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Transcona Grain Elevator Failure: Lessons on Bearing Capacity

The Canadian Pacific Railway Company in the year 1913 constructed the Transcona grain elevators of about 36400 m3 capacity to provide relief for the Winnipeg Yards during the months of peak grain shipment. The structure consisted of a reinforced concrete work-house, and an adjoining bin-house, which contained five rows of 13 bins, each 28 m in height and 4.4 m in diameter.

The bins were based on a concrete structure containing belt conveyors supported by a reinforced concrete shallow raft foundation (Puzrin et al, 2010). The reinforced concrete raft foundation was 600 mm thick, with dimensions of 23.5 x 59.5 m.

Excavation for the construction of the elevator foundations started in 1911, and the first 1.5m depth of soil at the site was rather soft. Beyond the soft clay layer was a relatively stiff blue clay, typical for that area, and locally known as the “blue gumbo”.

According to literature cited by Puzrin et al (2010), no borings or extensive geotechnical investigations were carried out prior to the construction. This is not a surprise, given the level of technology and knowledge about soil engineering at that time. However, an in-situ bearing capacity test (test loading applied using a specially constructed wooden framework) was performed at a depth of 3.7 m.

According to the literature cited by Puzrin et al (2010), the plate load test result indicated that the soil was capable of bearing a uniformly distributed load of at least 400 kPa. The maximum foundation pressure from the bins at maximum load was not expected to exceed 300 kPa, therefore the tests appeared satisfactory to the engineers. Furthermore, they assumed that the “blue gumbo” at the site had similar characteristics and a depth to that on which similar raft foundations of many heavy structures had been founded in the vicinity of Winnipeg. This eventually turned out not to be so.

After the structure was completed, the filling was begun and grain was distributed uniformly between the bins. On October 18, 1913, after the elevator was loaded to 87.5% of its capacity, settlement of the bin-house was noted. Within an hour, the settlement had increased uniformly to about 300 mm following by a tilt towards the west, which continued for almost 24 hours until it reached an inclination of almost 27 degrees (Puzrin et al, 2010).

 Transcona grain elevator
Fig 1: Transcona grain elevator (a) Before failure (b) After failure

Several wash-borings were made immediately after the failure, showing that the elevator was underlain by rather uniform deposits of clay. This finding was in agreement with the geological history of the area, according to which extensive fine-grained sediments were deposited in the waters of the glacial Lake Agassiz which came into being when the Wisconsin ice-sheet blocked the region’s northern outlet.

It was noted that no laboratory test was carried out on the samples collected during the wash borings, but classification was done based on visual observation. The wash-borings, therefore, confirmed the designers’ assumptions of uniform clay layer, and the failure of the Transcona Grain Elevator remained a mystery for another 40 years.

It was thought that, if the smaller-scale plate loading tests predicted a safety factor of more than 1.3 (400/300 = 1.33), and the soil profile is homogeneous, how could the foundation fail? The answer to this question was given by Peck and Bryant (1953) who, in 1951 (38 years later), made two additional borings, far enough from the zone of failure to be in material unaffected by the displacements (Puzrin et al, 2010).

They obtained undisturbed soil samples and performed unconfined compression strength tests (triaxial shear tests with zero confining stress), which produced some eye-opening results shown in Fig 2.

unconfined compression
Fig 2: Soil profile below the elevator (after Peck and Bryant, 1953) (a) classification; (b) unconfined compression strength

On observation of the unconfined compressive strength (qu) result of the site, there are two easily distinguishable layers (Fig. bb). The upper one, a 7.5 m thick stiff clay layer with qu = 108 kPa (undrained shear strength cu = qu/2 = 54 kPa), appears to be resting on a softer clay layer with qu = 62 kPa (cu = 31 kPa). This finding suggests that the elevator failure was most likely caused by the insufficient bearing capacity of its foundation.

In a bearing capacity failure, a failure mechanism is formed below the foundation (Fig. 3b). The settlement takes place much faster and without decrease of the soil volume. Therefore, the displaced soil has to find itself an exit, causing a ground heave in the vicinity of the structure. This ground heave is a distinctive feature of the failure of the Transcona Grain Elevator (Puzrin et al, 2010).

bearing capacity failure
Fig 3: Bearing capacity failure: (a) settlement; (b) failure; (c) the ground heave (Puzrin et al, 2010)

According to (Puzrin et al, 2010), the particular problem of the Transcona Grain Elevator was that the failure mechanisms of the plate loading tests were apparently confined to the upper stiffer clay layer, due to the relatively small size of the plates. The elevator foundation, however, developed a much deeper failure mechanism which entered the weaker clay layer, significantly reducing the bearing capacity.

This led the researchers to consider the two-layer model and other bearing capacity theories that are adequate to describe the situation on the site. Details of this can be found in Puzrin et al, (2010).

It was observed that the true failure contact pressure was 293 kPa. In the original design, it was assumed that the soil profile was homogeneous with the properties of the stiff upper layer uc1 = 54 kPa. In this case, the bearing capacity of the foundation was calculated as 386 kPa (applying the appropriate bearing capacity, shape, and depth correction factors). This was found to be close to the 400 kPa obtained from the plate load tests.

Note that if the soil was homogeneous but with the properties of the weaker lower layer (uc2 = 31 kPa), the resulting bearing capacity would be 251 kPa. If this value was available to designers, the result would actually not be that bad: not only the elevator would not fail, it would not even be too much overdesigned.

A more sophisticated analysis, based on a two-layer model, should produce more accurate predictions. First, the reseachers followed Peck and Bryant (1953) and used the approximate method based on the Prandtl solution using a weighted average of the undrained shear strength. This gave a bearing capacity of 321 kPa which is 10% larger than the failure pressure. While the Prandtl solution for a homogeneous soil is the exact solution, in the two-layer approximation, it was observed that it is not only inaccurate, but it is also not conservative and could lead to failure.

The scoop mechanism, in contrast, provides a remarkably good bearing capacity estimate of 297 kPa. Being an upper bound, this value, as expected, is higher than the true failure pressure of 293 kPa, but only marginally. It would provide an excellent estimate for the design of the elevator if only the Soil Mechanics was more mature in those days and the soil properties were properly determined.

Source:
The information in this article was majorly obtained from:
Puzrin A. M., Alonso E.E., Pinyol N. M. (2010): Geomechanics of Failures. Springer. DOI 10.1007/978-90-481-3531-8

Bridge Design Textbook

We are pleased to announce the release of our publication on the ‘Design of Reinforced Concrete Bridges‘. This publication is part of the proceedings of our January 2021 Webinar on Bridge Design.

The publication essentially contains full calculation sheets on the analysis and design of different structural components of a T-beam bridge (beam and deck slab bridge) such as;

  1. Deck slab
  2. Reinforced Concrete Girders (beams)
  3. Bearings
  4. Pier Cap
  5. Piers
  6. Abutment
  7. Pile Cap
bridge deck section
bridge design

Both computer and manual methods were adopted in the analysis presented, and the design considers the effect of wind, temperature, and secondary traffic loads such as braking, traction, etc on the bridge. Other elements such as parapets, precast planks (filigree slab) for receiving the in-situ topping of the deck slab, etc were also considered in the design. The design was carried out according to the British Standards.

To obtain the book for NGN 5,000 only, click HERE

BRIDGE SHOWING INFRASTRUCTRES DESIGNED BY RMC

Some excerpts from the publication are shown below;

bridge deck section
The typical bridge deck section
longitudinal section
Typical longitudinal view of the bridge
bending moment on bridge deck slab due to dead load
vertical shear in deck slab
main beam design
Temperature effect
bending moment in beams
bridge reinforcement calculation
piers loading
abutment design

To obtain the book for NGN 5,000 only, click HERE

bridge design

Deflection of Trusses | Worked Example

Trusses deflect when loaded. Under gravity loads, this is usually characterised by the sagging of the top and bottom chords, and the consequent movement of the web and diagonal members. In the design of trussed structures such as roofs and bridges, it is always important to keep the deflection of the trusses to a minimum in order to maintain the appearance and functionality of the structure. Deflection of trusses can be assessed manually using the virtual work method (unit load method), direct stiffness method, or finite element analysis.

