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Design of Strap Footing | Cantilever Footing

By
Ubani Obinna
-

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 + k1.σcp]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

By
Ubani Obinna
-

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 × ES5950/λLT2 = 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)


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

By
Ubani Obinna
-
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:

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Question of the day | 22-03-2021

By
Ubani Obinna
-

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

By
Ubani Obinna
-

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

By
Ubani Obinna
-

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

By
Ubani Obinna
-

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.

The Cost and Processes of Constructing a Raft Foundation in Nigeria

By
Ubani Obinna
-

Raft foundation is a type of shallow foundation that is provided in areas where the soils have a low bearing capacity, or where high superstructure load is anticipated such that individual pad foundations will overlap. A typical raft foundation is more expensive to construct than an equivalent pad foundation for a given area in a building. The aim of this article is to show the typical cost of constructing a raft foundation in Nigeria.

In the first place, there are different types of raft foundations such as flat raft foundation, beam and slab raft foundation, cellular raft foundation, etc. Each of these types of raft can be employed depending on the circumstances and design specifications, but the most common type of raft foundation employed for duplexes (most residential buildings) in Nigeria is the beam and slab raft foundation because of the favourable economic advantages it presents to the homeowner and the client.

The cost of constructing a raft foundation is influenced by;

  1. the cost of setting out
  2. environmental considerations
  3. the member sizes (depth and width of beams and slabs)
  4. the quantity of reinforcements required
  5. formwork requirements and the complexity of the construction
  6. the volume of excavation required
  7. backfilling the substructure
  8. Other substructure requirements such as compacting, hardcore (if required), damp proof membrane, etc
  9. availability and cost of labour, and
  10. the cost of construction materials in the area

The thickness of the raft slab and the depth and width of the ground beams will determine the volume of concrete required for executing the raft foundation. This is is actually a function of the design specifications, and will significantly affect the overall cost of the construction. Standard design drawings that will be economical and also satisfy all requirements can be provided by Structville Integrated Services Limited (info@structville.com). The cost of executing concrete works in Nigeria varies and is usually influenced by the price of cement, and aggregates.

Reinforcement requirements for a raft foundation are also determined by the design engineer, and the quantity provided will have a significant impact on the cost of the project too. The major cost impact will come from the cost of purchasing, cutting, bending, and installing the reinforcements according to the design drawing. The cost of executing reinforcement works will depend on the type of reinforcement, the quantity required, and the cost of labour.

The carpenters will also charge for the cost of fixing the formwork of the ground beams, and the edge formwork for receiving the raft slab. The cost of the marine boards, planks, 2″ x 3″ wood, props, nails, etc will also be borne by the client.

In areas of high water table, the cost of controlling groundwater so that construction can be carried out in the dry will also affect the cost of the raft foundation. This will also influence the cost of excavation of the trenches to receive the ground beams.

Let us show with an example, the typical cost of constructing the raft foundation of a duplex building in Nigeria. The design drawings are shown below;

typical layout of a raft foundation


The properties of the raft foundation are as follows;

Thickness of the raft slab = 200 mm
Dimensions of the ground beams = 1050 x 230 mm
Dimensions of the columns = 230 x 230 mm

section through raft foundation

The typical ground beam reinforcement details are shown below;

typical ground beam reinforcement details
typical slab reinforcement details

Typical Costing and Processes for Raft Foundation Construction

(1) Setting Out Works
Allow a lump sum of ₦150,000 (Note that this may vary depending on how challenging the setting out process is. The services of a surveyor may be required, and materials like pegs, 2 inches nails, 3 inches nails, 2″ x 3″ hardwood, lines, etc will be required. The professional fee of the carpenters, foreman, supervisors, resident engineers, consulting architects etc may also be required depending on the nature of the contract).

In this case, we are assuming that the contractor is accepting to do the setting out for the price stated above with his team. All professionals in the job have been paid by the client. The contractor is to provide all the materials needed for the setting out.

