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Analysis and Properties of Plastic Sections

In structural engineering designs, plastic section properties are used to analyze and design structural members that can withstand plastic deformation (such as structural steel). Plastic deformation is a permanent, non-reversible deformation that occurs when a material reaches its yield point.

According to clause 5.6 of EN 1993-1-1:2005, class 1 steel sections may be designed using plastic global analysis. The stress distributions in the elastic and fully plastic states for the general case of a steel section that is symmetrical about the plane of bending are shown in the picture below.

development of plastic hinge
Development of plastic hinge in an I-section

Tensile and compressive forces need to be equal for normal forces to be in equilibrium. This state is attained when the bending stress varies from zero at the neutral axis to a maximum at the extreme fibres. This is true especially when the neutral axis passes through the centroid of the section.

In a fully plastic state, equilibrium is attained when the neutral axis divides the section into two equal areas, since the stress at that state is equal to the yield stress of the material. Therefore,

Mpl = (first moment of area about the plastic neutral axis)fy

Where;
Mpl = plastic moment
fy = yield strength of the material

Key plastic section parameters

Usually, in the design of steel structures, plastic section properties are a set of properties that describe the ability of a cross-section to resist plastic bending. These properties include the plastic section modulus, the plastic moment of inertia, and the plastic rotation capacity.

(1) Plastic Moment (Mp): The plastic moment, denoted as Mp, is the maximum moment that a structural section can resist before it undergoes plastic deformation. It is a measure of the section’s capacity to resist bending without failing. The plastic moment can be calculated using the formula:

Mp = fyZp

where:
fy is the yield strength of the material.
Zp is the plastic section modulus, a property that quantifies the distribution of material around the section’s centroid.

(2) Plastic Section Modulus (Zp): The plastic section modulus, Zp, represents the ability of a structural section to resist plastic bending about a particular axis. The plastic section modulus is an important parameter in plastic design because it determines how much bending moment a section can sustain without undergoing plastic deformation.

(3) Plastic Neutral Axis (PNA): The plastic neutral axis is the location within a structural section where the moment arm is maximized, resulting in the maximum moment capacity.

(4) Plastic Section Shape Factor (α): The plastic section shape factor, denoted as α, is a dimensionless parameter that relates the plastic section modulus of a given shape to that of a reference shape (usually a rectangle). It is used to compare the plastic bending capacity of different section shapes while keeping the same material properties.

Solved Example

Determine the plastic section moduli about the y–y and z–z axes for the ‘I’ section shown in the figure below. The section is for a 457×191×98 kg/m Universal Beam with the root radius omitted.

plastic analysis i section

To determine the plastic section modulus about the y–y axis divide the section into A1 and A2 as shown in the figure below where;

equal area division of plastic section



A1 = (h/2) × tw = (467.2/2) × 11.4 = 2663.04 mm2
A2 = (bf − tw)tf = (192.8 − 11.4) × 19.6 = 3555.44 mm2

z1 = h/4 = 467.2/4 = 116.8 mm
z2 = (h – tf)/2 = (467.2 − 19.6)/2 = 223.8 mm

Plastic section modulus
Wply = 2(A1z1 + A2z2) = 2(2663.04 × 116.8 + 3555.44 × 223.8) = 2213501.088 mm3 = 2213.5 cm3

The value obtained from Section Tables is 2230 cm3 which is slightly greater because of the additional material at the root radius.

Similarly, for the plastic section modulus about the z–z axis divide the section into areas A3 and A4 as shown in the Figure below where;

plastic section z z

A3 = [(h – 2tf)tw]/2 = [(467.2 – 2×19.6) × 11.4]/2 = 2439.6 mm2
A4 = 2(bf/2)tf = 2(192.8/2) × 19.6 = 3778.88 mm2
y3 = tw/4 = 11.4/4 = 2.85 mm
y4 = bf/4 =192.8/4 = 48.2 mm

Plastic section modulus
Wplz = 2(A3y3 + A4y4) = 2(2439.6 × 2.85 + 3778.88 × 48.2) = 378189.752 mm3 = 378.189 cm3

The value obtained from Section Tables is 379 cm3 which is slightly greater because of the additional material at the root radius.

Design of Timber Shear Walls | Studs

Timber shear (stud) walls are an essential part of the structural system of timber structures/buildings. They are typically made up of a frame of vertical timber studs sheathed with plywood or other structural sheet material.

In the design of timber frame structures, one of the significant functions of walls is to provide load-bearing support and stability for vertical actions from the floor (decking) and roof (rafters or trusses), and also to provide strength and stability against the effects of lateral actions. These walls also accommodate various architectural features such as doors, windows, and built-in shelving.

details of stud walling
Figure 1: Details of a typical stud wall: (a) elevation; ( b) section; (c) typical fixing of top and bottom plates to studs

A timber stud wall is a type of wall that is made up of a frame of vertical timber studs (structural wooden posts), which are spaced evenly apart and held in place by horizontal plates at the top (header plates) and bottom (sole plates) of the wall. The studs are typically made of 75mm x 50mm or 100mm x 50mm sawn timber, and the spacing between them is typically between 450mm to 600mm.

Timber shear walls provide lateral resistance to wind and seismic forces, helping to keep the structure from undergoing failure due to lateral actions. The plywood sheathing helps to transfer the lateral forces to the studs, which then resists the forces by bending and shearing.

timber stud walling construction
Figure 2: Timber stud wall construction

Design of Stud Walls

As stated above, the vertical components in a timber shear wall system are commonly known as studs, while the walls are typically referred to as stud walls. These studs are aligned with their stronger axis (y-y) parallel to the wall’s surface and are securely positioned by the header and sole plates. Battens or noggins are utilized for in-plane stability, preventing lateral movement through diagonal or equivalent bracing members. These battens serve a dual purpose, both during construction and in persistent design scenarios.

In situations where the wall sheathing cannot provide sufficient lateral resistance, the use of diagonal or equivalent bracing (noggins) compensates for this. In such cases, the effective length of the stud around the z-z axis is determined by the longest span of stud between the plate and batten support.

When the sheathing material can provide adequate lateral restraint, concerns about stud buckling around the z-z axis can be disregarded. The sheathing material provided it is fixed to the studs and plates as per the manufacturer’s guidelines or design requirements, offers satisfactory lateral resistance. However, if sheathing is affixed to only one side of the wall, complete lateral restraint for the studs is not achieved, necessitating the use of a reduced effective length.

Stud walls are unlikely to exhibit any form of fixity at their ends. In the case of out-of-plane buckling around the y-y axis, the studs are deemed to be securely held in place and torsionally restrained by the fastenings to the header and sole plates. Nonetheless, these studs are allowed to rotate laterally at these points. Consequently, the effective length of the stud in relation to this axis is assumed to be equivalent to the height of the stud wall.

Timber stud analysis and design example (EN1995-1-1:2004)

In accordance with EN1995-1-1:2004 + A2:2014 incorporating corrigendum June 2006 and the UK national annex.

Stud details
Description;  47 x 125 C24 timber studs
Restraint in plane of panel; Sheathing
Stud spacing;  sStud = 600 mm
Stud height;  lStud = 2800 mm
Panel height; lPanel = lStud + 2b = 2894 mm

timber stud walling

Forces input on Stud
Lateral wind load;  HW_Z0_Stud = 1.50 kN/m2
Permanent distributed load on top rail; LG_Stud = 5.60 kN/m
Imposed distributed load on top rail; LQ_Stud = 8.40 kN/m

Stud loading details

Lateral wind load; 
pW,h0 = HW_Z0_Stud × sStud = 0.90 kN/m

Total vertical permanent point load (2800 mm);        
PG_1 = LG_Stud × sStud = 3.36 kN

Total vertical imposed point load (2800 mm);            
PQ_1 = LQ_Stud × sStud = 5.04 kN

Member Loads

The summary of the member loads is shown in the Table below.

MemberLoad caseLoad TypeOrientationDescription
MemberPermanentPoint loadGlobalZ3.36 kN at 2.8 m
MemberImposedPoint loadGlobalZ5.04 kN at 2.8 m
MemberWindUDLGlobalX0.9 kN/m at 0 m to 2.8 m

Loading Combinations

The load combinations used in the analysis of the structure are shown in the Table below.

Load combinationPermanentImposedSnowWind
1.35G + 1.5Q (Strength)1.351.500.000.00
1.35G + 1.5Q + ψS1.5S + ψW1.50W (Strength)1.351.500.750.75
1.0G + 1.5W (Strength)1.000.000.001.50
1.00G + 1.0Q (Service)1.001.000.000.00
1.00G + 1.0Q + ψS1.0S +ψW1.0W (Service)1.001.000.500.50
1.0G + 1.0W (Service)1.000.000.001.00

Support reactions

The support reactions from the load combinations are shown below.