The deflection of roof trusses is often limited to span/240. On the other hand, a span to depth ratio of 15 is often found adequate for all practical purposes. For bridges, AASHTO design code states that the maximum deflection of a truss bridge due to live load should not exceed span/800. The principle of virtual work can be used to compute the maximum deflection of the truss, which is then compared to the allowable deflection.

In this article, we are going to explore how to determine the deflection of trusses using the virtual work method (unit load method). In the virtual work method, the truss is analysed for the axial forces due to the externally applied load. Subsequently, the external forces are removed and replaced with a unit virtual load at the node where the deflection is to be obtained. The direction of the unit load should be in the same direction with where the deflection is sought. The truss is re-analysed for the axial forces in the members due to the virtual unit load.

The summation of the deflection of the individual members using the formula below gives the deflection of the roof truss at the point of interest.

δ = ∑niNiLi/EiAi

Where;
n = axial force in member i due to virtual unit load
N = axial load in member i due to externally applied load
L = length of member
E = Modulus of elasticity of member
A = Area of member

Worked Example

A truss is loaded as shown below. Obtain the vertical deflection at point C due to the externally applied load (AE = constant).

deflection of trusses

Support Reaction due to externally applied load
Let MG = 0;
9Ay – (6 × 6) – (4 × 3) + (10 × 3) = 0
Ay = 2 kN↑

Let MA = 0;
9Gy – (10 × 3) – (4 × 6) – (6 × 3) = 0
Gy = 8 kN ↑

Let Fx = 0
Ax = 10 kN ←

Internal forces due to the externally applied loads

Joint A

JOINT A

Fy = 0
2 + FABsin45 = 0
FAB = -2.8 kN (compression)

Fx = 0
-10 + FAC + FABcos45 = 0
-10 + FAC – 2.8cos45 = 0
FAC = 12 kN (tension)

Joint B

JOINT B


Fx = 0
FBD – FBAcos45 = 0
FBD – 2.8cos45 = 0
FBD = -2 kN (compression)

Fy = 0
-6 – FBC – FBAsin45 = 0
-6 – FBC + 2.8sin45 = 0
FBC = -4 kN (tension)

JOINT C

JOINT C

Fy = 0
FBC + FCDsin45 = 0
-4 + FCDcos45 = 0
FCD = 5.656 kN (tension)

Fx = 0
-FCA + FCDcos45 + FCE = 0
-12 + 5.656cos45 + FCE = 0
FCE = 8 kN (tension)

JOINT G

JOINT G

Fy = 0
8 + FGF = 0
FGF = -8 kN (compression)

Fx = 0
FGE = 0

JOINT F

JOINT F


Fy = 0
-FFG – FFEsin45 = 0
-(-8) – FFEsin45 = 0
FFE = 11.31 kN (tension)

Fx = 0
-FFD – FFEcos45 + 10 = 0
-FFD – 11.31cos45 + 10 = 0
FFD = 2 kN (tension)

JOINT E

JOINT E

Fy = 0
FED + FEFsin45 = 0
FED + 11.31sin45 = 0
FED = -8 kN (compression)

Having obtained the internal forces in all the members due to the externally applied loads, let us remove all the loads and replace them with a unit vertical load at point C as shown below;

TRISS WITH VIRTUAL LOAD

Support Reaction due to virtual load
Let MG = 0;
9Ay – (1 × 6) = 0
Ay = 6/9 = 0.667 ↑

Let MA = 0;
9Gy – (1 × 3) = 0
Gy = 3/9 = 0.333 ↑

Let Fx = 0
Ax = 0

When analysed using the same procedure as above;

FAC = 0.665 (tension)
FAB = -0.941 (compression)
FBC = 0.665 (tension)
FCE = 0.334 (tension)
FCD = 0.47 (tension)
FBD = 0.665 (compression)
FDE = 0.331 (compression)
FDF = 0.333 (compression)
FEF = 0.47 (tension)
FEG = 0
FGF = 0.333 (compression)

We can now form a table as follows;

MembernN (kN)L (m)nNL/AE
AC+0.665+12.003.00023.940
AB-0.940-2.804.24311.167
BC+0.665-4.003.000-7.980
CE+0.334+8.003.0008.016
CD+0.470+5.654.24311.267
BD-0.665-2.003.0003.990
DE-0.331-8.003.0007.944
DF-0.331+2.003.000-1.986
EF+0.470+11.314.24322.554
EG003.0000.000
GF-0.333-8.003.0007.992
∑niNiL = 86.904/AE

Therefore the vertical deflection at point C is 86.904/AE metres.

Design of Strap Footing | Cantilever Footing

Strap footings or cantilever footings are a special form of combined footings. They consist of two separate bases that are connected (balanced) by a strap beam. In the design of strap footing, it is assumed that the strap beam is rigid and does not transfer any load by bearing on the soil at its bottom contact surface.

Strap footing is necessary when the foundation of a column cannot be built directly under the column or when the column should not exert any pressure below. It is then necessary to balance it by a cantilever arm rotating about a fulcrum and balanced by an adjacent column (or a mass of concrete or by piles) in case of where footings cannot be built.

balanced bases

As was described above, balanced footings consist of two separate footings connected by a strap beam. In the design of balanced bases, uniform soil pressure can be assumed if we can make the centre of the areas of the system coincide with the centre of gravity of the loads. When these two centres do not coincide, we have the vertical load and moment acting on the system due to the eccentricity. As a result, the distribution of base pressure will not be uniform but can be assumed to be linearly varying.

The difference between balanced footings and cantilever footings can be described as follows. In a balanced footing, we make the centre of gravity of the loads and the centre of the areas coincide. Hence, the ground pressure will be uniform. In cantilever footings, in general, the two centres may not coincide, so we have a moment in addition to the vertical loads. Hence, the ground pressure will be varying.

Worked Example

Two columns with the following loading conditions are spaced 4 m apart. Due to site boundary constraints, design a strap footing for the columns, if the safe bearing capacity of the soil is 150 kN/m2; fck = 25 N/mm2; fyk = 500 N/mm2

ColumnSize (mm)Service Load (kN)Ultimate Load (kN)
C1300 x 300450617
C2300 x 300600822

Solution
We can either dimension each of the footings so that the CG of the areas and loads coincide, resulting in uniform pressure, or adopt
suitable base dimensions and then check the resulting ground pressure taking the unit as a whole, which may not be uniform, and design for non-uniform pressure. We will adopt the first method.

Design of strap footing

Step 1: Preliminary sizing of the footing of the exterior column
Base area (A1) needed for column C1 = Service load/Safe bearing capacity = 450/150 = 3m2
Try a rectangular footing of size 1.75m wide x 2 m long along the centre line (Area provided = 3.5 m2)

Hence, we fix the fulcrum at 2.0/2 = 1.0 m from the end near C1.

Therefore;
Distance of R1 from C1 = L1 = 1.0 – 0.3 = 0.7 m from C1
Distance of C2 from R1 = L2 = 4 – 0.7 = 3.3 m

Taking moment about C2;
3.3R1 = 450 x 4
R1 = 545.45 kN

Let the summation of the vertical forces be equal to zero;
R2 = 450 + 600 — 545.45 = 504.55 kN

Step 2: Check the factor of safety (FOS) against overturning using characteristic loads
FOS = C2L2/C1L1 = (600 x 3.3)/(450 x 0.7) = 6.285 > 1.5 (Okay).

