British Verification Problem

7/31/2019 British Verification Problem

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Section 7 British Verification Problems

This section includes discussion on the following topics:

British Verification Problem 1





7.1 British Verification Problem 1

Reference

Reinforced Concrete by T.J.Macgingley & B.S.Choo, Page 333 and Example: 11.1.

Problem

A column 400mm X 400mm carries a dead load of 800 kN and an imposed load of 300 kN. The safe

bearing pressure is 200 kN/m2. Design a square base to resist the loads. The concrete is grade 35 andthe reinforcement is grade 460.

Solution

Figure 7.1

Page 1 of 347 British Verification Problems

11/05/2012file://C:\Documents and Settings\albert\Local Settings\Temp\~hhB594.htm


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Figure 7.2

Figure 7.3

Size of base:

Self-weight of footing = 2.5 x 2.5 x 0.5 x 25 = 78.125 kN.

Therefore, Service load = Dead load + Imposed load + Self weight

= (800 + 300 + 78.125) kN

= 1178.125 kN.




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Area required = 1178.125 / 200 m2 = 5.890625 m2.

Make the base 2.5 m x 2.5 m.

Moment Steel:

Ultimate load = (1.4 x 800) + (1.6 x 300) = 1600 kN.

Ultimate pressure = 1600 / (2.5 x 2.5) = 256 kNm2

The critical section YY at the column face is shown in Figure 6.1.

MYY= 256 x (2.5 / 2 - 0.4 / 2) x 2.5 x 0.525 = 352.8 kNm.

Try an overall depth of 500 mm with 20 mm bars.

Effective depth = 500 40 20 10 = 430 mm.

Therefore z = 0.95d,

.

= 1976.31 mm2.




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Minimum area of steel = 0.0015 x B x d = 0.0012 x 2500 x 430 = 1625 mm2 < AS (Hence Safe)

Let us provide 10 nos. 16 mm bars, AS = 2010.62 mm2.

One Way Shear:

The critical section Y1 Y1 at d = 430 mm from the face of the column is shown in Figure 6.2.

Design shear force, VU = 256 x (2.5 /2 0.43 0.4 / 2) x 2.5 = 396.8 kN

Design shear stress, v = =0.369 N/mm2

Now, vC1 = min(0.8 ,5) = 4.7328 N/mm2

> v (Hence Safe)

= 0.395 N / mm2 > v (Hence Safe)

Hence no shear reinforcement is required.

Punching Shear:

Punching shear is checked on a perimeter 1.5d = 625.5 mm from the column face. The critical




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perimeter is shown in Figure 6.3.

Perimeter = 1690 x 4 = 6760 mm.

Shear = 256 x (2.52 1.692) = 868.8 kN.

v = = 0.3 N / mm2 < VC (Hence Safe).


Spacing:

We provided 10 nos. 16 mm bars, AS = 2010.62 mm2.

Spacing = = 267.11 mm.

Comparison

Table 7.1


Reference

Reinforced Concrete by T.J.Macgingley & B.S.Choo, Page 340 and Example: 11.2.

Value Of ReferenceResult

STAAD.foundationResult

Difference inPercent

Effective Depth 430 mm 430 mm None

Governing Moment

352.8 KN-m

352.8 KN-m

None

Area of Steel 1976.31 1976.31 NoneShear Stress (One-

Way) 0.369 N/mm2 0.369 N/mm2 None

Shear Stress (Two-Way) 0.3 N/mm

2 0.3 N/mm2 None




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Problem

The characteristic loads for an internal column footing in a building are given in the following table.The proposed dimensions for the column and base are shown in Figure 6.4. The safe bearing

pressure of soil is 150 kN / m2. The materials to be used in the foundation are grade 35 concrete andgrade 460 reinforcement.

Table 7.2

Figure 7.4

Solution

Self-weight of footing = 0.5 x 3.6 x 2.8 x 24 = 120.96 kN.

Total axial load = 770 + 330 + 120.96 = 1220.96 kN.

Total moment = 78 + 34 = 112 kN-m.

Vertical Load (kN) Moment (kN m)

Dead Load 770 78

Imposed Load 330 34




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Base area = 2.8 x 3.6 = 10.08 m2.

Section modulus = (I / y) = = 6.048 m3.

Maximum pressure = = 139.65 kN / m2 < 150 kN / m2 (Hence Safe).

Factored axial load = (1.4 x 770) + (1.6 x 330) = 1606 kN.