 Load case/combinationFx (kN)Fz (kN)
1.35G + 1.50Q (Strength)012.2
1.35G + 1.50Q + ψS1.50S + ψW1.50W (Strength)-1.912.2
1.0G + 1.50W (Strength)-3.83.4
1.00G + 1.00Q (Service)08.5
1.00G + 1.00Q + ψS1.00S + ψW1.00W (Service)-1.38.5
1.0G + 1.00W (Service)-2.53.4

Member Envelope (Strength)

Position (m)Shear force (kN)Moment (kNm)
01.9 (max abs)00 
1.40 1.3 (max)0
2.80-1.90 

Partial factor for material properties – Table 2.3; 
γM = 1.300

Member details
Load duration – cl.2.3.1.2; Short-term
Service class – cl.2.3.1.3;  2

Timber section details
Number of timber sections in member; N = 1
Breadth of sections;   b = 47 mm
Depth of sections; h = 125 mm
Timber strength class – EN 338:2016 Table 1; C24

timber section properties

Bearing length;   Lb = 100 mm

Modification factors
Duration of load and moisture content – Table 3.1; kmod = 0.9
Deformation factor – Table 3.2;  kdef = 0.8
Depth factor for bending – Major axis – exp.3.1; kh,m,y = min((150 mm / h)0.2, 1.3) = 1.037
Bending stress re-distribution factor – cl.6.1.6(2);  km = 0.7
Crack factor for shear resistance – cl.6.1.7(2); kcr = 0.67
Load configuration factor – cl.6.1.5(4); kc,90 = 1
System strength factor – cl.6.6;  ksys = 1.1

Check compression parallel to the grain – cl.6.1.4

Design axial compression; Pd = 3.428 kN
Design compressive stress; σc,0,d = Pd / A = 0.583 N/mm2
Design compressive strength; fc,0,d = kmod × ksys × fc.0.k / γM = 15.992 N/mm2
σc,0,d / fc,0,d = 0.036

PASS – Design parallel compression strength exceeds design parallel compression stress

Compression perpendicular to the grain – cl.6.1.5

Design perpendicular compression – major axis; Fc,y,90,d = 1.89 kN
Effective contact length; Lb,ef = Lb = 100 mm

Design perpendicular compressive stress – exp.6.4; 
σc,y,90,d = Fc,y,90,d / (b × Lb,ef) = 0.402 N/mm2

Design perpendicular compressive strength;             
fc,y,90,d = kmod × ksys × fc.90.k / γM = 1.904 N/mm2
σc,y,90,d / (kc,90 × fc,y,90,d) = 0.211

PASS – Design perpendicular compression strength exceeds design perpendicular compression stress

Check shear force – Section 6.1.7

Design shear force; Fy,d = 1.89 kN
Design shear stress – exp.6.60; τy,d = 1.5Fy,d / (kcr bh) = 0.720 N/mm2

Design shear strength; 
fv,y,d = kmod × ksys × fv.k / γM = 3.046 N/mm2
τy,d / fv,y,d = 0.236

PASS – Design shear strength exceeds design shear stress

Compression or combined compression and bending – cl.6.3.2

Effective length for y-axis bending;
Le,y = 0.9 × 2800 mm  = 2520 mm

Slenderness ratio;                                                           
λy = Le,y / iy = 69.836

Relative slenderness ratio – exp. 6.21;
λrel,y = λy / π × √(fc.0.k / E0.05) = 1.184

Effective length for z-axis bending; Le,z = 0 mm

Slenderness ratio;                                                           
λz = Le,z / iz = 0

Relative slenderness ratio – exp. 6.22;                        
λrel,z = λz / π × √(fc.0.k / E0.05) = 0

λrel,y > 0.3 column stability check is required

Straightness factor; βc = 0.2

Instability factors – exp.6.25, 6.26, 6.27 & 6.28;         
ky = 0.5 × (1 + βcrel,y – 0.3) + λrel,y2) = 1.290
kz = 0.5 × (1 + βcrel,z – 0.3) + λrel,z2) = 0.470
kc,y = 1 / (ky + √(ky2 – lrel,y2)) = 0.556
kc,z = 1 / (kz + √(kz2 – lrel,z2)) = 1.064

Column stability checks – exp.6.23 & 6.24;                 
σc,0,d / (kc,y fc,0,d) = 0.066
σc,0,d / (kc,zfc,0,d) = 0.034

PASS – Column stability is acceptable

Check design 1400 mm along span (Mid-span)

Check bending moment – Section 6.1.6

Design bending moment; My,d = 1.323 kNm
Design bending stress;  σm,y,d = My,d / Wy = 10.809 N/mm2

Design bending strength;
fm,y,d = kh,m,y × kmod × ksys × fm.k / γM = 18.956 N/mm2
σm,y,d / fm,y,d = 0.57

PASS – Design bending strength exceeds design bending stress

Combined bending and axial compression – Section 6.2.4

Combined loading checks – exp.6.19 & 6.20;             
c,0,d / fc,0,d)2 + σm,y,d / fm,y,d = 0.572
c,0,d / fc,0,d)2 + km × σm,y,d / fm,y,d = 0.400

PASS – Combined bending and axial compression utilisation is acceptable

Compression or combined compression and bending – cl.6.3.2

Effective length for y-axis bending;
Le,y = 0.9 × 2800 mm  = 2520 mm

Slenderness ratio;
λy = Le,y / iy = 69.836

Relative slenderness ratio – exp. 6.21;                        
λrel,y = λy / π × √(fc.0.k / E0.05) = 1.184

Effective length for z-axis bending;
Le,z = 0 mm

Slenderness ratio;
lz = Le,z / iz = 0

Relative slenderness ratio – exp. 6.22;                        
λrel,z = λz / π × √(fc.0.k / E0.05)= 0

lrel,y > 0.3 column stability check is required

Straightness factor; βc = 0.2

Instability factors – exp.6.25, 6.26, 6.27 & 6.28;         
ky = 0.5(1 + βcrel,y – 0.3) + λrel,y2) = 1.290
kz = 0.5(1 + βcrel,z – 0.3) + λrel,z2) = 0.470
kc,y = 1 / (ky + √(ky2 – λrel,y2)) = 0.556
kc,z = 1 / (kz + √(kz2 – λrel,z2)) = 1.064

Column stability checks – exp.6.23 & 6.24;                 
σc,0,d / (kc,yfc,0,d) + σm,y,d / fm,y,d = 0.636
σc,0,d / (kc,zfc,0,d) + σm × sm,y,d / fm,y,d = 0.433
PASS – Column stability is acceptable

Check beams subjected to either bending or combined bending and compression – cl.6.3.3

Lateral buckling factor – exp.6.34; kcrit = 1.000

Beam stability check – exp.6.35;
m,y,d / (kcritfm,y,d))2 + σc,0,d / (kc,zfc,0,d)  = 0.359
PASS – Beam stability is acceptable

Conclusion

Timber shear wall design in accordance with the Eurocodes involves the structural analysis and design of timber-framed walls (stud walls) to resist gravity (such as floor and roof loads) and lateral loads (such as wind or seismic forces). Timber shear walls are typically made up of a frame of vertical timber studs, which are sheathed with plywood or other structural sheet material. The plywood sheathing helps to transfer the lateral forces to the studs, which then resist the forces by bending and shearing.

The structural design of timber shear walls involves the determination of the appropriate timber stud section, grade, spacing, and sheating material that will effecively and economically resist the anticipated design loading. The design must therefore take into account a number of factors, including the wind and seismic loads that the wall will be subjected to, the height and length of the wall, the type and thickness of the plywood sheathing, the size and spacing of the timber studs, and the type of connections between the studs and the plywood sheathing.

In addition to the design standards, the structural design of timber shear walls must also consider the following factors: the moisture content of the timber, the fire resistance requirements of the wall, the insulation requirements of the wall, and the cost of the wall. By carefully considering all of these factors, the structural engineer can design timber shear walls that are strong, stable, and cost-effective.

Analysis of Coupled Shear Walls


When two shear walls are connected intermittently with rigid beams or links, a coupled shear wall structure is formed. In coupled shear walls, the structural response of the two shear walls is linked to each other, influenced by the stiffness of the connecting beams. Coupled shear walls have found a variety of applications in the design of tall buildings, and other earthquake-resistant structural systems.

The effect of wind becomes more influential as a building gets taller. To effectively resist the force from wind, shear walls, frames, or a combination of both structural forms are normally employed in tall buildings. When shear walls have openings (pierced) which may serve for aesthetic, lighting or ventilation purposes, the behaviour of the shear wall is coupled to some degree.

The coupling beams, which connect the shear walls, play a crucial role in the behaviour of coupled shear walls. These beams are specifically designed to exhibit ductile inelastic behaviour, meaning they can undergo significant deformation without failure. This ductility allows them to dissipate energy during seismic events, effectively absorbing and distributing the lateral forces throughout the structure.

Coupled shear walls
Figure 1: Difference between shear wall and coupled shear wall

Furthermore, the dissipation of energy in coupled shear walls relies on the yield moment capacity and plastic rotation capacity of the coupling beams. When the yield moment capacity is excessively high, the beams experience limited rotations and dissipate minimal energy. Conversely, if the yield moment capacity is too low, the beams may undergo rotations exceeding their plastic rotation capacities.

Consequently, it is very important to design the coupling beams to have an optimal level of yield moment capacities, which are contingent upon the available plastic rotation capacity in the beams.

Advantages of Coupled Shear Walls over Planar Shear Walls

The arrangement of coupled shear walls provides several advantages over regular shear walls or other lateral load-resisting systems:

  1. Enhanced stiffness: The coupling of multiple shear walls significantly increases the overall stiffness of the structure, reducing lateral deflections and sway during wind or seismic events.
  2. Improved load distribution: The coupling beams ensure a more even distribution of lateral loads across the shear walls, preventing localized stress concentrations and improving the overall load-carrying capacity of the system.
  3. Architectural flexibility: Coupled shear walls offer architectural flexibility, allowing for larger open spaces and creative designs, as the lateral stability is not solely dependent on the positioning of walls.
  4. Seismic resilience: The ductile behaviour of the coupling beams helps dissipate seismic energy, making coupled shear walls particularly effective in earthquake-prone regions, where they can significantly enhance a building’s seismic resilience.
  5. Cost-effectiveness: Compared to other lateral load-resisting systems like moment frames or braced frames, coupled shear walls can provide a cost-effective solution while still offering excellent performance

This aericle explores the wind load analysis of coupled shear wall structures with a fully solved example, under the following assumptions;

(i) The entire wind load is transferred to the shear walls (the contribution of the frames have been neglected)
(ii) The wind load acts as a uniformly distributed force of F/H

The two main parameters that define​ the behaviour of pierced shear walls are alpha(α) and beta (β), which depend on the geometrical properties of the coupling/connecting media (beams or floor slabs) and the shear wall. For fairly symmetrical arrangements as shown below, the following relationships given below can be used for the purpose of analysis.