Step 3: Find the dimension of footing for R2 so that the CG of loads and areas coincides.
This is given by;
B = √(R2/SBC) = √(504.55/150) = 1.83 m
Adopt a footing 1.85 m x 1.85 m

Step 4: Recalculate the necessary breadth of footing F1 so that CG of loads and areas of footings coincides.
x1 = CG of loads = (C1 x 4)/(C1 + C2) = (450 x 4)/(450 + 600) = 1.714 m from C2

Let us find the CG of areas we have assumed.
Area of F2 = A2 = 1.85 x 1.85 = 3.4225 m2. Find Ax required for CG to be same as that of loads.

x2 = (A1 x 3.3)/(A1 + A2) = 1.714 (for a balanced base)
(A1 x 3.3)/(A1 + 3.4225) = 1.714
On solving; A1 = 3.698 m2

For a length of 2m, a width of 3.689/2 = 1.85 m is required.
Therefore slightly increase the width of A1 to 1.85 m.

Hence;
A1 = (1.85 x 2)m = 3.7 m2
A2 = (1.85 x 1.85)m = 3.4225 m2

Step 5: Calculate uniform pressure for factored load
qnet = (617 + 822)/(3.7 + 3.4225) = 202.03 kN/m2

You can verify independently;
At ultimate limit state; R1 = (545.45 x 617)/450 = 747.87 kN
Soil pressure under base 1 = 747.87/3.7 = 202.13 kN/m2

At umtimate limit state R2 = (504.55 x 822)/600 = 691.23 kN
Soil pressure under base 2 = 691.23/3.4225 = 201.967 kN/m2

Step 6: Design of footing F1
Let the width of the strap beam be 0.3 m. The maximum moment will occur at the face of the strap beam (overhang of the strap beam).

Length of overhang = (1.85 – 0.3)/2 = 0.775 m on both sides of the beam

MEd = ql2/2 = (202.13 x 0.7752)/2 = 60.7 kNm

Assuming a footing depth h = 400 mm and concrete cover of 50 mm;
Effective depth d = 400 – 50 – 8 = 342 mm.

Critical design moment at the face of the strap beam
MEd = 125 kNm/m
k = MEd/(bd2fck) = (60.7 x 106)/(1000 x 3422 x 25) = 0.0207
Lever arm = z = d[0.5 + √(0.25 – 0.882k)] = 0.95d
⇒ z = 0.95d = 0.95 x 342 = 325 mm
⇒ As = MEd/0.87fykz = (60.7 x 106)/(0.87 x 500 x 325) = 429 mm2/m
Asmin = 520 mm2/m

Provide H12 @ 200 c/c (Asprov = 565 mm2/m)

Beam shear
Check critical section d away from the face of the strap beam
VEd = 202.13 x (0.775 – 0.342) = 87.522 kN/m
vEd = 87.522/342 = 0.256 N/mm2

vRd, c = CRd, c × k × (100 × ρ1 × fck) 0.3333
CRd, c = 0.12
k = 1 + √ (200/d) = 1 + √ (200/342) = 1.76
ρ = 565/(342 × 1000) = 0.00165
vRd, c = 0.12 × 1.76 × (100 × 0.00165 × 25)0.333 = 0.338 N/mm2
=> vEd (0.256 N/mm2) < vRd,c (0.338 N/mm2) beam shear ok

Step 7: Design of the strap beam
Equivalent line load under column base 1 = 202.13 x 1.85 = 373.94 kN/m
Equivalent line load under column base 2 = 202.13 x 1.85 = 373.94 kN/m

loading of strap footing


The shear force values at the critical points are;
V1L = (0.15 x 373.94) = 56.09 kN
V1R = 56.09 – 617 = -560.9 kN
Vi = (373.94 x 2) – 617 = 130.88 kN
V2L = 130.88 + (373.94 x 0.925) = 476.774 kN
V2R = 476.774 – 822 = -345.226 kN
Vj = -345.225 + (373.94 x 0.925) = 0

shear force diagram of strap footing 2

The maximum bending moment will occur at the point of zero shear which is 1.5 m from column A. This can be easily obtained by using similar triangles.

Mmax = (373.94 x 1.652)/2 – (617 x 1.5) = -416.474 kNm (hogging moment).
The maximum sagging moment will occur under column C2;
Mmax = (373.94 x 0.9252)/2 = 159.976 kNm

Assume a total depth of 600 mm for the beam;
Effective depth = 600 – 50 – 10 – 10 = 530 mm
k = MEd/(bd2fck) = (416.474 x 106)/(300 x 5302 x 25) = 0.1976

Since k < 0.167, compression is required. This means that the beam will have to be designed as a doubly reinforced beam.
Area of compression reinforcement AS = (MEd – MRd) / (0.87fyk (d – d2))

MRd = 0.167fckbd2 = (0.167 × 25 × 300 × 5302) × 10-6 = 351.827 kNm

d2 = 50 + 10 + 10 = 70 mm

AS2 = ((416.474 – 351.827) × 106) / (0.87 × 500 × (530 – 70)) = 323 mm2
Asmin = 234 mm2
Provide 4H16 Bottom (Asprov = 804 mm2)
Confirm that this reinforcement can satisfy for bending under column C2.

Area of tension reinforcement As1 = MRd / (0.87fyk z) + AS2
Where z = d[0.5+ √(0.25 – 0.882K’)]
K’ = 0.167
z = d[0.5+ √((0.25 – 0.882(0.167))] = 0.82d
As1 = MRd/(0.87fyk z) + AS2 = (351.827 × 106) / (0.87 × 500 × 0.82 × 530) + 323 mm2 = 2184 mm2

Provide 8H20 Top (ASprov = 2512 mm2)

Shear Design
Using the maximum shear force for all the spans
Support A; VEd = 560.9 kN
VRd,c = [CRd,c.k. (100ρ1 fck)1/3 + k1cp]bw.d
CRd,c = 0.18/γc = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/530) = 1.61 > 2.0, therefore, k = 1.61

Vmin = 0.035k3/2fck1/2
Vmin = 0.035 × 1.613/2 × 251/2 = 0.357 N/mm2

ρ1 = As/bd = 1256/(300 × 530) = 0.007899 < 0.02;

VRd,c = [0.12 × 1.61(100 × 0.007899 × 25)1/3] × 300 × 540 = 84597.88 N = 84.597 kN
Since VRd,c (84.597 kN) < VEd (560.9 kN), shear reinforcement is required.

The compression capacity of the compression strut (VRd,max) assuming θ = 21.8° (cot θ = 2.5)
VRd,max = (bw.z.v1.fcd)/(cotθ + tanθ)
V1 = 0.6(1 – fck/250) = 0.6(1 – 25/250) = 0.54
fcd = (αcc fck)/γc = (0.85 × 25)/1.5 = 14.167 N/mm2
Let z = 0.9d
VRd,max = [(300 × 0.9 × 540 × 0.54 × 14.167)/(2.5 + 0.4)]× 10-3 = 384.619 kN

Since VRd,c < VRd,max < VEd
The beam is subjected to high shear load, we need to modify the strut angle.
θ = 0.5sin-1[(VRd,max /bwd)/0.153fck(1 – fck/250)]
(VRd,max /bwd) = 2.418 N/mm2
0.153fck(1 – fck/250) = 3.4425
θ = 0.5sin-1[(2.418/3.4425] = 22.31°

Since θ < 45°, section is OK for the applied shear stress

Hence Asw/S = VEd /(0.87fykzcotθ) = 560900/(0.87 × 500 × 0.9 × 530 × 2.437) = 1.109
Maximum spacing of shear links = 0.75d = 0.75 × 530 = 597.5
Provide 4 legs Y10 @ 250mm c/c as shear links (Asw/S = 1.256) Ok






Design of Gantry Crane Girders | BS 5950

Overhead cranes are usually required in industrial or storage buildings for the lifting and/or movement of heavy loads from one point to another. These overhead cranes can be manually operated (MOT) or electrically operated (EOT). A Gantry girder may therefore be defined as a structural beam section, with or without an additional plate or channel connected to the top flange to carry overhead electric travelling cranes.

