕ’ = 1.0, not used)

Partial Resistance factors
R2 γ_t = 1.1 (Total/combined compression)

F_c,d = 1.35Gk + 1.5Qk = (1.35 × 3600) + (1.5 × 1740) = 7470 KN

For a single pile;
R_c,d = R_c,k/γ_t = 3711.77/1.1 = 3374.336 KN

Assuming no pile group effect, for n piles,
Resistance = n × R_c,d
Hence,
n ≥ F_c,d/R_c,d= 7470/3374.336 = 2.21

Therefore number of piles required = 3 piles

Design Approach 3

Combinations of sets of partial factors
DA3 —– A1 + M1 + R3

Partial Resistance factors
R3 γ_t = 1.0 (Total/combined compression)

Since the R3 recommended partial resistance factor is 1.0, there is no margin for safety on the resistance provided. Therefore this cannot be used for the design.

Summary

3 Number of 750mm diameter piles is suitable for the load at ultimate limit state.

Thank you for visiting Structville today, and we will look at other aspects of pile design in our subsequent posts.

Meanwhile, have you read…
Solved example on elastic settlement of shallow foundations

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Analysis of Continuous Beams with Partially Distributed Load Using Force Method and Clapeyron’s Theorem

By

Ubani Obinna Uzodimma

-

January 6, 2018

For the continuous beam loaded as shown above, it is desired to find the bending moments at the critical points using force method (method of consistent deformations) and Clapeyron’s theorem (3 Moment Equation). We should however note that both methods are force methods (flexibility method) since we generally solve for unknown forces.

Solution

(1) By force method
Degree of static indeterminacy neglecting horizontal forces and reactions.

D = 2m + r – 2n
D = 2(3) + 4 – 2(4) = 2
Therefore the structure is indeterminate to the 2nd order.

Basic System
A basic system is a system that is statically determinate and stable. This obtained by removing the redundant supports at A and B, and replacing them with unit loads. See the figure below.

Case 1

Bending moment on the basic system due to vertical unit virtual load at support A

Case 2
Bending moment on the basic system due to unit virtual load at support B

Case 3
Bending moment due to externally applied load on the basic system;

Influence Co-efficients

δ₁₁

δ₁₁= [1/3 × 2L × 2L × 2L] + [1/3 × 2L × 2L × L] = 4L³

δ₂₂

δ₁₁= 2[1/3 × L × L × L] = 2L³/3

δ₂₁= δ₁₂

δ₁₁= [1/6 × L × L(4L + L)] + [1/3 × L × L × L] = 3L³/2

δ_1P

δ_1P= [1/6 × qL²/16 × (2L + 2L) × L/2]] + [15/12 × qL²/16 × L × L/2] = 13qL⁴/384

δ_2P

δ_2P= [1/6 × qL²/16 × (L + L) × L/2] + [15/12 × qL²/16 × L/2× L/2] = 13qL⁴/768

The appropriate cannonical equation is given by;

δ₁₁X₁ + δ₁₂X₂ + δ_1P = 0
δ₂₁X₁ + δ₂₂X₂ + δ_2P = 0

Therefore;

(4L³)X₁ + (3L³/2)X₂ = -13qL⁴/384
(3L³/2)X₁ + (2L³/3)X₂ = -13qL⁴/768

On sloving the above equations simultaneously;

X₁ = 13qL/1920 KN

X₂ = -13qL/320 KN

The final moment values can now be computed;

M_f = M₁X₁ + M₂X₂ + M_P

M_B = (13qL/1920 × L) + 0 + 0 = 13qL²/1920
M_C = (13qL/1920 × 2L) – (13qL/230 × L) + 0 = -13qL²/480
M_D = (13qL/1920 × L) – (13qL/230 × L/2) + qL²/16 = 47qL²/960

(2) By Clapeyron’s Theorem;

First of all, we draw the free bending moment diagram

By Clapeyron’s three moment equation (EI = constant, no sinking of support);

M_AL₁ + 2M_B + M_CL₂ + 6A₁X₁ + 6A₂X₂= 0

Geometrical Properties of the free moment diagram (centroid)

SPAN A – C
M_A = 0
2M_B(L + L) + M_CL = 0
4M_B L + M_CL = 0 ——————– (1)

SPAN B-D
M_D = 0
M_BL + 2M_C(L + L) = [-6 × 7qL³/192 × 13L/128]/L
M_B L + 4M_CL = -13qL³/128 ——————– (2)

Solving (1) and (2) simultaneously;
M_B = 13qL²/1920
M_C = -13qL²/480

Therefore, the bending moment due to externally applied load is given below;

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