N/B: This is a curtailed information; kindly consult specialist textbooks/publications for full knowledge of the subject.

shear%2Bwall%2Bterm%2Bdefinitions
formular
Formulars%2B2

The value of Kv can be obtained from the table below;

FACTOR%2BTABLE

Solved Example

The structural arrangement of a tall building is shown below. The building is 90m tall, with 30 storeys @3m storey height. We are to determine the internal forces induced in the shear walls due to the wind load.

TALL%2BBUILDING%2BFLOOR%2BPLAN
coupled shear wall structure

Building Data:
Height of building (H) = 90m
Storey height (h) = 3m
Thickness of shear wall (t) = 0.35m
Depth of connecting beam (d) = 900mm
Width of opening (b) = 2500mm
Centreline of connected walls (L) = 8000mm
Basic wind speed = 40 m/s

Wind Load Analysis (Eurocode 1 Part 4)

(a) Basic wind velocity

The fundamental value of the basic wind velocity Vb,0 is the characteristic 10-minute mean wind velocity irrespective of wind direction and time of the year, at 10 m above ground level in open-country terrain with low vegetation such as grass, and with isolated obstacles with separations of at least 20 obstacle heights.

The basic wind velocity Vb,0 is calculated from;

Vb = Cdir .Cseason .Vb,0

Where:
Vb is the basic wind velocity defined as a function of wind direction and time of the year at 10m above the ground of terrain category II
Vb,0 is the fundamental value of the basic wind velocity
Cdir is the directional factor (defined in the National Annex, but recommended value is 1.0)
Cseason is the season factor (defined in the National Annex, but recommended value is 1.0)

For the area and location of the building that we are considering;

Basic wind velocity Vb,0 = 40 m/s
Vb = Cdir .Cseason . Vb,0 = 1.0 × 1.0 × 40 = 40 m/s





(b) Mean Wind
The mean wind velocity Vm(z) at a height z above the terrain depends on the terrain roughness and orography, and on the basic wind velocity, Vb, and should be determined using the expression below;

Vm(z) = cr(z). co(z).Vb

Where;
cr(z) is the roughness factor (defined below)
co(z) is the orography factor often taken as 1.0

The terrain roughness factor accounts for the variability of the mean wind velocity at the site of the structure due to the height above the ground level and the ground roughness of the terrain upwind of the structure in the wind direction considered. Terrain categories and parameters are shown in the table below;

Wind%2BTerrain%2BFactor%2BTable

cr(z) = kr. In (z/z0) for zmin ≤ z ≤ zmax
cr(z) = cr.(zmin) for z ≤ zmin

Where:
Z0 is the roughness length
Kr is the terrain factor depending on the roughness length Z0 calculated using
Kr = 0.19 (Z0/Z0,II )0.07
Where:
Z0,II = 0.05m (terrain category II)
Zmin is the minimum height
Zmax is to be taken as 200 m
Kr = 0.19 (0.05/0.05)0.07 = 0.19
cr(z) = kr. In (z/z0) = 0.19 × In(90/0.05) = 1.424

Therefore;
Vm(90) = cr(z).co(z).Vb = 1.424 × 1.0 × 40 = 56.96 m/s

(c) Wind turbulence
The turbulence intensity Iv(z) at height z is defined as the standard deviation of the turbulence divided by the mean wind velocity. The recommended rules for the determination of Iv(z) are given in the expressions below;

Iv(z) = σv/Vm = k1/(c0(z).In(z/z0)) for zmin ≤ z ≤ zmax

Iv(z) = Iv.(zmin) for z ≤ zmin

Where:
kI is the turbulence factor of which the value is provided in the National Annex but the recommended value is 1.0
Co is the orography factor described above
Z0 is the roughness length described above.
For the building that we are considering, the wind turbulence factor at 90m above the ground level;

Iv(90) = σv/Vm = k1 / (c0 (z).In (z/z0)) = 1/(1 × In(90/0.05) ) = 0.1334

(d) Peak Velocity Pressure
The peak velocity pressure qp(z) at height z is given by the expression below;

qp(z) = [1 + 7Iv(z)] 0.5.ρ.Vm2(z)= ce(z).qb

Where:
ρ is the air density, which depends on the altitude, temperature, and barometric pressure to be expected in the region during wind storms (recommended value is 1.25kg/m3)

ce(z) is the exposure factor given by;
ce(z)= (qp(z))/qb
qb is the basic velocity pressure given by;
qb = 1/2.ρ.Vb2

qp(90m) = [1 + 7(0.1334)] × 1/2 × 1.25 × 56.962 = 3921.313 N/m2
Therefore, qp(90m) = 3.921 kN/m2

(e) Wind Pressure Coefficient

h/d = 90/22.5 = 4
Pressure coefficient for windward side (zone D) = +0.8
Pressure coefficient for leeward side (zone E) = -0.65 (from linear interpolation)

Net pressure coefficient = 0.8 – (-0.65) = 1.45

According to clause 7.2.2(2), the force generated from zone D and E simultaneously should be corrected for lack of correlation.

Since h/d = 4 in our calculation, we have to interpolate between h/d ≥ 5(1.0) and h/d ≤ 1.0 (0.85).

On interpolating;
Correction factor = 0.9625
Therefore, net wind pressure = 0.9625 × 1.45 × 3.921 = 5.472 kN/m2

IV Structural Analysis

The wind force of the structure;
F = 5.472 kN/m2 × 24m × 90m = 11819.52 kN (taking structural factor as 1.0 because building is less than 100m tall, and h/d = 4)

If we assume that this load is shared equally among the two shear walls, then the load on each shear wall = 5909.76 kN

Geometrical Properties of Elements
Area of shear walls = A1 = A2 = 0.35 × 5.5 = 1.925 m2
Second moment of area of walls = I1 = I2 = (0.35 × 5.53)/12 = 4.852 m4
Moment of inertia of connecting beam Ib = (0.35 × 0.93)/12 = 0.02126 m4

Ie

be = (2b + 5d)/3 = [(2 × 2.5) + (5 × 0.9)] / 3 = 3.167

beta%2Bsquare
alpha%2Bsquare

αH = 19.962
Therefore, the axial force (N) at the bottom of wall;

N

The bending moment at the bottom of the wall elements;

M

The shear force at the connecting beams;

vb

The bending moment on the connecting beams;

mb

These internal forces should be factored, and combined with other appropriate load cases for the purpose of design.

Thank you for reading.
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Design and Construction of Steel Staircase | Industrial Staircase

Steel staircase structures are usually considered secondary steelworks. Primary structures comprise the critical elements necessary for the strength and stability of the overall structural frame. This usually comprises the beams, columns, walls, slabs, etc. This frame serves as the support for all other building components. Any steelwork supported by the main structure, without needing to enhance its strength or stiffness, is called secondary steelwork.

The essential elements of a staircase include treads, risers, stringers, landings, and their supports. These components can be configured in different ways to create stairs with varying levels of functionality, from simple utility to prominent architectural features. BS 5395, the code of practice governing the design, construction, and maintenance of straight stairs and winders, offers the guidelines described in the sections that follow.

image 8
Figure 1: Parts of a steel staircase

Geometry of Steel staircase

The geometry of a staircase structure is usually determined by building regulation requirements. The relationship between rise, tread, and pitch must be such that the stair is safe and comfortable to use. It is essential to make all rises in a flight uniform, subject to the tolerances given in Clause 5.5 of BS 5359. The relationship between the rise and going for a stair should not change along the walking line, subject to the same tolerances.

The minimum clear width should be 600 mm for occasional one-way traffic, 800 mm for regular one-way traffic and occasional two-way traffic and 1000 mm for regular two-way traffic. Stairs that are often used by large numbers of people at the same time (assembly stairs in public buildings) should be designed with a large going and a small rise to achieve a maximum pitch of 33 degrees. Stairs that are used as means of escape may require a clear width greater than 1000mm.

All stairs are required to have a minimum of three and a maximum of 16 rises per flight and the clear width of all landings should never be less than the stair clear width. The length of a landing should be not less than the clear width of the stair or 850 mm, whichever is greater.

Industrial Steel Staircase
Figure 2: Industrial Steel Staircase

For steel staircases in industrial structures, there should be a change of line or direction of not less than 30° after 32 risers, for straight stairs, or 44 risers, for helical or spiral stairs. Landings at the head of a stair should be designed so that it is not possible to step from a platform or walkway onto the stair without a change in direction.

Typical dimensions for public stairs are:

  • Rise: 100 – 190 mm
  • Going: 250 – 350 mm
  • Pitch: maximum 38 degrees
  • Clear width: minimum 1000 mm.

The minimum pitch for straight stairs should be 30°
The maximum pitch for occasional access should be 42°
The maximum pitch for regular two-way traffic should be 38°

Treads

Treads should comply with the requirements for strength given in BS 4592 and should be slip-resistant or at least have a slip-resistant nosing not less than 25 mm wide. Treads on open riser stairs should overlap not less than 16 mm and have a nosing depth in the range of 25 mm to 50 mm to aid visibility.

Strings

Strings should be sufficiently robust to minimize lateral flexing of the structure and should not project more than 50 mm beyond the nosing of the bottom tread.

Loads and robustness

Industrial staircases need to be designed with consideration for accidental loads, particularly when they serve as emergency escape routes. It is important to ensure that these stairs can withstand potential damage caused by accidental loads without collapsing.

To achieve this, the connections between the stairs and the primary structure must be strong and well-designed, providing enough bearing area and tie resistance. When using individual treads, their design should account for the dynamic impact of repeated foot loading, and it is advisable to opt for a cautious and conservative approach to the design.

The dynamic response can be critical as steel stairs tend to have little inherent damping.

Other criteria – safety, slip resistance, durability, acoustic requirements and lighting requirements all influence stair design and are addressed in BS 5395.

external escape steel staircase
Figure 3: External escape steel structure

Dynamics of Steel Staircase Structures

Footfall excitation is the primary dynamic load that significantly affects the behaviour of a staircase. Staircases can experience different types of footfall excitation, including the impact of a single person walking, a group of individuals going up or down the stairs together, and the impulse load generated when someone jumps from a height to a lower step.