Factored moment = (1.4 x 78) + (1.6 x 34) = 163.6 kN-m.

Maximum pressure = = 186.38 kN / m2.

Minimum pressure = = 132.28 kN / m2.

Calculation of Reinforcement Along Shorter Span (X1-X1):

Average pressure for section X1X1 (as shown in Figure 6.5) = 159.33 kN / m2.

Moment (MY) = (159.33 x 1.175 x 3.6) x (1.175 / 2) = 395.955 kN-m.

Effective depth (d) = 500 40 10 = 450 mm.




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The minimum area of steel = 0.13 x 3600 x 500 / 100 = 2340 mm2 > calculated area of steel.

Provide minimum steel.




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Figure 7.5

Calculation of Reinforcement Along Longer Span (Y1-Y1):

Pressure at section Y1 Y1 (as shown in Figure 6.5) = 162.7 kN / m2.

Moment(MX) = (162.7 x 2.8 x 1.575)x(1.575 / 2)+(0.5 x 1.575 x (186.38 162.7) x 2.8)x(2 / 3 x

1.575) = 619.862 kN-m.

Effective depth (d) = 500 40 20 10 = 430 mm.

The minimum area of steel = 0.13 x 2800 x 500 / 100 = 1820 mm2 v (Hence Safe)

Let us consider 1.5 times shear enhancement.

Vce = 1.5 x 429.6 = 644.4 kN/m2 > v (Hence safe)





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Figure 7.6

Along Section X2-X2:

The critical section X2X2 at d = 450 mm from the face of the column is shown in Figure 6.6.

Average pressure for the required section = 159.33 kN / m2.

Design shear force, VU = 159.33 x 3.6 x 7.25 = 415.85 kN

Design shear stress, v = =256.698 kN / m2

Now, vC1 = min(0.8 ,5) = 4732.8 kN / m2 > v (Hence Safe)




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= 409.6 kN / m2.

Let us consider 1.5 times shear enhancement.

Vce = 1.5 x 409.6 = 614.4 kN/m2 > v (Hence safe)


Punching Shear:




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Figure 7.7

The punching shear will be calculated for an area outside the area enclosed by the rectangle at adistance 1.5d from the column face as shown in Figure 6.7.

Total pressure under the base = 2.8 x 3.6 x 132.28 + 0.5 x 3.6 x 2.8 x (186.38 132.28) = 1606.05kN.

Pressure under enclosed rectangle = (1.74)2 x 146.255 + 0.5 x (1.74)2 x (172.4 146.255) = 482.38kN

Punching shear force = 1606.05 482.38 = 1123.67 kN.

Critical perimeter = 1.74 x 4 = 6.96 m.

Punching shear stress = =375.46 kN / m2.

Comparison




Effective Depth (X-X) 430 mm 430 mm None

Effective Depth (Y-Y) 450 mm 450 mm None

Governing Moment

(My)395.955 KN-m 395.943 KN-m None

Governing Moment

(Mx)619.862 KN-m 619.909 kN-m None

Area of Steel(Along X-X) 2340.00 2340.00 None

Area of Steel(Along Y-Y) 3472.2334 3472.2334 None

Shear Stress (One-

Way) (Y1-Y1)473.388 kN/m2 473.37 kN/m2 None

Shear Stress (One-

Way) (X1-X1)256.698 kN/m2 256.698 kN/m2 None

Shear Stress (Two-Way) 375.46 kN/m

2 375.44 kN/m2 None




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Table 7.3


Reference

Reinforced Concrete by T.J.Macgingley & B.S.Choo, Page 366 and Example: 11.6

Problem

Design a four pile cap to support a factored axial column load of 3600 kN. The column is 400 mmsquare and the piles are 400 mm in diameter spaced at 1200 mm centers. The materials are grade 35concrete and grade 460 reinforcement.

Figure 7.8




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Solution

Pile reaction = 3600 / 4 = 900 kN.

Calculation of Bending Moment and Area of Steel:

Figure 7.9

Bending moment at the face of the column = 2 x 900 x 0.4 = 720 kN-m.

Let us assume effective depth as 598 mm.




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The minimum area of steel = 0.13 x 2200 x 598 / 100 = 1710 mm2 < calculated area of steel.

Calculation of One Way Shear:

Figure 7.10

Shear force at a section aV distance away from the face of the column (as shown in Figure 6.10)

Design shear stress, v = =1368.197 kN / m2




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Now = 280 mm.