The typical components of an overhead travelling crane are shown in the figure below;

crane
Fig 1: Typical components of overhead crane

Crane gantry girders are usually designed to resist unsymmetrical forces and moments from vertical loads and reactions, horizontal forces, longitudinal forces, fatigue, and impact forces. The design of a gantry crane girder, therefore, involves the selection of a suitable and workable steel model and section to satisfy the machine (crane) requirements, loading, equipment, etc without leading to any structural or service failure.

Normally, for medium-duty (say 25 to 30 t capacity) cranes, standard universal rolled I-beams are used. In the design of gantry girders with long spans supporting heavy vertical dynamic crane wheel loads with transverse horizontal crane surges, the standard universal rolled beam section is not adequate as a gantry girder; the built-up section of a plate girder is adopted instead.

built up plate girder section
Fig 2: Built-ip plate girder sections

Gantry girders are usually subjected to very high vertical impacts, transverse horizontal surges, and longitudinal horizontal surges depending on their lifting capacity and geometry.

The transverse horizontal force generated either of the two following factors or by a combination both of it;

  1. Thrust from sudden application of the brakes of the crab motor, causing abrupt stoppage of the crab and load when transversing the crab girders. This thrust is resisted by the frictional force developed between the crab wheels and crab girders, is then transferred to the crosshead girders of the crane, and finally transferred as point loads through the main wheels of the crane into the top flange of the crane girders.
  2. A crane often drags weights across the shop floor. If the weight is very heavy, this pulling action induces a transverse horizontal component of force ( a point load) on the crane girders through the crane wheels.

This transverse horizontal force will be transferred to the crane girders through the double-flanged crane wheels on the end carriages and the crane is designed to avoid the possibility of derailment. Due to the difficulties in determining this kind of force, the horizontal transverse force on each gantry girder is equal to 10% of the total load lifted.

Furthermore, during the traveling of the crane, the sudden application of brakes induces frictional resistance to the sliding of the locked wheels upon a rail fixed to the gantry girder. This frictional resistance, in turn, generates a horizontal force along the length of the gantry girder and finally transfers to the columns that support the gantry girder. This is usually the source of the longitudinal horizontal force.

Gantry girders may be simply supported or continuous over supports. Continuity over supports reduces the depth and cost, but any differential settlement of the supports may reverse the original design values of the moments in the sections, thus exceeding the allowable stress in the material, with consequent collapse of the member. So, we adopt simply supported gantry girders, particularly for very heavily loaded girders.

Design Example of a Gantry Crane Girder

Design a gantry to satisfy the manufacturer’s design data given below;

Crane capacity = 20 tonnes (200 kN)
Maximum load lifted = 200 kN.
Crane span = 13.0 m
Weight of crane bridge = 120 kN
Spacing of wheels = 1.2 m
End clearance of crane = 600 mm (minimum).
Minimum headroom from rail top = 4500 mm.
Weight of crab = (1/5 of maximum load lifted + 5 kN) = (1/5) × 200 + 5 = 45 kN.

Crane and Girder Details

Design of Gantry Crane Girders

Crane details
Self weight of crane bridge (excluding crab);  Wcrane = 120.0 kN
Self weight of crab; Wcrab = 25.0 kN
Crane safe working load (SWL); Wswl = 200.0 kN
Span of crane bridge; Lc = 13000 mm
Minimum hook approach; ah = 600 mm
No. of wheels per end carriage; Nw = 2
End carriage wheel centres; aw1 = 3000 mm
Class of crane; Q3
No. of rails resisting crane surge force; Nr = 1
Self weight of crane rail; wr = 0.5 kN/m
Height of crane rail;  hr = 100 mm

Gantry girder details
Span of gantry girder; L = 5000 mm
Gantry girder section type; Plain ‘I’ section
Gantry girder ‘I’ beam; UB 610x305x238
Grade of steel; S 275

Loading, Shear forces and Bending Moments

Unfactored self weight and crane rail UDL
Beam and crane rail self weight udl; wsw = (Massbm × gacc) + wr = 2.8 kN/m

Maximum unfactored static vertical wheel load
From hook load; Wh = Wswl × (Lc – ah)/(Lc × Nw) = 95.4 kN
From crane self weight (including crab);  Ws = [Wcrane/2 + Wcrab × (Lc – ah)/Lc]/Nw = 41.9 kN

Total unfactored static vertical wheel load; Wstat = Wh + Ws = 137.3 kN

Maximum unfactored dynamic vertical wheel load
From BS2573:Part 1:1983 – Table 4

Dynamic factor with crane stationary; Fsta = 1.30;
Dynamic wheel load with crane stationary; Wsta = (Fsta × Wh) + Ws = 165.9 kN
Dynamic factor with crane moving; Fmov = 1.25;
Dynamic wheel load with crane moving; Wmov = Fmov × Wstat = 171.6 kN
Max unfactored dynamic vertical wheel load;Wdyn = max(Wsta, Wmov) = 171.6 kN

Dynamic vertical wheel loads
Unfactored transverse surge wheel load
Number of rails resisting surge; Nr = 1
Proportion of crab and SWL acting as surge load;  Fsur = 10 %
Unfactored transverse surge load per wheel; Wsur = Fsur × (Wcrab + Wswl)/(Nw × Nr) = 11.3 kN

Surge wheel loads
Unfactored transverse crabbing wheel load
Unfactored transverse crabbing load per wheel; Wcra = max(Lc × Wdyn/(40 × aw1), Wdyn/20) = 18.6 kN

Unfactored longitudinal braking load
Proportion of static wheel load acting as braking load; Fbra = 5 %
Unfactored longitudinal braking load per rail; Wbra = Fbra × Wstat × Nw = 13.7 kN

Ultimate loads
Load Case 1 (1.4 Dead + 1.6 Vertical Crane)

Vertical wheel load; Wvult1 = 1.6 × Wdyn = 274.6 kN
Gantry girder self weight udl; wswult = 1.4 × wsw = 4.0 kN/m

Load Case 2 (1.4 Dead + 1.4 Vertical Crane + 1.4 Horizontal Crane)
Vertical wheel load; Wvult2 = 1.4 × Wdyn = 240.3 kN
Gantry girder self weight udl; wswult = 1.4 × wsw = 4.0 kN/m
Horizontal wheel load (surge); Wsurult = 1.4 × Wsur = 15.7 kN
Horizontal wheel load (crabbing); Wcrault = 1.4 × Wcra = 26.0 kN

Maximum ultimate vertical shear force
From load case 1; Vv = Wvult1 × (2 – aw1/L) + wswult × L/2 = 394.4 kN

Ultimate horizontal shear forces (load case 2 only)
Shear due to surge; Vsur = Wsurult × (2 – aw1/L) = 22.0 kN
Shear due to crabbing; Vcra = Wcrault = 26.0 kN

Maximum horizontal shear force; Vh = max(Vsur, Vcra) = 26.0 kN

Ultimate vertical bending moments and co-existing shear forces
Bending moment loadcase 1; Mv1 = Wvult1 × L/4 + wswult × L2/8 = 355.7 kNm
Co-existing shear force; Vv1 = Wvult1/2 = 137.3 kN

Bending moment loadcase 2; Mv2 = Wvult2 × L/4 + wswult × L2/8 = 312.8 kNm
Co-existing shear force;  Vv2 = Wvult2/2 = 120.1 kN

Ultimate horizontal bending moments (loadcase 2 only)

Surge moment; Msur = Wsurult × L/4 = 19.7 kNm
Crabbing moment; Mcra = Wcrault × L/4 = 32.5 kNm

Maximum horizontal moment; Mh = max(Msur, Mcra) = 32.5 kNm

Section Properties

Beam section properties
Area; Abm = 303.3 cm2
Second moment of area about major axis; Ixxbm = 209471 cm4
Second moment of area about minor axis; Iyybm = 15837 cm4
Torsion constant; Jbm = 785.2 cm4