The limitations in the design geometry of steel staircases lead to various approaches for different individuals when ascending or descending. Some people may choose to increase their step frequency to maintain their speed, while others may stick to the typical 25 step frequency.

Additionally, some individuals may take two or more steps at once, resulting in a higher impact load but reducing the step frequency. Due to these variations, it becomes challenging to provide a single or specific set of instructions for ascending or descending a staircase.

Slender staircases are characterized by their lightweight, leading to a low stiffness-to-mass ratio. As a result, these staircases have low natural frequencies, making the vibration serviceability criteria the primary design consideration. It is important to simulate the dynamic behaviour of a staircase using finite element analysis before construction to ensure that it meets the necessary dynamic and comfort requirements for users.

Design of Steel Staircase

Structural steel is the material commonly used in the design of industrial steel staircases. Grade S275 or S355 is commonly used. After the selection of the proper material, the next step is to design the individual components of the staircase. This includes the treads, risers, stringers, and landings.

The stringers are the beams that support the treads, risers, and landings, and they must be designed to carry the loads of the staircase. They are usually designed using I-beam sections, even though channel and rectangular hollow sections can also be used. They are usually subjected to bending moment, shear, axial, and torsional stresses.

The load from the tread can be transferred to the stringers as a series of concentrated loads or uniformly distributed loads depending on the manner of connection adopted. The stringer can be bolted or welded to a base plate or the primary structure.

The treads and risers are the parts of the staircase that people will step on, and they must be designed to provide a safe and comfortable walking surface. The treads are commonly designed using chequered structural steel plates to provide slip-resistant surfaces. Angle or channel sections can be used to enhance the rigidity of the plates.

The landings are the platforms that people will use to rest and change direction, and they must be designed to be level and stable.

Forms of Construction

Stair flights, consisting of treads and risers, are supported by stringers to create a staircase. Typically, two flights are required per storey height, arranged at 180 degrees to each other, and should occupy a footprint not exceeding 6m x 3m (although larger stairs are allowed in assembly buildings). Each end of the stair flight connects to a landing.

image 9
Figure 4: Forms of composite steel staircase construction

The most straightforward stair construction involves placing the staircase internally, within an opening in the primary floor structure. In this case, both the floor level and half-level landings can be directly supported by the primary structure, allowing the stair flight to span between landings. The floor level landing can be designed as part of the staircase or as part of the floor structure. Stair treads can be positioned above or in the plane of the stringers.

When the treads are located in the plane of the stringers, the stringer depth resulting from minimum planning dimensions will be structurally sufficient. Additionally, if folded steel plates are used for both the treads and risers, the staircase will inherently have enough rigidity to respond adequately to dynamic forces. This construction method proves to be highly efficient.

However, if a staircase is placed at the edge of a floor slab, the support of landings, especially the half-landings, becomes critical.

Conclusion

Steel staircases are important structural elements used for providing vertical circulation in a building. Steel offers several advantages, such as high strength, durability, and versatility. The design process involves considering various factors, including the intended use, building regulations, and safety standards.

Key considerations in the design include the geometry and arrangement of treads, risers, and stringers. The configuration of the staircase should provide a safe and efficient means of vertical circulation. Landings are essential for connecting stair flights and offering resting points for users. Structural engineers must account for dynamic loads caused by footfalls and other factors that can affect the staircase’s performance. Ensuring sufficient structural integrity to withstand accidental loading and to meet vibration serviceability criteria is crucial.

Proper connections to the primary structure are vital for ensuring stability and load-bearing capacity. The use of folded steel plates for treads and risers can enhance the staircase’s rigidity and dynamic response.

Summarily, the structural design of steel staircases requires careful planning, analysis, and adherence to building codes and safety regulations. By considering the specific requirements of each project, designers can create safe, efficient, and aesthetically pleasing staircases that meet the needs of users while enhancing the overall architectural design of the building.

Pylons in Cable-Stayed Bridges

In cable-stayed bridges, the main purpose of the pylon is to transfer the forces resulting from anchoring the cable stays to the foundation. As a result, these forces will significantly influence the design of the pylon. The tensile cable forces in cable-stayed bridges are part of a closed force system that balances these forces with the compression that occurred within the deck and the pylon. Ideally, the pylon should resist these forces through axial compression whenever feasible to minimize any uneven loading.

The pylon serves as the central element that defines the visual appearance of a cable-stayed bridge, offering a chance to impart a unique style to the overall design. Additionally, the pylon’s design must be adaptable to different stay cable layouts, accommodate the bridge site’s topography and geology, and support the forces efficiently and cost-effectively. The stability of cable-stayed bridges is dependent on the stability and stiffness of the pylons.

Pylons for cable-stayed bridges are predominantly constructed using structural steel, reinforced concrete, or composite sections.

Steel Pylons

Early designs of cable-stayed bridge pylons mostly involved steel boxes, like the Stromsmund Bridge (opened in 1956). The pylons resembled steel portal frames meant to offer transverse restraint to the stay system. However, it was subsequently observed that this restraint was unnecessary as the stay system itself could provide sufficient transverse restraint.

Stromsund bridge
Figure 1: Stromsmund Bridge

When a single mast supports each stay plane, any lateral displacement at the top of the mast results in a rotation of the stay plane. This rotation ensures that the resultant reaction from the main span and back span stay cables passes through the foot of the pylon. The weight of the pylon remains vertical, but the reaction from the stays dominates. Thus, the effective length of the mast in buckling is not twice the height (2H) of a simple cantilever, but equal to the height (H).

In the longitudinal direction, the main and back stay cables restrain the pylon against buckling as long as the deck, to which the stays are anchored, is properly restrained against longitudinal movement. If the deck is unrestrained, the pylon will behave as a cantilever with maximum bending at the base, resulting in an effective length of twice the height (2H) in buckling.

image 1
Figure 2: Longitudinal restraint of the pylon by the anchor stays (Farquhar, 2008)

Effective pylon restraint can be achieved when the deck is adequately connected to an abutment, another pylon, or an independent gravity anchorage. Earlier designs used a pin at the pylon foot to prevent large bending moments on the mast. Modern designs prefer a fixed-end cantilever mast, which is simpler and more stable during erection. The use of a frictionless bearing with a fixed-end mast is possible when the member is slender enough, causing the axial load to approach the buckling capacity of the mast in a free cantilever condition.

For single mast pylons supporting a single plane of stays, two methods have been used to connect the mast at its base:

(1) encastre construction into a transverse girder forming part of the deck, requiring bearings at the pylon foundation, and
(2) passing the mast directly through the deck to sit upon the pylon foundation, needing bearings only at each end of the transverse girder.

The second method (Method 2) is more efficient and has been widely adopted in recent designs.

Concrete Pylons

Concrete has become increasingly competitive for pylon construction due to advancements in concrete construction and formwork technology, despite its higher self-weight compared to steel. Concrete has proven to be particularly adaptable to complex pylon forms.

Various types of pylons have been developed to support both vertical and inclined stay layouts, including H-frame, A-frame, and inverted Y-frame pylons (shown in Figure 3).

image 2
Figure 3: H-frame, A-frame and inverted Y-frame pylons (Farquhar, 2008)

In the case of H-frame pylons, the stay anchors are usually located above a crossbeam. For modified fan arrangements of stays, this crossbeam would be positioned between mid-height and two-thirds of the pylon height above the deck. However, when adopting the harp arrangement of stays, the anchors are distributed over the full height of the pylon above the deck. In such cases, a crossbeam can only be practically provided below the deck level, as seen in the Øresund Bridge between Denmark and Sweden.

Oresund Bridge
Figure 4: Øresund Bridge

The deck section located at the pylon is typically subjected to the highest stresses, combining maximum negative moment and axial load. Connecting the stay directly between the pylon leg and an edge stiffening girder within the deck requires the pylon legs to be inset into the deck.

This creates a practical detailing problem and results in a zone of concentrated stress in an already highly stressed section. Several geometrical configurations can overcome this problem: widening the pylon and connecting the stays to the deck using out-stand brackets, or sloping the pylon leg outwards at its base.

The pylon leg can be inclined over its entire height, in which case the pylon must be designed to accommodate a small eccentricity arising from the stay cable reactions. Another approach is to maintain the upper section of the leg in a vertical plane and incline the pylon only from below the level of the bottom anchorage.

By locating the crossbeam at this change of direction, the stay force reaction can be efficiently transmitted as a direct thrust. Examples of this pylon geometry can be observed in the Annacis Bridge over the Fraser River, Canada (see Figure 5), and the Vasco da Gama Bridge (Capra and Leveille, 1998) over the Tagus River, Portugal.

alex fraser bridge
Figure 5: Annacis Bridge

Pylon Geometry

The A-frame pylon is well-suited for inclined stay arrangements and was first used in the Severins Bridge. Another variation is the inverted Y-frame, where the vertical leg, containing the stay anchors, extends above the bifurcation point. Examples of the inverted Y-frame can be seen in the Normandy Bridge over the River Seine, France, and the Rama VIII Bridge in Bangkok, Thailand.

However, the wide footprint of the inverted Y-frame can lead to excessive land usage when a high navigation clearance to the deck is necessary. To address this, some designs break the pylon legs at or just below the deck, creating inward-leaning legs to the foundation, forming a diamond configuration (Figure 6).

image 3
Figure 6: Diamond shape pylon geometry (Farquhar, 2008)

This modification reduces land usage but makes the pylon less stiff against transverse wind or seismic forces, resulting in increased deflection. To counteract this deflection, a substantial increase in stiffness is required in the lower section of the pylon leg below the deck.

