507 39 Solutions Instructor Manual Chapter 3

Design of RC Structures Dr. N. Subramanian

© Oxford University Press 2013. All rights reserved.

SOLUTIONS MANUAL

CHAPTER 3

EXERCISE 3.1

Measurement of loads on a floor slabs of residential buildings are given below in kN/m2

8 x 0.90, 12 x 1.1, 15 x 1.2, 30 x 1.4, 40 x 1.5, 15 x 1.6, 5 x 1.7

(8 x 0.90 means eight samples of 0.9 kN/m2 each)

Determine the characteristic load on the floors if acceptable probability of load not to exceed 5% of the specified

load.

Solution

The measured live loads are denoted by Qvi

Qvi = 8 x 0.90, 12 x 1.1, 15 x 1.2, 30 x 1.4, 40 x 1.5, 15 x 1.6, 5 x 1.7

Total number of samples, n = 8 + 12 + 15 + 30 + 40 + 15 + 5 = 125

The mean value, Qvm = 1.3832 kN/m2

Standard deviation = 0. 2027

Coefficient of variation Cv = / Qvm = 0.1465

(The above calculations can be obtained by using a scientific calculator)

The characteristic Live load, Qvk = Qvm + 1.65

= 1.3832 + 1.65 x 0.2027 = 1.717 kN/m2

EXERCISE 3.2

An office building was designed to resist a floor load of 4 kN/m2. After the building was constructed, measurements,

were taken on the various floors as

12 x 3.2, 10 x 3.8, 20 x 4, 15 x 4.2, 8 x 4.4

Determine the probability of the loads exceeding the specified load of 4 kN/m2

Hint: After calculating [(Q – Qm)/), the probability of exceeding the specified load (Q) can be found out by

referring to the statistical table for normal distribution, given in any standard book on Statistics.

Solution:

The measured live loads are denoted by Qvi

Qvi = 12 x 3.2, 10 x 3.8, 20 x 4, 15 x 4.2, 8 x 4.4

Total number of samples, n = 12 + 10 + 20 + 15 + 8 = 65

The mean value, Qvm = 3.917 kN/m2

Standard deviation = 0.3825

Coefficient of variation Cv = / Qvm = 0.098

(The above calculations can be obtained by using a scientific calculator)

The characteristic Live load, Qvk = Qvm + 1.65

= 3.917 + 1.65 x 0.3825 = 4.548 kN/m2

Thus, the characteristic value that should have been used in the design is 4.548 or 4.55 kN/m2, which is greater than

that assumed in the design.

The probability of the load exceeding the specified load of 4 kN/m2 :

Z value =

Referring to the statistical table for normal distribution, the probability of exceeding the specified load (Q) is 0.5843

or 58.43%



Example 3.3

A six storey building is to be used for residential purposes. Calculate the load on an interior column in the ground

floor, assuming that the columns are placed in a grid of 6 m x 4 m. Consider live load reduction as per IS 875 – part

3.

Fig. 3.1 Six storey building of Exercise 3.3

Solution

Assuming height of each storey as 3m,

Live load on each floor = 2000 N/m2

Live load on roof, with access = 1500 N/m2

Assuming slab of thickness 125 mm, dead load = 25 kN/m3 x 0.125 m = 3125 N/m

2

Dead load due to floor finish (assumed) = 1000 N/m2

Total dead load = 4125 N/m

2

Loads from the various floor levels are computed as below in Table 3.1. The live load has been reduced as per IS

875 (part 2).

Table 3.1 Imposed load reduction for 6 storey building

Column Floor Live load N/m2 Dead load N/m

2 Total load from

Floor N/m2

GF Roof 1500 4125 5625

FE 6th

floor 0.9 x 2000 =1800 4125 5925

ED 5th

floor 0.8 x 2000 =1600 4125 5725

DC 4th

floor 0.7 x 2000 =1400 4125 5525

CB 3rd

floor 0.6 x 2000 =1200 4125 5325

BA 2nd

floor 0.6 x 2000 =1200 4125 5325

Total load on column 33,450

Design load on column AB = 1.5(33,450) x 4 x (6+6)/2

= 1,204,200/1000 = 1204.2 kN

Note that the dead load on the roof in a real structure may be more due to the type of weathering course adopted.

EXERCISE 3.4

A tall building is proposed in Mumbai where there are some existing tall buildings. Use the following data:

a. Level ground



b. Design for a return period of 50 years

c. Basic wind speed = 44 m/sec

d. Size of the building = 30 m x 40 m and height = 60 m

Estimate the risk, topography, terrain coefficients and compute design wind speed and pressure.