Section properties of top flange only
Elastic modulus;  Ztf = Tbm × Bbm2/6 = 507.5 cm3
Plastic modulus; Stf = Tbm × Bbm2/4 = 761.2 cm3

Steel design strength
From BS5950-1:2000 – Table 9
Flange design strength (T = 31.4 mm); pyf = 265 N/mm2
Web design strength (t = 18.4 mm); pyw = 265 N/mm2

Overall design strength; py = min(pyf, pyw) = 265 N/mm2

Section classification (cl. 3.5.2)
Parameter epsilon; ε = (275/py)1/2 = 1.019;
Flange (outstand element of comp. flange);ratio1 = Bbm/(2 × Tbm) = 4.959;
Web (neutral axis at mid-depth); ratio2 = dbm/tbm = 29.348;

Flange Classification = Class 1 plastic;
Web Classification = Class 1 plastic;
Overall Section Classification = Class 1 plastic

Shear buckling check (cl. 4.2.3)
Ratio d/t;    d/t = dbm/tbm = 29.348;
PASS – d/t ≤ 70ε – The web is not susceptible to shear buckling

Design Checks

Vertical shear capacity (cl. 4.2.3)
Vertical shear capacity of beam web; Pvv = 0.6 × py × tbm × Dbm = 1860.1 kN
UF1 = Vv/Pvv = 0.212
PASS – Vv ≤ Pvv – Vertical shear capacity adequate (UF1 = 0.212)

Loadcase 1 – Vv1 ≤ 0.6Pvv – Beam is in low shear at position of max moment
Loadcase 2 – Vv2 ≤ 0.6Pvv – Beam is in low shear at position of max moment

Horizontal shear capacity (cl. 4.2.3)
Horizontal shear capacity of beam flange; Pvh = 0.6 × py × 0.9 × Tbm × Bbm = 1399.2 kN
UF2 = Vh/Pvh = 0.019
PASS – Vh ≤ Pvh – Horizontal shear capacity adequate (UF2 = 0.019 – low shear)

Vertical bending capacity (cl. 4.2.5)
Vertical bending capacity of beam;                            
Mcxz = 1.2 × py × Zxxbm = 2095.4 kNm
Mcxs = py × Sxxbm = 1983.8 kNm
Mcx = min(Mcxz, Mcxs) = 1983.8 kNm

UF3 = Mv1/Mcx = 0.179
PASS – Mv1 ≤ Mcx – Vertical moment capacity adequate (UF3 = 0.179)

Effective length for buckling moment (Table 13)
Length factor for end 1; KL1 = 1.00
Length factor for end 2;  KL2 = 1.00
Depth factor for end 1; KD1 = 0.00
Depth factor for end 2;KD2 = 0.00

Effective length;Le = L × (KL1 + KL2)/2 + Dbm × (KD1 + KD2)/2 = 5000 mm

Lateral torsional buckling capacity (Annex B.2.1, 2.2 & 2.3)

Slenderness ratio; λ = Le/ryybm = 69.2;
Slenderness factor; v = 1/[1 + 0.05 × (λ/xbm)2]0.25 = 0.899;
Section is class 1 plastic therefore; βw = 1.0
Equivalent slenderness;  λLT = ubm × v × l × √(βw) = 55.1;
Robertson constant;  αLT = 7.0
Limiting equivalent slenderness; λL0 = 0.4 × (p2 × ES5950/py)0.5 = 35.0;

Perry factor; ηLT = max(αLT × (λLT – λL0)/1000, 0) = 0.141;
Euler buckling stress; pE = π2 × ES5950LT2 = 665.5 N/mm2
Factor phi; ϕLT = [py + (ηLT + 1) × pE]/2 = 512.3 N/mm2
Bending strength; pb = pE × py/[ϕLT + (ϕLT2 – pE × py)0.5] = 218.9 N/mm2
Buckling resistance moment; Mb = pb × Sxxbm = 1638.8 kNm
Equivalent uniform moment factor; mLT = 1.0
Allowable buckling moment;  Mballow = Mb/mLT = 1638.8 kNm

UF4 = Mv1/Mballow = 0.217
PASS – Mv1 ≤ Mballow – Buckling moment capacity adequate (UF4 = 0.217)

Horizontal bending capacity (loadcase 2 only) cl. 4.2.5
Horizontal moment capacity of top flange; Mctf = min(py × Stf,1.2 × py × Ztf) = 161.4 kNm
UF5 = Mh/Mctf = 0.202
PASS – Mh ≤ Mctf – Horizontal moment capacity adequate (UF5 = 0.202)

Combined vertical and horizontal bending (loadcase 2 only)
Cross section capacity (cl. 4.8.3.2)
Section utilisation; UF6 = Mv2/Mcx + Mh/Mctf = 0.359;
PASS – Section capacity adequate (UF6 = 0.359)

Member buckling resistance (cl. 4.8.3.3.1)
Uniform moment factors;                                              
mx = 1.0
my = 1.0

Case 1; UF7 = mx × Mv2/(py × Zxxbm) + my × Mh/(py × Ztf) = 0.421;
Case 2;   UF8 = mLT × Mv2/Mb + my × Mh/(py × Ztf) = 0.433;
PASS – Buckling capacity adequate (UF7&8 = 0.433)

Check beam web bearing under concentrated wheel loads (cl. 4.5.2.1)
End location
Maximum ultimate wheel load; Wvult1 = 274.6 kN
Stiff bearing length (dispersal through rail); b1 = hr = 100 mm
Bearing capacity of unstiffened web; Pbw = [b1 + 2 × (Tbm + rbm)] × tbm × py = 954.7 kN
UF9 = Wvult1/Pbw = 0.288
PASS – Wvult1 ≤ Pbw – Web bearing capacity adequate (UF9 = 0.288);

Check beam web buckling under concentrated wheel loads (cl. 4.5.3.1)
End location – top flange not effectively restrained rotationally or laterally

Maximum ultimate wheel load; Wvult1 = 274.6 kN
Stiff bearing length (dispersal through rail); b1 = hr = 100 mm
Effective length of web; LEweb = 1.2 × dbm = 648 mm

Buckling capacity of unstiffened web;                        
Pxr = 1/2 × 25 × ε × tbm/[(b1 + 2 × (Tbm + rbm)) × dbm]1/2 × 0.7 × dbm/LEweb × Pbw
Pxr = 401.3 kN
UF10 = Wvult1/Pxr = 0.684
PASS – Wvult1 ≤ Pxr – Web buckling capacity adequate (UF10 = 0.684)

Allowable deflections
Allowable vertical deflection = span/600; dvallow = L/limitv = 8.3 mm
Allowable horizontal deflection = span/500; dhallow = L/limith = 10.0 mm

Calculated  vertical deflections
Modulus of elasticity;E = ES5950 = 205 kN/mm2
Due to self weight; dsw = 5 × wsw × L4/(384 × E × Ixxbm) = 0.1 mm

Due to wheels at position of maximum moment;    
dv1 = Wstat × L3/(48 × E × Ixxbm) = 0.8 mm

Total vertical deflection; dv = dsw + dv1 = 0.9 mm

PASS – dv ≤ dvallow – Vertical deflection acceptable (Actual deflection = span/5641)

Calculated horizontal deflection
Due to surge (wheels at position of max moment);  dhs = Wsur × L3/(48 × E × Iyybm/2) = 1.8 mm
Horizontal crabbing deflection; dhc = Wcra × L3/(48 × E × Iyybm/2) = 3.0 mm

Maximum horizontal deflection; 
dh = max(dhs, dhc) = 3.0 mm
PASS – dh ≤ dhallow – Horizontal deflection acceptable (Actual deflection = span/1676)


.