Nevertheless, this diamond configuration was favoured for the pylons of the Tatara Bridge in Japan, currently the world’s longest cable-stayed span at 890m, and the Industrial Ring Road Bridges in Bangkok. The same configuration was also used for the twin cable-stayed crossing of the Houston Ship Channel, where the twin diamonds were connected and tied together at the deck level to form a strong truss, transmitting transverse wind loads to the foundations.

image 4
Figure 7: Tatara Bridge

Another architectural feature includes inclining the pylon in the longitudinal direction, resulting in visually exciting structures. However, this design introduces inclined thrust from the pylon that must be carried by the foundation, generating a significant horizontal component. In rock foundations, these horizontal reactions are easily resisted with only minor foundation displacement. In contrast, in typical estuarine soil conditions, foundation costs may represent a significant portion of the overall project cost.

Cable Stay Connection to the Pylons

In early designs, the connection between the stays and the pylon resembled that of suspension bridges, where the cables were laid in a deviator saddle and carried through the pylon. However, this method had limitations. As an alternative, modified fan and harp arrangements were introduced, with stays anchored over the upper section of the pylon leg. This led to the use of separate stays for the main span and back span.

The most straightforward anchoring method involves attaching the stay socket or anchorage plate directly to the pylon wall. In concrete pylons, the horizontal component of the cable forces can cause the shaft to split vertically, requiring transverse pre-stressing to resist these forces. Different layouts were developed to address this, as seen in Figures 8 and 9.

image 5
Figure 8: Pre-stressing layout for stay connection to concrete pylon (Farquhar, 2008)
image 7
Figure 9: Alternative layout for stay connection to concrete pylon (Farquhar, 2008)

To simplify pylon construction and ensure accurate placement of steel formers, anchor pre-stress, and reinforcement within the concrete walls, steel fabricated anchorage modules have been utilized. These modules define the required stay anchorage geometry and are incorporated into the concrete shaft during construction.

Adequate shear connection, typically in the form of shear studs, allows the anchorage forces in the fabrication to be transferred to the concrete shaft. Examples of this construction method, with the fabricated anchorage module centrally located within the concrete shaft, can be seen in the Normandy Bridge and the Stonecutters Bridge. The Ting Kau Bridge in Hong Kong also used a similar concept but connected the fabricated anchor modules on the outside of the concrete core.

It is very important to accurately model any eccentricity of the stay anchor within the pylon during structural analysis. The inclination of the back span stays and main span stays are usually different, requiring the anchors to be located at different levels to maintain the same intersection line on the pylon centerline.

Alternatively, keeping the levels of the two anchors the same may slightly eccentrically place the vertical resultant of the stay forces to the pylon. This approach simplifies the anchor zone detailing but requires careful consideration of the pylon moments arising from this small eccentricity (Figure 10).

image 6
Figure 10: Stay anchor pylon geometry (Farquhar, 2008)

Conclusion

In conclusion, pylons play a crucial role in the structural integrity and aesthetic appeal of cable-stayed bridges. The use of pylons in these bridges has evolved over time, with advancements in concrete construction and formwork technology making concrete pylons increasingly competitive, despite their higher self-weight compared to steel alternatives.

Various types of pylons have been developed to accommodate different stay arrangements, including H-frame, A-frame, and inverted Y-frame pylons. The modified fan and harp arrangements have allowed for separate stays for the main span and back span, enabling more adaptable and efficient designs.

Pylon geometry has been a subject of innovation and consideration in recent designs. Solutions such as breaking the pylon legs to create inward-leaning legs or using steel-fabricated anchorage modules have been employed to reduce land usage and simplify construction processes. However, these modifications must be carefully designed to maintain sufficient stiffness and resist transverse wind or seismic forces.

The design of pylons in cable-stayed bridges must take into account various factors, such as the inclination of the back span and main span stays, the connection between the stays and the pylon, and the accurate modelling of stay anchor eccentricities. These considerations are critical to ensuring the stability and performance of the cable-stayed bridge.

In summary, pylons in cable-stayed bridges continue to undergo refinement and innovation, striving to achieve optimal efficiency, safety, and aesthetics. The choice of pylon type, arrangement, geometry, and design are essential elements in the successful construction of cable-stayed bridges, contributing to their functionality, longevity, and architectural distinctiveness in modern infrastructure projects.

References:
Farquhar, D. J. (2008). Cable-stayed bridges. ICE Manual of Bridge Engineering. Published by the Institution of Civil Engineers (ICE) UK

Implementing AI and Machine Learning in Bridge Inspections

All over the world, there is growing concern that environmental degradation, overloading, initial construction defects/imperfections, natural and man-made hazards, and other factors have contributed to structural deterioration and deficiencies in highway bridges. The deterioration of bridges usually occurs in response to external loads and environmental disturbance, and various maintenance plans are usually adopted in various states, countries, and jurisdictions to prolong the service life of bridges.

This article will comprehensively explore the implementation of Artificial Intelligence (AI) and Machine Learning (ML) in bridge inspections. As technological advancements continue to reshape various industries, these innovative technologies have revolutionized how we assess and maintain bridges. By leveraging the power of AI and ML, drone bridge inspection has become more efficient, cost-effective, and safer for both inspectors and the public.

According to Xia et al. (2021), the primary components of highway bridge inspection include;

  • geometric parameter inspection,
  • mechanical performance assessment,
  • interior inspection, and
  • appearance inspection.

The Transformation of Bridge Inspections

AI and Machine Learning technologies have brought about a profound transformation in bridge inspections. The traditional methods, often labour-intensive and time-consuming, have given way to automation, enabling faster and more accurate assessments of bridge conditions. This transformation has paved the way for a new era of infrastructure management, where data-driven decision-making and AI-driven analytics play a pivotal role.

It is important to note that other recently developed technologies have substantially improved the precision and effectiveness of bridge inspection work. The location of damages in bridges in three-dimensional space has been determined using a variety of technologies, including point cloud techniques, unmanned aerial vehicles (UAV), and terrestrial laser scanning techniques. Ground-penetrating radar has also been used to locate the spatial and temporal variations of concrete bridges. Infrared thermography techniques have also been utilized to inspect thermal abnormalities using thermal cameras on UAVs.

The structural state of bridges can also be evaluated using satellite-based remote sensing techniques. Persistent Scatterer Interferometry (PSI), a satellite remote sensing technique, has been used to assess the displacement of bridges. The long-term displacements of the Hong Kong-Zhuhai-Macao Bridge (HZMB) have been studied utilizing PSI and InSAR technology.

Automated Image and Data Collection

One of the key advantages of implementing AI and Machine Learning in bridge inspections is the ability to automate image and data collection. Inspectors can efficiently gather detailed images and data points without the need for risky and time-consuming under-bridge walks or expensive snooper trucks by employing drones equipped with high-definition cameras and AI algorithms. The automated process accelerates the inspection timeline and enhances the reliability and accuracy of the data collected.

Typical model for bridge inspection with an unmanned airborne vehicle
Typical model for bridge inspection with an unmanned airborne vehicle (Karim et al., 2020)

AI-Driven Defect Detection

AI and ML techniques have found extensive application in diverse areas concerning structural safety, including predicting conditions and detecting damages. For instance, the neural network (NN) is well-suited for addressing large-scale data challenges as it can effectively extract multidimensional features and recognize non-linear relationships within the input data.

AI and ML algorithms analyze vast amounts of data swiftly and accurately. In bridge inspections, these algorithms can detect even the most subtle signs of wear, corrosion, cracks, and other defects that might not be readily noticeable to the human eye. Early identification of potential issues enables proactive maintenance decisions, reducing the risk of sudden and catastrophic failures. Moreover, AI-driven defect detection enhances the inspector’s ability to prioritize and focus on critical areas that require immediate attention.

Enhanced Structural Analysis

AI and Machine Learning have revolutionized structural analysis in bridge inspections. By leveraging the data collected during inspections, advanced analytics can provide a comprehensive assessment of a bridge’s overall health and performance. This includes evaluating stress distributions, load-bearing capacities and predicting the bridge’s response to different environmental conditions over time.

By gaining deeper insights into the structural integrity of bridges, engineers can make more informed and data-driven decisions about maintenance and repairs, which empowers them to ensure the safety of the bridges.

Predictive Maintenance and Lifecycle Management

One of the most significant benefits of AI and Machine Learning in bridge inspections is the implementation of predictive maintenance strategies. Through the analysis of historical data and predictive analytics, authorities can gain a thorough understanding of the deterioration patterns of bridges.

This valuable information allows for developing comprehensive maintenance plans, optimizing resources, and extending the lifespan of these critical infrastructure assets. Predictive maintenance shifts the focus from reactive repairs to proactive and strategic management, resulting in long-term cost savings and enhanced bridge performance.

Overcoming Challenges and Limitations

While AI and Machine Learning offer significant advantages in bridge inspections, there are challenges to address to ensure their successful implementation. Accurate data collection is paramount, as the data quality directly impacts the effectiveness of AI algorithms.

Additionally, training AI models requires a vast and diverse dataset to detect various bridge defects accurately. Moreover, integrating AI and Machine Learning technologies with existing inspection protocols and standards demands meticulous planning and consideration to ensure seamless adoption.

drones for bridge inspection

Building Trust in AI-Driven Inspections

The successful integration of AI and Machine Learning in bridge inspections relies on building trust among engineers, inspectors, and the public. Demonstrating the effectiveness and reliability of AI-driven technologies is essential to gain acceptance and confidence in automated systems.

Rigorous testing, validation, and transparency are crucial steps in proving the capabilities and accuracy of AI algorithms. Building trust will foster greater acceptance and encourage further adoption of these technologies in infrastructure management.

Advancements in AI Technology for Bridges

The future holds exciting possibilities for AI and Machine Learning in bridge inspections. Ongoing advancements in computer vision, sensor technologies, and AI-driven robotics are expected to drive further innovation. As these technologies evolve, we can anticipate even more sophisticated applications in bridge inspections, including fully autonomous systems that can operate with minimal human intervention. The potential for greater efficiency, accuracy, and safety in bridge inspections will continue to grow as AI technology progresses.