Solution:

Basic wind speed (as per Appendix A of IS 875 (Part 3):1987), Vb = 44 m/sec

Design wind speed, Vz = Vbk1k2k3

From Table 1 of IS 875 (Part 3):1987, with assumed design life of 100 years,

k1 = 1.07

The value of k2 is obtained from Table 2 of IS 875 (Part 3):1987. With Category 3 and height > 50 m (Class C), we

get the values of k2 as shown in Table 3.2.

As it is level ground, k3 =1.0

The wind speed, Vz, and wind pressure, pz = 0.6 (Vz)2, are calculated at various heights and are shown in Table 3.2

Table 3.2 Design wind speed and pressure at different heights of the building

Height k2 Vz (m/sec) pz (N/m2)

0-10 m 0.82 38.61 894.4

10-15 m 0.87 40.96 1006.6

15-20 m 0.91 42.84 1101.2

20-30 m 0.96 45.20 1225.8

30-50 m 1.02 48.02 1383.5

50-60 m 1.04 48.96 1438.2

EXERCISE 3.5

Compute the design wind pressure and design forces on walls and roofs of a two storey building having a height of

6.5 m and size of 10 m x 30 m. Assume there are 6 openings on each floor of size 1.2 m x 1.2 m in the wall of length

30 m and 2 similar openings in each floor in the 10 m length wall. The building has a flat roof and supported on load

bearing walls (see Fig. 3.2).

Fig. 3.2

Solution

Let us assume that the building is in Bangalore and has a design life of 100 years. Also assume that the terrain is

category 2.

Basic wind speed in Bangalore [from wind zone map or Appendix A of IS 875 (Part 3)]

Vb = 33 m/s

As the building is to be designed for a 100 year life, the risk coefficient from Table 1 of IS 875 (Part 3) is

k1 = 1.05

The terrain factor for category 2 and class B (Width < 30 m) from Table 2 of IS 875 (Part 3), for a height of 6.5m is

k2 = 0.98

The ground is assumed to be plain, hence the topography factor is:

k3 = 1



Design wind speed = Vz = Vbk1k2k3

= 33 x 1.05 x 0.98 x 1 = 33.96 m/s

Wind pressure pz = 0.6 Vz2 = 0.6(33.96)

2

= 692 N/m2 = 0.692 kN/m

2

Permeability of the building:

Area of the walls = 6.5(2 x 10 + 2 x 30) = 520 m2

Area of all the openings = 32 x 1.2 x 1.2= 46.08 m2

% opening area = 8.86%, between 5% and 20%. Hence the building is of medium permeability.

Wind load calculations

F = (Cpe – Cpi) A pd

Internal pressure coefficient (Table 3.6)

Cpi = ±0.5

External pressure coefficient

On Roof: Using Table 5 of IS 875 (Part 3), with roof angle 0o without local coefficients, for h/w = 0.217, the

following coefficients are obtained (see Table 3.3).

Table 3.3

Portion of

roof

Wind incidence Angle

0o 90

o

E -0.8 -0.8

F -0.8 -0.4

G -0.4 -0.8

H -0.4 -0.4

Design pressure coefficients for walls

For h/w = 0.217 and l/w = 3, Cpe for walls, using Table 4 of IS 875 (Part 3), we get the coefficients as shown in

Table 3.4.

Table 3.4

Wall Wind Incidence Angle

0o 90

o

Wall A +0.7 -0.5

Wall B -0.25 -0.5

Wall C -0.6 +0.7

Wall D -0.6 -0.1

The above values have to be combined with the internal pressure coefficients Cpi = ± 0.5

Thus

Cpnet for walls A or B

= 0.7 – (– 0.5) = +1.2 pressure

= –0.5 – (+ 0.5) = –1.0 suction

Cpnet for walls C or D (See Fig. 3.3)

= 0.7 – (– 0.5) = + 1.2 pressure

= – 0.6 – (+ 0.5) = – 1.1 suction



Fig.3.3 Net roof pressure coefficients for different zones and combinations

Design pressure for walls

For long walls, F = Cpnet x pd

F = 1.2 x 0.692 = 0.830 kN/m2 pressure

= –1 x 0.692 = –0.692 kN/m2 suction

For short walls

F = 1.2 x 0.692 = 0.830 kN/m2 pressure

= –1.1 x 0.692 = –0.692 kN/m2 suction

For roof

F = 1.3 x 0.692 = 0.90 kN/m2 pressure

= –0.1 x 0.692 = –0.07 kN/m2 suction

Calculation of Force due to Frictional Drag

Since 30/6.5 = 4.6 > 4.0 (even though 30/10 = 3.0), the frictional drag due to wind has to be considered. This will

act in the longitudinal direction of the building along the wind. Here h < b, and hence as per Clause 6.3.1 of IS 875