It very important that the architect and design engineer verify the setting out before excavation can commence. All setbacks and airspace should be confirmed, including the squareness of the building. The deviation of dimensions on the profile board should not exceed 5 mm.

(2) Excavation
From the design drawings and quantity take-off, the following quantities have been verified;
Total length of excavation = 121 m
Depth of excavation = 500 mm
Width of excavation = 700 mm
Volume of excavation = 42.218 m3

Allow for excavation using direct manual labour @ ₦300 per linear metre.
Cost of excavation = 300 x 121 = ₦36,300
Allow 30% to cover for contractors profit and overhead = 1.3 x 36,300 = ₦47,190

(3) Blinding of excavation
The thickness of blinding = 50 mm
The volume of concrete required for blinding = 4.3 m3 (Using M15 concrete)
For mixing, placing, and consolidating grade 15 concrete (cement price at ₦3,500) = ₦30,800/m3
Cost of blinding = 30800 x 4.3 = ₦132,440
Allow 20% to cover for contractor’s profit and overhead = 1.2 x 132,440 = ₦158,928

(4) Reinforcement works – Ground beam
Y20 required = 596.772 kg
Y16 required = 573.177 kg
Y12 required (side bars) = 644.688 kg
Y8 required (links) = 462.655 kg
Y16 required (column starter bars)= 297.8 kg
Y8 required (as links for column starter bars)
Total quantity of reinforcement required for ground beams = 2277.292 kg = 2.278 tonnes
Allow 5% for waste and laps = 1.05 x 2.278 = 2.391 tonnes

Total cost of reinforcements and binding wire = ₦825,120
Cost of labour = ₦60,000
Cost of materials and labour for reinforcement works (ground beam) = ₦885,120
Allow 25% to cover for contractor’s profit and overhead = 1.25 x 885,120 = ₦1,106,400

(5) Formwork – Ground beam
The marine board to be purchased will be reused for the decking of the first-floor slab. Therefore, we will utilise full boards for the ground beam formwork. Marine board can be reused about 5 times before the quality deteriorates. Let us assume that the quantity to be purchased at this stage should be able to do half of the ground beams.

The total area of formwork = 288 m2
Quantity of marine board required for half of the formwork = 55 pieces @ ₦9000/board = ₦495,000
2″ x 3″ softwood required = 380 pieces @ ₦400/length = ₦152,000
3″ Nails reuired = 100 kg @ ₦300/kg = ₦30,000
Labour @ ₦500/m2 = ₦144,000

Cost of ground beam formwork (materials and labour) = ₦821,000
Allow 20% to cover for contractor’s profit and overhead = 1.2 x 821,000 = ₦985,200

(6) Concrete Works – Ground Beam
Volume of concrete required = 23.46 m3
For mixing, placing, and consolidating grade 25 concrete (cement price at ₦3,500) = ₦44,100/m3
Cost of concrete for ground beams = 44,100 x 23.46 = ₦1,034,586
Allow 20% to cover for contractor’s profit and overhead = 1.2 x 1,034,586 = ₦1,241,503

(7) Backfilling and compaction
The volume of filling sand required = 86.775 m3 = 145 tonnes of filling sand
Cost of filling sand = ₦181,250

Cost of labour for filling and compacting (direct manual labour) = @ ₦800/m3 = ₦69,420
Total cost for filling = ₦250,670
Allow 20% to cover for contractor’s profit and overhead = 1.2 x 250,670 = ₦300,804

(8) Levelling and installation of damp proof membrane

damp proof membrane


Allow a lump sum of ₦70,000

(9) Blinding to receive raft slab
Thickness of blinding = 50 mm
The volume of concrete required for blinding = 9.35 m3 (Using M15 concrete)
For mixing, placing, and consolidating grade 15 concrete (cement price at ₦3,500) = ₦30,800/m3
Cost of blinding = 30800 x 9.35 = ₦287,980
Allow 20% to cover for contractor’s profit and overhead = 1.2 x 287,980 = ₦345,576