Embracing the Future of Bridge Inspections

The successful use of AI and Machine Learning in bridge inspections marks a pivotal step towards safer, more efficient, and cost-effective infrastructure management. By embracing these innovative technologies, engineers and inspectors can focus on critical analysis and decision-making, armed with comprehensive and actionable data. As AI and Machine Learning are developing and advancing, overcoming challenges, building trust, and driving further innovation will be instrumental in ensuring our bridges’ continued safety and longevity.

Conclusion

Integrating AI and Machine Learning in drone bridge inspections has ushered in a transformative era in assessing and maintaining critical infrastructure. Automation, AI-driven defect detection, enhanced structural analysis, and predictive maintenance have streamlined the inspection process, increased safety, and reduced costs. As technology advances, the potential for AI and Machine Learning in drone bridge inspections is boundless.

By addressing challenges, building trust, and leveraging data-driven decision-making, the future of drone inspections holds great promise. As these technologies continue to evolve, engineers and inspectors can embrace the opportunities presented by AI and Machine Learning to create resilient and reliable bridges that serve as vital links in connecting communities for generations to come.

References

Xia Y., Lei X., Wang P. and Sun L. (2021). Artificial Intelligence Based Structural Assessment for Regional Short- and Medium-Span Concrete Beam Bridges with Inspection Information. Remote Sens. 2021, 13, https://doi.org/10.3390/rs13183687

Karima M.M., Daglia C.H. and Qina R. (2020). Modeling and Simulation of a Robotic Bridge Inspection System. Procedia Computer Science 168 (2020) 177–185. DOI. 10.1016/j.procs.2020.02.276

Uplift Capacity of Piles

In the design of pile foundations, it is usually very common to express the load carrying capacity in terms of axial compression. However in some cases, piles are subjected to uplift forces, and must be designed to resist such forces in order to avoid the pull out of the foundation or the structure. Piles that are subjected to uplift forces are also called tension or anchor piles. Uplift forces are developed due to hydrostatic pressure or overturning moments.

Forces due to earthquake, wind, or waves, are the major actions that can induce uplift forces in piles. Structures such as transmission towers, mooring systems for ocean surface or submerged platforms, tall chimneys, jetty constructions, etc., all have their foundations subject to uplift loads.

Factors Affecting Uplift Capacity

Several factors influence the uplift capacity of piles and they are briefly described below.

  1. Soil Characteristics: Soil properties, including the cohesion and friction angle, significantly affect uplift capacity. Clay soils will rely on the cohesion or unit adhesion of the soil with the pile material for uplift resistance, while granular soils will rely on frictional forces (dependent on the angle of internal friction) for uplift resistance.
  2. Pile Geometry: The shape, size, and length of the pile influence its uplift capacity. Longer piles provide a larger surface area in contact with the soil, thus increasing resistance against uplift forces. Additionally, the pile diameter and shape can affect the distribution of uplift forces along the pile shaft. It is also important to note that pile cap dimensions and group effects can also influence the uplift capacity of piles.
  3. Pile Material: The material used for pile construction plays a role in uplift capacity. Steel piles are often preferred due to their high strength and ductility, providing better resistance against uplift. However, other factors, such as corrosion potential, must also be considered. Furthermore, scholars have also determined that the angle of wall friction between the pile and the surrounding soil is dependent on the pile material.
  4. Pile Installation Method: The uplift capacity is typically higher in driven piles than bored piles.
  5. Groundwater Conditions: The level of groundwater significantly affects uplift capacity. In saturated soil conditions, the presence of water can reduce the soil’s effective stress, potentially reducing the uplift resistance of piles. Conversely, dewatering or densification techniques can improve uplift capacity.

Methods for Determining Uplift Capacity

Uplift capacity of piles may be assessed using theoretical analyses (classical equations developed using soil properties), field tests, and numerical modeling. The following are commonly used approaches:

  1. Analytical Methods: Several analytical methods, such as the Terzaghi’s equation and the Vesic’s method, provide simplified solutions to estimate uplift capacity. These methods typically consider soil parameters, pile geometry, and groundwater conditions to calculate the uplift resistance.
  2. Field Load Tests: Load tests conducted in the field help validate the design assumptions and provide valuable data on the actual uplift capacity. Common load test methods include the pile load test, plate load test, and pile uplift test. These tests involve applying controlled loads to the pile and measuring its response to determine uplift capacity.
  3. Numerical Modeling: Finite element analysis (FEA) and other numerical modeling techniques allow engineers to simulate complex soil-pile interactions and evaluate uplift capacity. Numerical models consider soil behavior, pile-soil interaction, and other relevant factors to provide detailed insights into the uplift response.

Generally, the equation for uplift capacity of a single pile may be written as;

Pul = Wp + Asfr ——– (1)

where,
Pul = uplift capacity of pile,
Wp = weight of pile,
fr = unit resisting force
As = effective surface area of the embedded length of pile.

Uplift capacity of pile in clay

For piles embedded in clay, equation (1) may be written as;

Pul = Wp + Asαcu ——– (2)

where,
cu = average undrained shear strength of clay along the pile shaft,
α = adhesion factor (= ca/cu),
ca = average adhesion

image 1
Figure 1: Relationship between adhesion factor α and undrained shear strength cu

Figure 1 gives the relationship between α and cu based on pull out test results as collected by Sowa (1970). As per Sowa (1970), the values of ca agree reasonably well with the values for piles subjected to compression loadings.

Uplift capacity of pile in sand

The ultimate uplift capacity of a vertical pile for piles embedded in sandy soil depends on the skin resistance created between the pile shaft and the soil.

According to Murthy (2012), adequate confirmatory data are not available for evaluating the uplift resistance of piles embedded in cohesionless soils. Ireland (1957) reports that the average skin friction for piles under compression loading and uplift loading are equal, but data collected by Sowa (1970) indicate lower values for upward loading as compared to downward loading especially for cast-in-situ piles. A study by Ramasamy et al. (2004) showed that the pull out shaft friction is significantly less than the push in shaft friction.

Poulos and Davis (1980) suggest that the skin friction of upward loading may be taken as two-thirds of the calculated shaft resistance for downward loading.

According to Verma and Joshi (2010), the net uplift capacity of piles embedded in sand can be given by the equation (3) as:

Qu = ½ Ks σv tanδ πdZc + Ks σv tanσ π(L – Zc) ——– (3)

Where;
Ks = coefficient of earth pressure
σv = effective vertical stress at a depth of Zc = γdZc
d = diameter of pile
γd =dry unit weight of soil
δ = soil-pile friction angle
L = Length of pile
Zc or Lcr = critical depth of embedment

According to Verma and Doshi (2010), the value of coefficient of earth pressure in Equation (3) has a large range of 0.3 to 4.0, which implies that there are many implications to the value of Ks.

Solved Example

A 450 mm diameter pile is embedded in a homogenous medium dense sand, determine the net pullout capacity (FOS = 3.0). Given: L = 12 m, φ = 38°, Ks = 1.5, and δ = 25°, γ(average) = 17 kN/m3
The water table is at great depth, and take the critical depth to be 15D.

Solution

Critical depth = 15 × 0.45 = 6.75 m
σv = 6.75 × 17 = 114.75 kN/m2

fs = 114.75 × 1.5 × tan 25 = 80.263 kN/m2

fs1 = ½ fsπdZc = ½ × 80.263 × π × 0.45 × 6.75 = 382.9 kN
fs2 = fsπd(L – Zc) = 80.263 × π × 0.45 × (12 – 6.75) = 595.71 kN

Total skin resistance = fs1 + fs2 = 382.9 + 595.71 = 978.61 kN

Qa = 978.61/3 = 326.2 kN (Factor of safety of 3)

After Poulos and Davis (1980), pull out capacity = 2/3 × 326.2 = 217.46 kN

Conclusion

The uplift capacity of piles is an important factor in the design of structures that are subjected to uplift loads. Geotechnical design of piles must ensure that the uplift capacity of the piles is sufficient to resist the expected loads. There are a number of equations that can be used to estimate the uplift capacity of piles, but the final design should be based on the specific soil conditions and the design requirements.

References

Ireland, H.O. (1957). “Pulling Tests on Piles in Sand,” Proc. 4th Int. Conf. SM and FE, Vol. 2.
Murthy V.N.S (2012): Geotechnical Engineering. Marcel Decker Inc. New York
Poulos, H.G., and Davis, E.H. (1980). Pile Foundation Analysis and Design, John Wiley & Sons, New York.
Ramasamy, G., Dey, B. and Indrawan, E. (2004): “Studies on skin friction in piles under tensile and compressive load”, Indian Geotechnical Journal, Vol. 34, No. 3, pp. 276-289.
Sowa, V.A. (1970). “Pulling Capacity of Concrete cast-in-situ Bored Pile,” Can. Geotech. J., Vol. 7.
Verma A. K. and Joshi R. K. (2010): Uplift Load Carrying Capacity of Piles in Sand. Indian Geotechnical Conference – 2010, GEOtrendz pp 857 – 860

Column Splice Connection Design

Column splice connection design involves joining two or more column sections to create a longer column or to provide additional strength. It is used when the length of a column is too long to be fabricated as a single piece. They are also used for connecting columns of different sections as a building goes higher. Similar to beam splices, column splices are typically designed to transfer axial load, shear, and moment.

Properly designed column splice connections are important in ensuring easier buildability, transportation, handling, and maintaining the overall stability and safety of various construction projects in steel structures. The primary function of a column splice connection is to transfer axial loads and moments between the connected column sections. The design should effectively distribute the applied loads through the connection without compromising the structural integrity.

In this article, we will explore the design of bolted column splice connection.

Types of Column Splices

There are two main types of column splices: bearing splices and non-bearing splices. Bearing splices rely on full bearing contact between the two column sections to transfer the axial load. In this case, the loads from the upper column are directly transferred to the lower column through the use of division plates or direct contact between the two sections.