(Part 3), we get

( )

( )

= 0.01 (30 – 4 x 6.5) 10 x 0.692 + 0.01 (30– 4 x 6.5) 2 x 6.5 x 0.692

= 0.2768 + 0.3598 = 0.637 kN/m2

This above frictional drag will act along the longer side of the building.

EXERCISE 3.6

Consider a four storey office building shown below, located in Shillong. (seismic zone V). The soil condition is

medium stiff and the entire building is supported on raft foundation. The RC frames are infilled with brick masonry.

The lumped weight due to dead loads is 12 kN/m2 on floors and 10 kN/m

2 on roof. The floors carry a live load of 4

kN/m2 and roof 1.5 kN/m

2. Determine design seismic load on the structure by the equivalent static method. Assume

that the frames are moment resisting frames with R = 5 .



Fig. 3.4 Frame of Exercise 3.5

Solution:

Depth of building ( in X Direction)= 4 6 = 24 m

Depth of building ( in Y Direction)= 2 4.5 + 6 = 15 m

h = 4.2 + 3 3.2 = 13.8 m

Total dead load of the structure = WDL = (12 )( ) + 10 = 16560 kN

Total live load of the structure to be considered for earthquake, as per Clause 7.3.1 and 7.3.2 of IS 1893(Part 1):2002

[50% of L.L. (for L.L. > 3 kN/m2) on floors and zero L.L. on roof] = WLL = (4 )( ) = 2160 kN

Total seismic weight = 2160 + 16560 = 18,720 kN

For seismic zone V, the zone factor is 0.36 [Table 2 of IS 1893(Part-1)]. Being an office building the importance

factor is 1.0 (Table 6 of IS 1893). Response Reduction Factor, R (given) =5.

Fundamental Period: The lateral load resistance is provided by moment resisting frames infilled with brick masonry

panels. Hence for EL in X direction,

(clause 7.6.2 of IS 1893) 0.254 sec

From Fig.2 of IS 1893 for T =0.254 sec, Sa/g = 3.5

Ah=ZI(Sa/g)/(2R) (clause 6.4.2 of IS 1893)

= 0.36 x 1 x 2.5 / (2 x 5) = 0.09

Design Base shear VB = AhW

= 0.09 x 18720 = 1684.8 kN

EL in Y direction

(clause 7.6.2 of IS 1893) 0.32 sec

Therefore Sa/g = 2.5 and Ah= 0.09

Hence for this building, the design seismic force in Y direction is same as that in the X direction

EXERCISE 3.7

A three storey building and the seismic weights acting on it are shown below. Assuming that the building is in

seismic zone IV and supported by soft soil determine the design seismic load on the structure by

a. Equivalent static method



b. Response spectrum method

Fig. 3.

The free vibration properties of this building are shown in Table 3.5

Table 3.5

Natural period (see) Mode 1 Mode 2 Mode 3

0.114 0.040 0.021

Mode shapes

Roof 1.000 1.000 1.000

2nd

Floor 0.726 -0.583 -3.377

1st Floor 0.340 -1.146 1.378

Solution

a. Equivalent static method

For seismic zone IV, the zone factor is 0.24 [Table 2 of IS 1893(Part-1)]. Assuming office building, the importance

factor is 1.0 (Table 6 of IS 1893). Assuming special moment resisting frame, as per Table 7 of IS 1893, R=5.

Total seismic weight of the structure = Wi = 355 + 355 + 415 = 1125 kN

h = 4.875 + 3.65 2 = 12.175 m

Assume depth of building in the other direction as 10 m.

Fundamental Period: The lateral load resistance is provided by moment resisting frames infilled with brick masonry

panels. Hence for EL in X direction,

(clause 7.6.2 of IS 1893) 0.347 sec

From Fig.2 of IS 1893 for T =0.23 sec, Sa/g = 2.5

Ah=ZI(Sa/g)/(2R) (clause 6.4.2 of IS 1893)

= 0.24 x 1 x 2.5 / (2 x 5) = 0.06

Design Base shear VB = AhW

= 0.06 x 1125 = 67.5 kN

Force distribution with building height: The design base shear is distributed with height as per clause 7.7.1 and the

relevant calculations are shown in Table 3.6.