Enhanced shear VCE = = 1368.29 kN/m

2

> v (Hence Safe)

Calculation of Punching Shear:

Figure 7.11

Here, pile spacing


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Design shear stress vC = =3762.542 kN/m2.

Allowable shear stress = =4732.874 kN/m2 > vC (Hence safe).

Comparison

Table 7.4


Reference

Reinforced Concrete Design by S.N.Sinha, Problem 11-13.

Problem

Design a pile cap with the given data: Load FY = 2000KN, MY = 300 KN-m, Spacing = 900 mm,

Pile in Pile Cap = 75 mm, Bottom Cover = 100 mm, Edge Distance = 275 mm, No. of Pile = 9, fc =

15 MPa, fy = 415 MPa, Column Dimension = 500mm x 500mm, Ultimate Load Factor = 1.5.




Effective Depth 598 mm 598 mm None

Governing Moment 720 KN-m 720 KN-m None

Area of Steel 2900.188 mm2 2900.188 mm2 None

Shear Stress (One-Way) 1368.197 kN/m2 1368.197 kN/m2 NoneShear Stress (Two-

Way)3762.542

kN/mm2 3762.542 kN/mm2 None




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Figure 7.12

Solution

Pile reactions are calculated as follow,

P1 = = 416.667 kN,

P2 = = 416.667 kN,




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P3 = = 416.667 kN,

Calculation of Moment and Reinforcement Along Length (Y1-Y1):




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Figure 7.13

The critical bending moment will act at the face of the column.

Hence MYY = 3 x 416.667 x 0.65 = 812.5 kN-m.


The minimum area of steel = 0.13 x 2350 x 928 / 100 = 2835.04 mm2 > calculated area of steel.

Provide minimum area of steel.

Calculation of Moment and Reinforcement Along Width (X1-X

1):

The critical bending moment will act at the face of the column.

Hence MYY = (416.667 + 333.333 + 250) x 0.65 = 650 kN-m.





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The minimum area of steel = 0.13 x 2350 x 928 / 100 = 2835.04 mm2 > Calculated area of steel.

Provide minimum area of steel.

Calculation of One Way Shear (for section X1-X1 & X2-X2):

aV




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Figure 7.14

Shear force at a section aV

distance away from the face of the column (as shown in Figure 6.14)

Now force on the section X1-X1 = 3 x 416.667 = 1250 kN.

And force on the section X2-X2 = 3 x 250 = 750 kN.

Hence design shear stress, v = = 573.18 kN / m2




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Calculation of One Way Shear (for section Y1-Y1 & Y2-Y2):




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Figure 7.15

Shear force at a section aV

distance away from the face of the column (as shown in Figure 6.15)

Now force on the section Y1-Y1 = (416.667 + 333.333 + 250) = 1000 kN.

And force on the section Y2-Y2 = (416.667 + 333.333 + 250) = 1000 kN.

Hence design shear stress, v = = 458.547 kN / m2




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Calculation of Punching Shear:

Figure 7.16

Here, pile spacing > 3 x pile diameter.

Hence, design shear force at the face of the column=(3 x 416.667 + 2 x 333.333 + 3 x 250)=2666.667 kN,




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Hence we will check for punching shear at the crtical perimeter as shown in Figure 6.16.

(BS 8110:Part1:1997,Clause 3.11.4.5)

Critical perimeter = 4 x (2 x aV +column dimension) = 4 x (2 x 575 + 500) = 6600 mm.

Design shear stress vC = =435.388 kN/m2.

Comparison

Value Of

Reference

Result

STAAD.foundation

Result

Difference in

PercentEffective Depth (X-

X) 928 mm 928 mm None

Effective Depth (Y-Y) 928 mm 928 mm None

Governing Moment

(My)812.503 KN-m 812.5 kN-m m None

Governing Moment

(Mx)650 KN-m 650 KN-m None

Area of Steel(Along X-X) 2835.04 2835.04 None

Area of Steel(Along Y-Y) 2835.04 2835.04 None

Shear Stress (One-Way) (Y1-Y1)

458.547 kN/m2 458.547 kN/m2 None

Shear Stress (One-573.18 kN/m2 573.186 kN/m2 None




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Table 7.5

7.5 British Verification Problem 5Reference

Reinforced Concrete by T.J.Macgingley & B.S.Choo, Page 351 and Example: 11.4

Problem

Design a rectangular base to support two columns carrying the following loads:

Column 1 Dead load 310 kN, imposed load 160 kN

Column 1 Dead load 310 kN, imposed load 160 kN

The columns are each 350 mm square and are spaced at 2.5 m centers. The width of the base is not to

exceed 2.0 m. the safe bearing pressure on the ground is 180 kN/m2. The materials are grade 35concrete and grade 460 reinforcement.