(10) Raft slab reinforcement works
Y12 – 2922 kg
Y10 – 300 kg
Total quantity of reinforcement required = 3222 kg = 3.222 tonnes
Allow 5% for waste and laps = 1.05 x 3.222 = 3.383 tonnes

Cost of reinforcement and binding wire = ₦1,132,560
Cost of labour = ₦84,575
Cost of materials and labour for reinforcement works (raft slab) = ₦1,217,135
Allow 25% to cover for contractor’s profit and overhead = 1.25 x 1,217,135 = ₦1,521,419

(11) Edge formwork for raft slab
Allow a labour cost of = ₦30,000

(12) Concrete Works – Raft Slab
The volume of concrete required = 37.4 m3
For mixing, placing, and consolidating grade 25 concrete (cement price at ₦3,500) = ₦44,100/m3
Cost of concrete for ground beams = 44,100 x 37.4 = ₦1,649,340
Allow 20% to cover for contractor’s profit and overhead = 1.2 x 1,649,340 = ₦1,979,208

Summary
(1) Setting Out – ₦150,000
(2) Excavation – ₦47,190
(3) Blinding of excavation = ₦158,928
(4) Reinforcement works – Ground beam – ₦1,106,400
(5) Formwork – Ground beam – ₦985,200
(6) Concrete Works – Ground Beam – ₦1,241,503
(7) Backfilling and Compaction – ₦300,804
(8) Levelling and installation of damp proof membrane – ₦70,000
(9) Blinding to receive raft slab – ₦345,576
(10) Raft slab reinforcement works – ₦1,521,419
(11) Edge formwork for raft slab – ₦30,000
(12) Concrete Works – Raft Slab – ₦1,979,208

The total typical cost of constructing a raft foundation in Nigeria = ₦7,936,288

Remember to add 7.5% tax.

For the design, construction, and project management of your building projects in Nigeria, contact;

Structville Integrated Services Limited
E-mail: info@structville.com
Phone: +2348060307054
WhatAapp: +2347053638996

Elastic Deflection of Cantilever Beams Using Vereschagin’s Rule

By
Ubani Obinna
-

Beams deform when loaded. This deformation is the displacement of the beam section from its original position, and it is usually quantified using two parameters known as slope and deflection. When loaded, the neutral axis of the beam becomes a curved line which is referred to as the elastic curve.

The vertical distance between the elastic curve and the original neutral axis of the beam is known as the deflection, while the angle (in radians) that the original neutral axis makes with the elastic curve is known as the slope.

slope and deflection of beams

A cantilever is a beam that is rigidly fixed at one end and free at the other. In structural designs, cantilevers are the most sensitive to serviceability issues such as deflection and vibration. There are many ways of assessing the elastic deflection of cantilever beams such as;

  • Double integration method
  • Moment Area method
  • Virtual work method
  • Conjugate beam method
  • Strain energy method
  • Castigliano’s theorem
  • Finite element method
  • Vereschagin’s method, etc
slope and deflection of cantilevers
Typical deflection of a cantilever beam

The aim of this article is to demonstrate the application of Vereschagin’s rule to the evaluation of the deflection of cantilever beams. Vereschagin’s rule is based on the famous Maxwell-Mohr’s integral, but instead of carrying out the actual integration, the bending moment diagram due to the externally applied load is combined with the bending moment due to a unit concentrated virtual load (graph multiplication) to obtain the deflection at that point. Vereschagin’s rule is the graphical method of Maxwell-Mohr’s integral.