Non-bearing splices use bolts or welds to transfer the axial load. In this case, there is no bearing between the interconnected columns and they are usually detailed with a gap in between the column stacks.

Design of column Splices

The design of a column splice must consider the following factors:

  • The type of splice (bearing or non-bearing)
  • The size and strength of the column sections
  • The axial load, shear, and moment applied to the splice
  • The material properties of the column sections and the splice plates

Design Example

Design a column splice connection for two similar column sections (UC 254x254x89) subjected to an axial dead load of 528kN and a design moment of 178 kN. The ends are not prepared for contact in bearing.

Section Details
Upper Stanchion;  UC 254x254x89 (S275)
Lower Stanchion; UC 254x254x89 (S275)
Flange splice plate; 120 mm × 690 mm × 15 mm (S275)
Flange plate bolts; M22 (Grade 8.8 Black Bolts)
Web splice plate; 137 mm × 390 mm × 15 mm (S275)
Web plate bolts;  M22 (Grade 8.8 Black Bolts)

Splice Details

Flange plates
Number of bolt rows above and below splice; nfpr = 4
Number of bolt columns above and below splice;  nfpc = 2
Dist from end of upper column to first row of bolts; eucfend = 60 mm
Dist from end of lower column to first row of bolts; elcfend = 60 mm 
Bolt pitch;  pfpb = 75 mm
Distance from outer bolts to plate end;  efpend = 60 mm

Web plates
number of bolt rows above and below splice; nwpr = 2
number of bolt columns above and below splice;  nwpc = 2
Dist from end of upper column to first row of bolts; eucwend = 60 mm
Dist from end of lower column to first row of bolts; elcwend = 60 mm
Bolt pitch;  pwpb = 75 mm
Distance from outer bolts to plate end; ewpend = 60 mm

image
Column splice connection details

Loading Details
Axial compressive force due to dead load; Fcd = 528.0 kN
Moment on splice; M = 178.0 kNm

Design Loading
Design load; 
Ft = max(0 kN, abs(M) / min(Du – 2Tu – tfp, Dl – 2Tl – tfp) – Fcd / 2) = 580.8 kN

Check 1 – recommended detailing requirements – column splice, flange plates on inside

Upper Column
Overall depth;  Du = 260 mm
Overall width;  Bu = 256 mm
Flange thickness; Tu = 17.3 mm
Web thickness; tu = 10.3 mm
Root radius; ru = 12.7 mm

Lower Column
Overall depth; Dl = 260 mm
Overall width;  Bl = 256 mm
Flange thickness; Tl = 17.3 mm
Web thickness;  tl = 10.3 mm
Root radius;  rl = 12.7 mm

Flange Plates
Upper projection;                                                             
Lfpu = eucfend + (nfpr – 1) × pfpb + efpend = 345 mm

Lower projection;                                                             
Lfpl = elcfend + (nfpr – 1) × pfpb + efpend = 345 mm

Detailing Requirements 
The upper projection of the flange plate meets the detailing requirements
Lower projection of flange plate meets detailing requirements
The width of the flange plate meets the detailing requirements
The thickness of flange plates meets the detailing requirements
The width of web plates meets the detailing requirements

Flange plate packing;                                                     
tfpack = abs((Du / 2 – Tu) – (Dl/2 – Tl)) = 0.0 mm

Web plate packing;                                                         
twpack = abs(tu – tl) / 2 = 0.0 mm

Sufficient web plate bolts provided
Bearing plate not required

Check 2 – the presence of tension due to axial load and bending moment – internal

Tension Force;                                                                 
Ft = M / min(Du – 2Tu – tfp, Dl – 2Tl – tfp) – Fcd / 2 = 580.8 kN

Check 3 – tensile capacity of flange cover plate – internal

Grade of cover plate;                                                      
pyfp = 275 N/mm2

Net area of flange cover plates (one flange);             
Anetfp = 2(dfp – nfpc / 2Dhfb ) × tfp = 2880 mm2

Net area coefficient;                                                        
Kefp = 1.2

Gross area of flange cover plates (one flange);         
Agrossfp = 2dfptfp = 3600 mm2

Design area of flange cover plates (one flange);        
Afp = min(Kefp × Anetfp, Agrossfp) = 3456 mm2

Tensile capacity;                                                              
Ptfp = pyfp × Afp = 950.4 kN

Utilisation factor;                                                              
Ucheck3 = Ft / Ptfp = 0.611

Tension capacity of flange cover plate: PASS

Check 4 – Shear capacity of bolt group connecting flange cover plate to column flange

Grade of bolts;                                                                 
pbsfp = 460 N/mm2

Shear capacity of top pair of bolts;                               
Psfb1 = min(Psfb, 0.5 × efpend  × tfp × pbsfp) = 113.6 kN

Shear capacity of other bolts; Psfb = 113.6 kN

Joint length; Lj = (nfpr -1) × pfpb = 225 mm

Amended bolt capacity;                                                  
Psfba = 113.6 kN

Reduction factor;                                                             
Rf = min(9dfb / (8dfb + 3tfpack), 1) = 1.000

Shear capacity of bolt group;                                        
Psfbsum = Rf × nfpc × (Psfb1 +  (nfpr – 1) × Psfba) = 909.0 kN

Force on splice plate; Ft = 580.8 kN

Utilisation factor;                                                              
Ucheck4 = Ft / Psfbsum = 0.639

Shear capacity of flange plate bolt group :  PASS

Check 5 – Bearing capacity of flange cover plate connected to column flange

Bearing strength of the end plate;                                
pbsfp = 460 N/mm2

Bearing capacity of the flange plate per bolt (outer); 
Pbsfp1 = min(dfb × tfp  × pbsfp, 0.5 × efpend × tfp × pbsfp) = 151.8 kN

bearing capacity of the end plate per bolt (other);     
Pbsfp = dfb × tfp × pbsfp = 151.8 kN

Capacity of bolt group;                                                   
Pbsfpsum = nfpc × (Pbsfp1 + (nfpr – 1) × Pbsfp) = 1214.4 kN

Bearing force on bolt group;                                          
Ft = 580.8 kN

Utilisation factor;                                                              
Ucheck5 = Ft / Pbsfpsum = 0.478

Bearing capacity of flange plate: PASS

What is the Correct Position for Column Rebar Laps?

Columns are compression members used for transferring superstructure loads to the foundation. In reinforced concrete detailing and construction, the lapping of reinforcements is almost unavoidable due to logistics and handling issues during construction. As a result, design codes all over the world have guidelines on how to properly lap reinforcements in columns.

Improper detailing of reinforced concrete columns can lead to concrete spalling, cracking, and other forms of failure, which may compromise the structural integrity of a column. A poorly arranged reinforcement can also lead to premature buckling and bending that can reduce the load-carrying capacity of reinforced concrete columns.

Here are some important detailing guidelines provided for column laps;

(1) When the diameter of both bars at the lap exceeds 20mm and the cover is less than 1.5 times the size of the smaller bar, transverse links shall be provided throughout the lap length.

(2) At the lap, the least link diameter should be at least one-quarter the size of the smaller bar. 

(3) Link spacing shall not exceed;
(a) 12 x diameter of the smallest bar
(b) 0.6 x dimension of the smallest side of the column
(c) 240 mm


(4) Compression lap length should be at least 25% greater than the compression anchorage length.

(4) For cranked bars, the minimum slope of the crank should be maintained at 1:10. However, this value may be increased up to 1:20 for smooth load transfer.

However, the code appears to be silent on the actual location where columns should be lapped. What do you think? Kindly comment below.

Analysis of Soil Liquefaction Potential

Liquefaction refers to the process in which a granular material changes from a solid to a liquid state due to increased pore-water pressure and decreased effective stress. This phenomenon occurs during ground shaking when loose or medium-compact granular soils experience pore space shrinkage, causing the pore water to be squeezed and leading to loss of bearing capacity.

If the pore water cannot easily drain, it leads to a significant increase in pore-water pressure (u) and a reduction in effective stress (𝜎′). The effective stress (𝜎′) is related to the total stress (𝜎) by the equation;

𝜎′ = 𝜎 − u —— (1)

When the effective stress drops below a certain value, the soil loses contact between grains and starts behaving like a liquid. This is very dangerous for structures supported on such soils.

image 11
Figure 1: Saturated sand condition during liquefaction (Xiao, 2015)

In effect, when loose or medium-compact granular soils experience ground shaking, the pore spaces shrink and squeeze the pore water. If the pore water cannot easily drain, the pore-water pressure significantly increases, leading to a reduction in effective stress. When the effective stress drops to a certain value, the soil loses its grain-to-grain contact and behaves like a liquid.

image 14
Figure 2: Mechanism of liquefaction

Liquefaction can have severe consequences for structures and the surrounding environment. It can result in reduced bearing capacity, causing buildings and infrastructure to sink or collapse. Lateral spreads of liquefied soils can cause large settlements and horizontal displacements. These effects can lead to significant damage and pose risks to human safety.

Figure 2 illustrates the mechanism of liquefaction, which can result in reduced bearing capacity, significant settlement, and horizontal displacement due to lateral spreads of liquefied soils. This can manifest as sand boils or lateral spread of surface soils.

Liquefaction has been observed in various earthquakes, including the 1964 Niigata, 1964 Alaska, 1971 San Fernando, 1985 Mexico City, 1994 Northridge, 1994 Kobe, 1999 Taiwan, 1999 Turkey, 2010 Baja California, 2011 Tōhoku, and 2018 Palu Sulawesi Indonesia earthquakes.

Building failure due to liquefaction
Figure 3: Building failure due to liquefaction

Certain conditions must be met for liquefaction to occur:

  1. The soil deposit consists of sandy or silty soil.
  2. The soil is saturated or nearly saturated (usually below the groundwater table).
  3. The soil is loose or moderately compacted.
  4. The soil is subjected to seismic action (e.g., from an earthquake).