Bentley Systems Calls for Nominations for the ‘2021 Going Digital Awards in Infrastructure’

Image courtesy of Bentley Systems

Leading infrastructure engineering software company Bentley Systems has announced its call for nominations for the 2021 Going Digital Awards in Infrastructure program. Bentley Systems is an infrastructure engineering software company that provides innovative software to advance the world’s infrastructure. Their software solutions are used by professionals, and organizations of every size, for the design, construction, and operations of roads and bridges, rail and transit, water and wastewater, public works and utilities, buildings and campuses, and industrial facilities. 

Formerly known as the Year in Infrastructure Awards, this global awards program, judged by independent juries of industry experts, recognizes infrastructure projects for digital innovations that improve project delivery and/or asset performance. The deadline for nominations is May 21, 2021.

Read Also;
Bentley Systems Announces Winners of the 2020 Year in Infrastructure Awards

According to the information on their official website, The Going Digital Awards are an integral part of Bentley’s annual Year in Infrastructure Conference. The conference brings together infrastructure professionals and industry thought leaders from around the world to share best practices and learn about the latest advances in technology that will improve infrastructure project delivery and asset performance. Winners will be announced during the awards ceremony at the culmination of the conference.

Users of Bentley software are therefore invited to nominate their projects in the Going Digital Awards program, no matter which phase the project is in – planning/conception, design, construction, or operations. The three finalists chosen for each awards category will get a global platform to present their projects before the judges, industry thought leaders and media members.

Every project nominated for an award receives recognition across the global infrastructure community. Through the Going Digital Awards program, participants:

  • Gain global recognition by having their infrastructure projects profiled in Bentley’s Infrastructure Yearbook, which is distributed to media, government, and industry influencers around the world. All winning and finalist projects are also featured on bentley.com.
  • Enhance their competitive edge by demonstrating to existing and potential clients the value the participants add to projects through their digital innovations.
  • Receive coverage from global media and support from the Bentley team in marketing and promoting their respective projects to the media.

The 2021 Going Digital Awards will recognize outstanding achievements for infrastructure projects and assets in the following categories:

  • Bridges
  • Buildings and Campuses
  • Digital Cities
  • Digital Construction
  • Geotechnical Engineering
  • Land and Site Development
  • Manufacturing
  • Mining and Offshore Engineering
  • Power Generation
  • Project Delivery Information Management
  • Rail and Transit
  • Reality Modeling
  • Roads and Highways
  • Road and Rail Asset Performance
  • Structural Engineering
  • Utilities and Communications
  • Utilities and Industrial Asset Performance
  • Water and Wastewater Treatment Plants
  • Water, Wastewater and Stormwater Networks

Question of the day | 22-03-2021

For the section of a frame cut as shown in the figure above;

1. Generate the equation for bending moment along the beam.
(A) Mx = 5x + 27.5 – 10x2
(B) Mx = 27.5x – 5x2 – 15
(C) Mx = 27.5x – 10x2 + 15
(D) Mx = 5x + 27.5x2 + 10x

2. Generate the equation for shear force on the beam.
(A) Qx = 5x + 27.5 – 10x
(B) Qx = 27.5x + 10x
(C) Qx = 5x – 10
(D) Qx = 27.5 – 10x

3. What the axial force on column AB?
(A) 27.5 kN (compression)
(B) 10 kN (Tension)
(C) 5 kN (Compression)
(D) 17.5 kN (Tension)

Internal Stresses in Structures

When a structure is acted upon by a force, it undergoes deformation which increases gradually. During the process of deformation, the material of the structure develops some resistance against the deformation. When the material of the structure takes over the influence of the load, the structure becomes stable. The internal resistance which the body develops against the load is referred to as stress. When the configuration of the body cannot resist deformation, it is called a mechanism and no longer a structure.

Types of stresses:

  1. Direct stress
    (a) Tension (b) Compression (c) Shear
  2. Indirect Stress
    (a) Bending (b) Torsion
  3. Combined stress
    Possible combinations of 1 and 2 above.

Internal Stresses

Let us consider a straight two–force member in tension that is shown in Figure 1 below.

straight bar in axial tension
Fig 1: Axial force in a straight bar

We know that for the member in Figure 1(a) to be at equilibrium, the forces P and –P must be directed along AB in opposite direction and have the same magnitude P. If we cut the member at point C, and if equilibrium must be maintained, we must apply at CA a force P which is equal and opposite to –P, and to CB a force –P which must be equal and opposite to P. Since the member was at equilibrium before the member was cut, internal forces equivalent to these new forces must have existed. This internal force is the axial force in the member.


If we consider the frame that is loaded in Figure 2(a) as shown below;

internal stresses in a frame
Fig 2: Internal stresses in a frame

When we cut a section at point K between member BC (Figure 2a) and consider the free body diagram of section KC (Figure 2c), we will realise that for the body to be at equilibrium, the following conditions must be met;

  1. Application of a force Q at K which is equal and opposite of the force P
  2. Application of the force N at K which is equal and opposite of the force S
  3. Application of a moment M at K to balance the moment of P about K

We again, therefore, conclude that the forces must have existed in the member before the section was cut. The force Q is called the shear force, the force N is called the axial force, and moment M is called the bending moment at point K. The knowledge of these internal stresses is so important that in any given structure, we can use it to determine the critical stress in any section, so as to use it to compare with the permissible stress of the material of the structure before failure occurs. It is also necessary when we compute the deformations at any point in the beam, so as to satisfy the requirements for the effective functionality of the structure (serviceability requirements).

Even though real life structural problems are subjected to various and complex systems of loading, the internal forces induced are usually in form of any of the following:

  1. Shearing forces
  2. Bending moments
  3. Axial forces
  4. Torsion

Shear Forces

When a beam which is in a state equilibrium and subjected to a system of static loading is cut at a section X from the left, it is still expected that the section cut remains at equilibrium, which means that a force must act at the section that is cut. Before the section is cut, this force is provided by the adjacent material in the beam section; hence it acts tangentially to the section from which it is cut.

To ascertain the value of this force, all we need to do is to add up the algebraic value of the forces acting from the left to the point where the section is cut. Internal forces acting tangentially at sections of a beam in equilibrium are known as the shear forces.

Shear force in beams
Fig 3: Illustration of shear force in a beam

The shear force at section x-x of the beam loaded as shown in figure 3(a) is simply given by the summation of all the vertical forces acting just to the left of the section. In this case, it is denoted by Qx which is the force that balances all the external loads acting to the left of the structure;

Qx is given by;
Qx = Ay – P1 – P2

Bending Moment

Bending moment is defined as the rotational tendency of a force. It is basically given by the force, multiplied by the perpendicular distance. If we still consider the equilibrium at the left of the section we cut from the beam, we will still discover that there will be no resultant moment.

In other words, there will be a moment that will balance the bending moments produced by the other forces if the system is to remain at equilibrium. This is the bending moment of the beam at section X and will have the same value, whether you decide to come from the left or the right of the beam system. Always note that moment is maximum where the value of the shear force is equal to zero.

bending moment in beams
Fig 4: Illustration of bending moment in a beam

The bending moment is given by the algebraic sum of the moment that is acting just to the left or to the right of the section. Coming from the left of the structure, it is given by;

Mx = (Ay × x) – P1(x – L1) – P2[x – (L1 + L2)]

The value of bending moment in a structure can be negative or positive. When the value of the bending moment is positive, it is said to be a sagging moment and it implies that the beam is in tension in the lower fibre of the material. Otherwise, when it is negative, it is a hogging moment and the tension zone is in the upper fibre of the material (See Figure 5 below).

sagging and hogging moment
Fig 5: Illustration of sagging and hogging on a beam element

Axial Forces

Axial loads are loads that are applied along the longitudinal or centroidal axis of the member. These loads are common in trusses, columns, and stanchions, and are sometimes accompanied by some rotation and moment due to the eccentricity of the load or from another external load.

cantilever beam subjected to axial pull
Fig 6: Axial pull on a cantilever beam

When the load causes extension in the length of the member, it is called a tensile axial force, and when it causes decrement in length, it is a compressive force. The loads on columns and stanchions are usually compressive.