This method also forms the backbone of analysing structures using the force method. The charts for combining different types of bending moment diagrams are available in most standard structural engineering textbooks. In the year 2016, I also wrote an article on how to develop the equations on the charts from the first principle. You can download the article from the link below;

Formulation of diagram combination equations based on Vereschagin’s rule

Let us show an example of how this is done using the cantilever beam that is loaded as shown below.

cantilever beam

Using the Double Integration Method

Let us cut a section at an arbitary point x-x;

section through a cantilever beam

Mx = -5(4 – x)2/2 = -5(4 – x)2/2 —— (1)
EI(∂2y/∂x2) = -Mx = 5(4 – x)2/2 —— (1a)

On integrating equation (1a);
EI(∂y/∂x) = [5(4 – x)3]/6 + C1

At the fixed end A, we know that the slope is equal to zero. Therefore;

At x = 0; ∂y/∂x = 0
⇒ C1 = -53.333

The general equation for the slope of the beam at any point is therefore given by;

EI(∂y/∂x) = [5(4 – x)3]/6 – 53.333 —— (2)

On integrating equation (2), we can obtain the equation for the deflection of the beam at any point;

EI(y) = [-5(4 – x)4]/24 + 53.333x + C2

At the fixed end A, we know that the deflection is equal to zero. Therefore;
At x = 0, y = 0
C2 = -53.333

Hence, the general equation for deflection is;

EI(y) = [-5(4 – x)4]/24 + 53.333x – 53.333 —— (3)

Using equations (2) and (3), the slope and deflection at any point along the beam can be obtained.

At the free end (point C) where x = L = 4 m;

The slope at point C is given by;
EIϴC = [5(4 – 4)3]/6 – 53.333 = -53.333
ϴC = -53.333/EI radians

The deflection at point C is given by;
EIyc = [-5(4 – 4)4]/24 + 53.333x – 53.333 = 53.333(4) – 53.333 = 160
yc = 160/EI metres

Similarly, we can attempt to obtain the deflection at point B using equations (1) and (2);

At point B where x = 3 m;

The slope at point B is given by;
EIϴB = [5(4 – 3)3]/6 – 53.333 = -52.5
ϴB = -52.5/EI radians

The deflection at point B is given by;
EIyB = [-5(4 – 3)4]/24 + 53.333x – 53.333 = -0.208 + 53.333(3) – 53.333 = 106.457
yB = 106.457/EI metres

Using Vereschagin’s Rule (Graphical Method)

Step 1
The first step of the process is to draw the bending moment diagram due to the externally applied load.

A little calculation will show that the maximum moment will occur at the fixed end of the cantilever and it is given by MA = -qL2/2 = (5 × 42)/2 = -40 kNm

BENDING MOMENT DIAGRAM OF A CANTILEVER

Step 2
The second step is to place a vertical unit concentrated load at the point we want to obtain the deflection and draw the bending moment diagram due to the unit load. To obtain the slope at any point, we apply a unit moment (rotation) instead of a unit vertical load.

For instance, if we want to obtain the vertical deflection at point C, we remove the externally applied load and replace it with a unit vertical load at point C as shown below. The bending moment diagram obtained from this step is expected to be linear.

M BAR

Step 3
The next step is to combine the bending moment diagram due to the externally applied load with the bending moment diagram due to the unit load. In principle, this combination involves multiplying the area of the principal bending moment diagram with the ordinate that the linear bending moment diagram makes with the centroid of the principal diagram. This is shown below;

vty


You may not really have to stress yourself determining the area and centroid of different shapes because tables are readily available for most of the diagrams encountered during analysis.

For the shapes shown above, the combination equation is given by;

EIyc = 1/4 × M × Ṁ × L = 1/4 × 40 × 4 × 4 = 160
yc = 160/EI metres

You can confirm that this is similar to the answer obtained using the double integration method.

To obtain the slope at point C, we apply a unit rotation at point C and plot the moment diagram as shown below;

hju

The diagram combination will now be;

b5

From structural engineering tables;
EIϴC = -1/3 x 40 x 4 x 1 = -53.333
ϴC = -53.333/EI radians

As you can see, the Vereschagin’s method is the quickest way of assessing the deflection of statically determinate structures, provided that the combination tables for bending moment diagrams are available.