Soils that are relatively well-drained, such as well or poorly graded gravels (GW, GP), are less susceptible to liquefaction compared to sand or silty sand (SW, SP, or SM). Denser granular soils are less prone to liquefaction than loose soils. Liquefaction is less likely to occur in granular soils under higher initial confining effective stress (e.g., deeper soils). Based on case histories, liquefaction typically occurs within a depth of 15 meters.

Cohesive soils generally do not experience liquefaction. To qualitatively evaluate cohesive soils, the Chinese criteria defined by Seed and Idriss (1982) can be used. Liquefaction in cohesive soils can only occur if all three of the following conditions are met:

  1. The clay content (particles smaller than 75μm) is less than 15% by weight.
  2. The liquid limit is less than 35%.
  3. The natural moisture content is greater than 0.9 times the liquid limit.

Screening investigations should also consider the possibility of a locally perched groundwater table, which may occur due to changes in local or regional water management practices, leading to a significant rise in the groundwater table.

Assessment of Liquefaction Hazard

Liquefaction hazard assessments are commonly conducted by employing a factor of safety (Equation 2). This factor represents the relationship between the available resistance to liquefaction, measured in terms of the cyclic stresses required to induce liquefaction, and the cyclic stresses generated by the design earthquake.

Both of these stress parameters, the resistance and the generated stresses, are typically normalized with respect to the effective overburden stress at the specific depth under consideration. They are known as the cyclic resistance ratio (CRR) and cyclic stress ratio (CSR) respectively.

FS = CRR/CSR —— (2)

The recommended guidelines provide guidance on selecting an appropriate factor of safety. For instance, the National Earthquake Hazards Reduction Program (NEHRP) Recommended Provisions for Seismic Regulations for New Buildings and Other Structures (2004) suggest that a factor of safety ranging from 1.2 to 1.5 is generally suitable for building sites. The actual value chosen depends on factors such as the significance of the structure and the potential for ground displacement.

Other authorities also provide recommendations on acceptable levels of risk. The California Geological Survey (1997) proposes that a factor of safety greater than 1.3 can be considered an acceptable level of risk. The DOD Handbook on Soil Dynamics and Special Design Aspects (1997) specifies that a factor of safety of 1.2 is appropriate for engineering design. Empirical methods are commonly used in practice to calculate the factor of safety and are widely adopted.

Evaluation of CSR

Seed and Idriss (1971) presented the following equation for the calculation of the cyclic stress ratio (CSR), and this equation is still the most widely used empirical method:

CSR = 𝜏av/𝜎′vo = 0.65 (amax/g)⋅(𝜎vo/𝜎′vo)⋅rd —— (3)

where:
𝜏av = average cyclic shear stress induced by design ground motion,
𝜎′vo = initial vertical effective stress at the depth under consideration in static condition,
𝜎vo = initial vertical total stress at the depth under consideration in static condition,
amax = peak horizontal acceleration at the ground surface generated by the earthquake,
rd = stress reduction coefficient.

The NCEER workshop (1997) recommended the following equations by Liao and Whitman (1986a) for routine practice and noncritical projects:

rd = 1.0 − 0.00765z (for z ≤ 9.15m) ——-(4a)
rd = 1.174 − 0.0267z (for 9.15m < z ≤ 23m) ——-(4b)

The NCEER workshop (1997) cautioned users that there is considerable variability in rd, and the rd calculated from Equation (4) is the mean of a wide range of possible rd, and the range of rd increases with depth.

Evaluation of CRR

The standard penetration test (SPT), the cone penetration test (CPT), shear wave velocity measurements, and the Becker penetration test (BPT) are the four field tests that are typically utilized by empirical methods for the evaluation of the CRR.

The SPT and CPT methods are often favoured because of the more extensive database and previous experience, although the other tests may be done at sites underlain by gravelly sediment or where access to large equipment is limited. This preference is due to the fact that the SPT and CPT methods have been around longer.

SPT Method of CRR Assessment

The CRR is graphically determined from the SPT blow count as shown in Figure 4. This CRR curve – the SPT clean-sand base curve – is for fines content ≤ 5% under magnitude 7.5 earthquakes. The SPT blow count is first corrected to consider overburden stress, equipment used to conduct the SPT, and the fines content (>5%); then Figure 4 is used to derive the CRR, which in turn is corrected for other earthquake magnitudes.

image 12
Figure 4: SPT clean-sand (percent fines ≤ 5%) base curve for magnitude 7.5 earthquakes (Xiao, 2015)

Step 1: Corrections to overburden stress and various SPT equipment
To account for the effect of overburden stress and various equipment used for SPT, the following equation is used:

(N1)60 = NmCNCECBCRCS ——– (5)

Nm = measured standard penetration blow count,
CN = correction factor based on the effective overburden stress, 𝜎′vo,
CE = correction for SPT hammer energy ratio (ER)
CB = correction factor for borehole diameter,
CR = correction factor for SPT rod length,
CS = correction factor for samplers with or without liners.

Step 2: Corrections to fines content
The corrected (N1)60 in Equation (5) is further corrected for the fines content (FC) in the soil.

(N1)60CS = 𝛼 + 𝛽(N1)60 ——– (6)

where:
(N1)60CS = the (N1)60 for equivalent clean sand; (N1)60CS is used in Figure 4 to find the CRR under magnitude 7.5 earthquakes,
(N1)60 = corrected SPT blow count calculated in Equation (5),
𝛼 and 𝛽 = coefficients determined from the following relationships:

𝛼 = 0 for FC ≤ 5%
𝛼 = exp[1.76 − (190∕FC2)] for 5% < FC < 35%
𝛼 = 5.0 for FC ≥ 35%

𝛽 = 1.0 for FC≤ 5%
𝛽 = [0.99 + (FC1.5∕1000)] for 5% < FC < 35%
𝛽 = 1.2 for FC ≥ 35%

Step 3: Magnitude scaling factors (MSFs)
A magnitude scaling factor (MSF) is used to correct the factor of safety (FS) when the earthquake magnitude is not 7.5:

FS =(CRR7.5/CSR)⋅ MSF ——- (7)

where:
CRR7.5 = the cyclic resistance ratio for a magnitude 7.5 earthquake.
The NCEER workshop (1997) summarized the MSFs proposed by various investigators (Figure 5).

image 13
Figure 5: Magnitude scaling factors derived by various investigators (Xiao, 2015)

Solved Example

Liquefaction Analysis Using SPT Method (After Xiao, 2015)

It is proposed to build a new bridge across a river. The construction site contains poorly graded sandy soil with fines content (passing #200 sieve) of 18%. The soil deposit of the riverbed is fully saturated with 𝛾sat =19.5 kN/m3. The nearby Foot Hill fault system could generate a peak (horizontal) ground acceleration, amax, of 0.25 g at this construction site. Caissons are used as the bridge foundation.

The bottom of the caissons is at a depth of 5m below the riverbed. SPT were performed in a 10.2-cm (4-inch) diameter borehole using a safety trip hammer with a blow count of 6 for the first 15 cm (6 inches), 7 blows for the second 15cm (6 inches), and 9 blows for the third 15cm (6 inches) of driving penetration.

During the design earthquake of magnitude 6.0, will the saturated sand located at the bottom of the caisson liquefy?

Solution:

(1) Calculate CSR:
CSR = 𝜏av/𝜎′vo = 0.65 (amax/g)⋅(𝜎vo/𝜎′vo)⋅rd
where:
Total overburden stress: 𝜎vo = 19.5 × 5 = 97.5 kN/m2
Effective overburden stress: 𝜎′vo = 97.5 − 9.81 × 5 = 48.5 kN/m2
Peak horizontal acceleration: amax = 0.25 g
As z = 5m ≤ 9.15m,
Stress reduction resistance: rd = 1.0 − 0.00765 × 5 = 0.965

So: CSR = 0.65 × 0.25g/g × 97.5/48.5 × 0.965 = 0.304

(2) Calculate CRR using the SPT method:
First, calculate the corrected SPT blow count:
(N1)60 = NmCNCECBCRCS

Nm is the measured SPT blow count, which is the total blow count of the second and third six-inch penetrations.
So Nm = 7 + 9 = 16.
The correction factor on the basis of effective stress (note: Pa = 1 atm = 101.3 kN/m2):
CN = 2.2/(1.2 + 𝜎′vo/pa) = 2.2/(1.2 + 48.5/101.3) = 1.31

Given the automatic trip hammer, the borehole diameter of 102 mm (4 inch), and the rod length of 5m (15 ft),

the correction factor for SPT hammer energy ratio CE =1.0,
the correction factor for borehole diameter CB =1.0,
the correction factor for SPT rod length CR = 0.85, and
the correction factor for samplers with liner CS =1.0.

So ∶ (N1)60 = 16 × 1.31 × 1.0 × 1.0 × 0.85 × 1.0 ≈ 18

Then calculate the equivalent clean-sand SPT blow count (N1)60:
(N1)60CS = 𝛼 + 𝛽(N1)60
As fines content (FC) = 18%,
𝛼 = exp[1.76 − (190∕FC2)] = exp [1.76 − (190∕182)] = 3.234
𝛽 = [0.99 + (FC1.5∕1000)] = [0.99 + (181.5∕1000)] = 1.066

(N1)60cs = 3.234 + 1.066 × 18 = 23

From Figure 4, find CRR7.5 = 0.26.

(3) Calculate MSF:
From Figure 5, at M = 6.0, MSF is between 1.76 and 2.1. On the basis of the critical nature of the project (the foundation of a bridge), the MSF is chosen as the lower bound, 1.76.

(4) Factor of safety against liquefaction:
FS = (CRR7.5/CSR) × MSF = (0.26/0.304) × 1.76 = 1.5 > 1.3

Conclusion: The SPT analysis concludes that the site will not liquefy under the design earthquake.

References:
Xiao M. (2015): Geotechnical Engineering Design (1st Edition). ISBN: 9780470632239 John Wiley & Sons, UK.