Torsion

Torsion is very much like a torque that is applied to a beam structure. In other words, torsional stresses produce a kind of twisting effect on the structure. In real-life problems, it is not normally necessary to design for torsion in reinforced concrete structures, since adequate resistance is provided by the nominal shear reinforcements. However, in some structures such as roof gutter or beams supporting canopy slabs, it is necessary to design the beam to resist torsion. In the diagram shown below (Figure 7), the primary beam can be designed to resist torsion T and a point load P.

cantilever beam subjected to torsion
Fig 7: Torsion on a cantilever beam

Relationship between load, bending moment, and shear force

When a beam is subjected to an arbitrary system of loading, its analysis is facilitated if certain relations exist between the applied load, and the internal stresses.

relationship between between bending moment and shear force
Fig 8: Relationship between force, bending moment, and shear force in a beam

Let us consider a beam AB that is carrying a distributed load as shown in figure 8(a). Let CC’ be two points on the beam at a distance ∆x from each other. The shear force and bending moment at C is denoted by Q and M respectively and at point C’ by Q + ∆Q and M + ∆M. Let us now detach points CC’ and draw the free body diagram as shown in figure 8(b).

Let the summation of the vertical forces be equal to zero.
∑Fy = 0

Then we have; Q – (Q + ∆Q) – q∆x = 0
∆Q = – q∆x; Dividing both sides by ∆x and in the limit allowing ∆x to go to zero, we obtain;

dQ/dx = -q —– (1)

Equation (1) indicates that for a beam loaded as shown in figure 8, the slope of dQ/dx is negative. The equation is valid for only distributed loads, for concentrated loads, the formula is not valid. The absolute value of the gradient at any point is equal to the load per unit length.

Integrating equation (1) between points C and B, we obtain;

QB – QC = -∫q dx —— (2)
This implies that the load between B and C is equal to the area under the load curve between points B and C.

When we consider the sum of the moment for the free body diagram at figure 8(b);

Let the summation of the moment about C’ be equal to zero.

∑Mc’ = 0
(M+ ∆M) – M – (Q∆x ) + q∆x(∆x/2) = 0
∆M = Q∆x – q(∆x)2/2 .

Dividing both sides by ∆x and let ∆x go to zero in the limits;

We obtain;
dM/dx = Q —— (3)

Equation (3) shows that the derivative of the bending moment equation yields the equation for the shear force. This is valid for both when concentrated and distributed load is applied on the beam under consideration.

How to Determine the Bearing Capacity of Soils from Plate Load Test

The plate load test or ‘plate bearing test’ is one of the quickest ways of determining the bearing capacity and settlement characteristics of soils on site. This test is essentially useful especially for the design of shallow foundations such as pad footings.

It basically consists of loading a rigid plate at the foundation level and increasing the load in arbitrary increments. The settlement corresponding to each load increment is recorded using at least two or three dial gauges with a least count of 0.02 mm. The gauges should be placed separately at 120° or 90° respectively. The test load is gradually increased till the plate starts to settle at a rapid rate. The load-settlement curve is plotted from which the settlement and bearing capacity of the soil can be determined.

The total value of the load on the plate divided by the area of the steel plate gives the value of the ultimate bearing capacity of soil. A factor of safety is applied to give the safe bearing capacity of soil.

The apparatus required for carrying out a plate load test are;

  • Counterweight such as box or platform with heavy material such as concrete, steel, etc. The total counterweight should be at least 10% greater than the anticipated maximum test load.
  • Hydraulic jack for applying the load
  • Proving ring, 1 kg accuracy, for measuring the load
  • Bearing Plate, 350mm, 450mm, and 600mm diameter
  • Four dial gauges
  • Reference beams.
typical set up of plate load test
Typical plate load test set up (Venkatramaiah, 2006)

The procedure for carrying out plate load test according to BS 1377 part 9 are as follows;

  • A circular plate having a maximum diameter of 300 – 600mm shall be used.
  • Excavate to the test level as quickly as possible to minimise the effects of stress relief, particularly in cohesive fills. A mechanical excavator may be used provided that the excavator bucket does not have teeth and the last 100mm depth of excavation is carried out carefully by hand. If the test is performed in a test pit, the width of the pit should be at least 4 to 5 times of plate diameter.
  • Carefully trim off and remove all loose material and any embedded fragments so that the area for the plate is generally level and as undisturbed as possible.
  • Protect the test area and the apparatus from moisture changes, sunlight, and the effects of adverse weather as soon as the test level is exposed and throughout the test.
  • The plate shall be placed on a thin layer (10 to 15 mm thick) of clean dry sand to produce a level surface on which to bed the plate.
  • Set up the loading and deflection, measuring systems so that the load is applied to the plate without eccentricity and the deflection system is outside the zone of influence of the attachments. During these operations a small seating load may be applied to the plate to enable adjustments to be made: this seating load shall be less than 5 kN/m2.
  • The load shall be applied in five increments. Settlement reading will be taken at 0.50 minute intervals for the first 2 minutes, and 1 minutes intervals thereafter, until the detectable movement of the plate has stopped, i.e. until the average settlement rate is less than 0.02 mm per 5 minute interval.
  • At each increment, the pressure shall be maintained as near as possible constant.
  • After the final test increment has been completed, the pressure in the hydraulic pump shall then be released and the settlement of the plate allowed to recover. When the recovery is essentially complete, the residual settlement value shall be recorded.

According to Venkatramiah (2006), great care shall be taken when interpreting the results from plate load test load-settlement curves. Typical curves obtained from load-settlement curves of plate load tests are shown in the figure below;

typical load settlement curve of plate load tests
Typical load-settlement curves from plate load tests (Venkatramaiah, 2006)

Curve I is typical of dense sand or gravel or stiff clay, wherein general shear failure occurs. The point corresponding to failure is obtained by extrapolating backward (as shown in the figure), as a pronounced departure from the straight-line relationship that applies to the initial stages of loading is observed. (This coincides approximately with the point up to which the range of proportionality extends).

Curve II is typical of loose sand or soft clay, wherein local shear failure occurs. Continuous steepening of the curve is observed and it is rather difficult to pinpoint failure; however, the point where the curve becomes suddenly steep is located and treated as that corresponding to failure.

Curve III is typical of many c – φ soils which exhibit characteristics intermediate between the above two. Here also the failure point is not easy to locate and the same criterion as in the case of Curve II is applied.

Thus, it is seen that, except in a few cases, arbitrary location of failure point becomes inevitable in the interpretation of load test results.

However, it is important to know that the plate load test has some drawbacks such as size effects, and does not take into account the possibility of consolidation settlement, especially in cohesive soils. Furthermore, it is reported that the load test results reflect the characteristics of the soil located only within a depth of about twice the width of the plate.

In this article we are going to show how to make computations from plate load test.

Example
A plate load test was conducted on a uniform deposit of sand at a depth of 1.5m below the natural ground level and the following data were obtained;

Pressure (kPa)050100200300400500
Settlement (mm)024.510173050

The size of the plate was 600 mm × 600 mm and that of the pit 3.0 m × 3.0 m × 1.5 m.
(i) Plot the pressure-settlement curve and determine the failure stress.
(ii) A square footing, 1.5m × 1.5 m, is to be founded at 1.5 m depth in this soil.

Assuming the factor of safety against shear failure as 3.0 and the maximum permissible settlement as 25 mm, determine the allowable bearing pressure.

(iii) Design of footing for a load of 600 kN, if the water table is at a great depth.