Optimized Design of Passenger Aerial Ropeway Support Structures

By
Ubani Obinna
-

The use of passenger ropeways (cable car) is a viable alternative for public transportation in urban areas with challenging topography. Aerial ropeway transportation is a type of cable railway mass transit system in which rail cars are hauled from one location to another by the use of moving cables. The design of the supporting structures and the cables for strength and stability is the duty of a structural engineer.

According to researchers from the Institute of Fundamental and Applied Research, Petrovskii Bryansk State University, Bryansk, Russia, the construction of ropeways has a rather high cost and requires taking into account a significant number of restrictions associated with the features of the existing urban development and the placement of urban infrastructure. As a result, they carried out research to develop optimization models that minimize the total cost of modular intermediate height.

typical ropeway transport system
Typical ropeway transport system

The models developed were for discretely variable height and a rope system, and involves the optimal placement and selection of the height of these towers, taking into account the features of the surface topography and urban development.

Furthermore, the proposed modular principle for the construction of intermediate towers is expected to reduce the cost of construction. The study was published in the journal Urban Rail Transit (Springer) in the year 2020.

The cost of constructing a ropeway transit system depends on so many factors such as the location of the line, the ground in the interval between the terminal stations, the parameters of intermediate tower structures, the characteristics of the carrying and traction ropes, etc. During design, these factors can be manageable within certain limits, thereby managing the cost of ropeway construction.

Read Also;
Analysis of moving load on cable-stayed bridges

According to the authors, previous research works have shown that the cost of optimal variants of passenger aerial ropeways is significantly influenced by the height of intermediate towers. Therefore, the optimal design of the ropeway along the surface with a heterogeneous terrain results in the optimal variant requiring the installation of intermediate towers of individual height.

In order to solve the technical and economic problem of optimal ropeway design with intermediate towers of discretely variable height, they advised the use of two optimization models which involve the model of optimization of the installation step of the modular intermediate tower, and the model of optimization of the ropeway in general.

ropeway design parameters
Calculation diagram of the ropeway section between the adjacent intermediate towers (Lagerev and Lagerev, 2020)

After the development of the models, calculations were made for a number of possible variants of the aerial passenger ropeway using the computer program ‘‘Optimization of the ropeway lines with unified towers’’. The results showed that the technical and economic indicators of the optimal variant of installation of unified intermediate towers depend on the step of unification, the cost of the tower itself and the foundation structures, the cost of the process equipment, and the terrain inclination angle.

According to the authors, the developed design method for passenger aerial ropeways, based on minimizing the construction cost, can be recommended for use at the initial stage of a design. The analysis of the terrain along the axis of the ropeway allows one to build a height profile for the installation of intermediate towers, to determine the angles of inclination to the horizon of separate sections of this profile and identify the maximum angle among them.

terrain profile
Typical variants of ropeway lines in Bryansk City (Lagerev and Lagerev, 2020)

It is advisable to use the method when analyzing the following design situations:

• the location of the ropeway line on the ground has already been pre-selected;
• the location of the ropeway line on the ground has not yet been pre-selected, and the designer is considering several alternative options.

To obtain more information about the research findings and the models developed, download the article from the reference below.

Reference
Lagerev A. V. and Lagerev I. A. (2020): Designing Supporting Structures of Passenger Ropeways of Minimum Cost Based on Modular Intermediate Towers of Discretely Variable Height. Urban Rail Transit (2020) 6:265–277 https://doi.org/10.1007/s40864-020-00137-0

Disclaimer
This research article has been reproduced in part on www.structville.com because it is an open-access article licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution, and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made (To view a copy of this license, visit http://creativecommons.org/licenses/by/4.0/.). The contents of this article belong to the original copyright owners.

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