Figure 7.17

Way) (X1-X1)

Shear Stress (Two-Way) 435.388 kN/m

2 435.388 kN/m2 None




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Solution

Let us assume the self weight of the base is 130 kN.

Total vertical load = (310 + 160 + 430 + 220 + 130) = 1250 kN.

Area of base = 1250 / 160 = 7.81 m2 (considering safe base pressure as 160 kN/m2).

Length of base = 7.81/2 = 3.91 m

Let the dimension of the mat is as follows,

Width = 2 m,

Length = 4.5 m,

Depth = 0.6 m.

Hence self-weight of mat = 2 x 4.5 x 0.6 x 24 = 129.6 kN.

Hence, total vertical load = (310 + 160 + 430 + 220 + 129.6) = 1249.6 kN.

Actual base pressure = 1249.6 / (2 x 4.5) = 138.84 kN/m2

.

Load Case I (DL & IL on both the column):

The ultimate loads are,

Column 1 load = 1.4 x 310 + 1.6 x 160 = 690 kN,

Column 2 load = 1.4 x 430 + 1.6 x 220 = 954 kN.




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The distance of center of gravity from column 1 is checked for service load case 1:

x = (954 x 2.5)/(690 + 954) = 1.45 m.

The soil pressure is checked for service loads for case 1:

Base area = 4.5 x 2 =9.0 m2,

Base modulus = 2 x 4.52 / 6 = 6.75 m3.

Direct load = 310 + 160 + 430 + 220 + 129.6 = 1249.6 kN.

The moment about the centerline of the base is,

M = (430 + 220) 1.05 (310 + 160) 1.45 = 1.0 kN-m.

Maximum pressure = 1249.6 / 9 + 1 / 6.75 = 138.9 kN / m2 < 180 kN / m2 (Hence safe)

Total uniformly distributed upward load = (690 + 954) / (4.5 x 2 / 2) = 365.33 kN / m.

Load Case II (DL & IL on column 1; DL on column 2):


Column 1 load = 1.4 x 310 + 1.6 x 160 = 690 kN,

Column 2 load = 1.0 x 430 = 430 kN.

Direct load = 310 + 160 + 430 + 129.6 = 1029.6 kN.


M = 430 X 1.54 (310 + 160) 0.96 = 230 kN-m.




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Load Case III (DL on column 1; DL & IL on column 2):


Column 1 load = 1.0 x 310 = 310 kN,

Column 2 load = 1.4 x 430 + 1.6 x 220 = 954 kN.

Direct load = 310 + 430 + 220 + 129.6 = 1089.6 kN.


M = (430 + 220) 1.05 310 X 1.45 = 233 kN-m.





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Figure 7.18

Design of Longitudinal (Bottom) Steel:

The maximum negative BM from Figure 6.18 is 221.7 kN-m

Assuming 40 mm clear cover and 20 mm bar diameter, effective depth = 600 40 20/2 = 550 mm.




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Provide minimum steel.

Provide 16 bars 12 mm in diameter at 125 mm centers to give area of 1808 mm2.

Design of Transverse (Top) Steel:

The maximum positive BM from Figure 6.18 is 99.7 kN-m.

Assuming 40 mm clear cover and 20 mm bar diameter, effective depth = 600 40 20/2 = 550 mm.




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Provide minimum steel as above.

Calculation of Vertical Shear:

The maximum vertical shear from case 1 is V = 250.8 kN,

V = = 0.228 N/mm2.

Punching Shear Check:

The critical perimeter for punching shear check is at 1.5 d distance from the face of the column. Here

the perimeter crosses the boundary of the base on two sides. Hence punching shear is less criticalthan the vertical shear in this case.

Comparison




Max BendingMoment (-) 221.7 kN-m 201.204 kN-m 9.2

Max BendingMoment (+) 99.7 kN-m 108.94 kN-m 9.2

Area of StealRequired 780 mm

2/m 780 mm2/m None

Base Pressure 138.84 kN/m2 136 kN/m2 2


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