Solution
(1) The pressure-settlement curve is shown in the figure below. The failure point is obtained as the point corresponding to the intersection of the initial and final tangents. In this case, the failure pressure is 335 kN/m2.

plate load test settlement curve

The ultimate bearing capacity from the plate load test qult,bp = 335 kN/m2

Applying correction for sandy soil deposit and a footing of width 1.5m;
qult,f = qult,bp x (Width of foundation)/(Size of the base plate) = 335 x (1.5/0.6) = 837.5 kN/m2

Applying a factor of safety of 3.0 against shear failure;
qa = qult,f/FOS = 837.5/3 = 279.16 kN/m2

Alternatively;
Equate the value of qult,bp to 0.5γbpNγ

Where;
bp = size of the base plate = 600 mm
γ = density of soil (say 18.5 kN/m3)
Nγ = Bearing capacity factor (to be determined)

335 = 0.5 x 18.5 x 0.6 x Nγ
On solving, Nγ = 60.36
This reflects to an angle of internal friction (Φ) of about 36.5° using Terzaghi’s theory. The corresponding value of Nq is 50.48.

For a square footing of width (B) and depth (Df) 1.5m founded on sand;

qult = qNq + 0.4γBNγ = (18.5 x 50.48) + (0.4 x 18 x 1.5 x 60.36) = 1585.768 kN/m2
qa = qult/FOS = 1585.768/3 = 528.589 kN/m2

From settlement consideration;

Sp = S[bp(b + 0.3)/b(bp + 0.3)]2
Sp = 25[0.6(1.5 + 0.3)/1.5(0.6 + 0.3)]2 = 16 mm

From the load settlement curve, this settlement corresponds to a pressure of 290 kN/m2

For this particular case study, settlement will govern the design.

The maximum allowable service column load on a 1.5m x 1.5 m square pad footing will therefore be (1.5 x 1.5 x 290) = 652.5 kN. This shows that a column load of 600 kN can be safely supported on a footing of 1.5 m x 1.5m on the soil.

References
(1) BS 1377-9:1990 – Methods for test for soils for civil engineering purposes – In-situ tests. British Standard Institution
(2) Venkatramaiah C. (2006): Geotechnical Engineering (3rd Edition). New Age Publishers, New Delhi, India

The Art and Science of Structural Engineering

Structural engineering is an aspect of civil engineering that is concerned with the conceptualisation, modelling, design, and verification of engineering structures such as buildings, bridges, towers, etc for stability and satisfactory performance under the action of direct and indirect forces. Generally, the stability of man-made structures is the primary responsibility of a structural engineer.

However, the duties of a structural engineer extend way beyond the definition given above. This is because there are so many parts to the development of a structure that goes beyond conceptualisation and design. For instance, a structural engineer is also concerned about the materials to be utilized in a construction project and may go out of his way to develop new products or modify existing ones in order to obtain the desired result.

A structural engineer is also a manager who must take into account the availability of resources, and how best to utilize them in a safe, economical, and efficient manner. Whenever you see an output of a structural engineer, it is not just a sequence of lines and curves on a piece of paper or computer model, but a combination of constructive thinking, data processing, intricate piece of mathematics, physical sciences, environmental sciences, safety considerations, economics, and arts. A structural engineer combines different fields of arts and sciences in order to reach his goal.

typical structural engineering model
A typical structural engineering model of a building

The field of structural engineering is vast, developed, challenging, and interesting. A structural engineer is a highly technical and people-oriented personality who solves problems in a manner that you may not have previously imagined. He understands different structural systems and materials and knows how best to apply them for any given situation. Structural elements such as beams, columns, plates, trusses, shells, arches, domes, cables, etc are combined and produced using different materials such as concrete, steel, timber, aluminum, glass, etc to form a wholistic structural system that can resist loads.

The structure so designed must be able to resist forces coming from their own self-weight, other imposed loads due to storage and occupancy, indirect forces such as temperature difference and sinking of supports, and other environmental loads such as water waves, wind, and earthquake. These loads develop internal forces such as axial tension, compression, bending, shear, twisting, etc in the structural members which may cause collapse or failure. A structural engineer must assess the magnitude of these forces and design the structure to resist failure or collapse as a result of these forces.

internal forces in structures
Typical internal forces in a structure

While the laws of physics and mathematics are required to check that the selected system is stable, it is an art to ensure that the connection of the members, their alignment, and interaction maintains the elegance of the structure in a non-disruptive, economical, and efficient manner. It is also important to ensure that there is a sense of balance in the design and that the final output is aesthetically pleasing. A structural engineering work should not look haphazard or like the product of an afterthought.

A structural engineer is expected to have a detailed knowledge of the design codes of practice in his country, as well as good knowledge of engineering mechanics and structural analysis. In this modern era, the knowledge of different types of engineering design and drafting software such as Staad Pro, ETABS, Tekla, Revit Structures, AUTOCAD etc is very important. Furthermore, one of the most important tools of a structural engineer is experience.

With improvements in the field of material sciences and the quest for more elegant and environmentally friendly buildings, infrastructures, and general development, the intelligence, creativity, and skills of a structural engineer are constantly called upon. There are different kinds of structures for which the services of a structural engineer are required such as high-rise buildings, dams, bridges, earthworks, railways, pipelines, power stations, towers, water retaining structures, earth retaining structures, tunnels, roadways, offshore structures, culverts, transmission lines, reservoirs, etc.

Cable stayed bridges are works of structural engineering
Cable stayed bridges are works of structural engineering

Furthermore, the services of a structural engineer are also required in mechanical structures such as cranes, boilers, pressure vessels, elevators and escalators, carriages, marine vessels, hulls, etc.

A structural engineer’s primary concern is safety, and he must achieve this in an efficient and economical manner. In structures such as buildings and bridges, safety consideration ensures that the structure will not suffer any form of collapse or failure while in service. He must also ensure that the structure will perform satisfactorily by not vibrating excessively, swaying, or cracking when being used by the occupants. Some specialties in the field of structural engineering such as fire engineering, wind engineering, earthquake engineering, etc may be required depending on the peculiarities of the project.

Apart from the performance of the finished structure, a structural engineer is also concerned about the safety and good performance of the structure while it is still under construction. This involves the safety of the workers, ease of executing the design, environmental considerations, level of expertise available, and the cost and/or availability of the materials he is recommending. Generally, he ensures that the cost of the structure remains friendly to the client while satisfying other very important requirements.

incremental launching in bridge construction
Bridge under construction must be stable and safe for workers

In a typical building project, the structural engineer will not work alone but will be involved with other professionals such as architects, geotechnical engineers, surveyors, electrical engineers, interior decorators, mechanical engineers, etc. The technical coordination of these professionals is usually required during the design and construction stage. Therefore, a structural engineer should have good human relations skills.

Moreover, he should be able to communicate effectively using reports, emails, drawings, PowerPoint presentations, computer models, and orally. As a critical thinker whom a lot of persons and the environment depend on for safety, his training and development is vital and must be carefully observed.

A structural engineer is expected to obtain a bachelor’s degree in the field of structural or civil engineering from an approved university or institution of higher learning. He is then expected to work as a pupil or graduate engineer in a reputable engineering firm for a number of years in order to gather practical experience before applying for a ‘chartered’ or ‘registered/professional’ engineer status. This usually requires professional exams, presentations, and interviews from senior engineers and professional bodies. Engineers can be licenced as Civil Engineers, Structural Engineers or both .

In Nigeria, the criteria to be called a ‘structural engineer’ is to be issued a seal or license by COREN as ‘structural engineer’ or to be issued a license as a ‘civil engineer’ and belonging to the Nigerian Institution of Structural Engineers (NIStructE) as a professional member. You can only become a member of NIStructE by attempting and passing the part 3 exam of the professional body.

Some of the biggest structural engineering professional bodies in the world are the Institution of Structural Engineers (IStructE) and the International Association for Bridge and Structural Engineering (IABSE). Belonging to these institutions usually gives someone global recognition as a structural engineer.

In essence, structural engineering is a prestigious profession that has the responsibility of driving the infrastructural development of the world. Their roles in making our universe a better place cannot be overemphasized. Whenever you set your eyes on amazing structures such as skyscrapers, bridges, towers, and other infrastructures, appreciate the efforts of the structural engineer in making such structures safe, stable, and usable.