Table 3.6 Lateral load distribution as per static method



Storey

level

Wi ,kN hi ,m Wihi2 (Wihi

2)/(Wihi

2) Lateral force at i-th

level for EL in direction,

kN

X Y

3 355 12.175 52,622 0.596 40.23 40.23

2 355 8.525 25,800 0.292 19.71 19.71

1 415 4.875 9863 0.112 7.56 7.56

1125 - 88,285 1.000 67.5 67.5

EL in Y direction (clause 7.6.2 of IS 1893) 0.496 sec

Therefore Sa/g = 2.5 (soft soil) and Ah= 0.09

Hence for this building, the design seismic force in Y direction is the same as that in the X direction

b. Response spectrum method

Table 3.7 illustrates the calculation of modal mass and modal participation factors. It is seen that the first two modes

exite 94.93% of the total mass, which is greater than the 90% specified by the code.

Table 3.7 Calculation of modal mass and modal participation factor (Clause 7.8.4.5 of IS 1893)

Storey

Level

Weight

(kN)

Mode 1 Mode 2

ϕi Wi ϕi Wi(ϕi)2 ϕi Wi ϕi Wi(ϕi)

2

3 355 1.000 355 355 1.000 355 355

2 355 0.726 257.7 187.1 -0.583 -207 120.7

1 415 0.340 141.1 48 -1.146 -475.6 545

1125 - 753.8 590.1 - -327.6 1020.7

( )

963 kN/g = 98165 kg 105 kN/g=10718 kg

Percentage of total weight 85.6 9.33

1.277 -0.321

The lateral load Qik acting at ith floor in the kth mode is

Qik=Ah ik Pk Wi

With soft soil,

Mode 1

T1 = 0.114 sec

Sa/g = –2.50 (clause 6.4.5. of IS: 1893)

Ah= Ah1=ZI (Sa/g)/(2R)

= 0.24 x 1 x 2.5 /(2 x 5)



= 0.06

Qi1 = 0.06 x ik x 1.277 x Wi

= 0.07662 ikWi

Mode 2

T2 = 0.04 sec

Sa/g = 1.6

Ah= Ah2=ZI (Sa/g)/(2R)

= 0.24 x 1 x 1.6 /(2 x 5)

= 0.0384

Qi2 = 0.0384 x ik x (-0.321) x Wi

= -0.0123 ikWi

The storey shear force Vik = Qik is calculated in Table 3.8

Table 3.8 Lateral load calculation by modal analysis method (earthquake in X direction)

Floor

level

Weight

Wi, kN

Mode1 Mode 2

i1 Qi1 Vi1 i2 Qi2 Vi2

3 355 1.000 27.2 27.2 1.000 -4.37 -4.37

2 355 0.726 19.75 46.75 -0.583 2.55 -1.82

1 415 0.340 11.0 57.75 -1.146 5.85 4.03

Since the two modes are well separated (see clause 3.2 of IS:1893) the contribution of different modes can be

combined by SRSS (Square Root of the Sum of the Square) method.

V3 = [27.22 + 4.37

2]

0.5 = 27.55 kN

V2 = [46.752 + 1.82

2]

0.5 = 46.79 kN

V1 = [57.752 + 4.03

2]

0.5 = 57.89 kN

The externally applied design loads are then obtained by

Q3 = V3 = 27.55

Q2 = V2 – V3 = 46.79 – 27.55 = 19.24

Q1 = V1 – V2 = 57.89 – 46.79 = 11.10

Q = 57.89

Clause 7.82 requires that the base shear obtained by dynamic analysis (VB = 57.89 kN) be compared with that

obtained from empirical fundamental period as per clause 7.6. If VB is less than that obtained from empirical

equation, the response quantities are to be scaled up.



Base shear for this frame as per empirical fundamental period is 67.50 kN. Thus the base shear as per dynamic

analysis of 57.89 is lower. Hence all respond quantities are to be scaled up in the ratio (67.50 / 57.89 = 1.166). Thus

the revised values of applied design loads are

Q3 = 27.55 x 1.166 = 32.12

Q2 = 19.24 x 1.166 = 22.43

Q3 = 11.10 x 1.166 = 12.94

EXERCISE 3.8

Calculate the gap required between two parts of a building for thermal expansion. Check it with the gap required to

avoid pounding as per Clause 7.11.3 of IS 1893(Part-1):2002. Assume coefficient of thermal expansion, α = 12 x 10-

6 mm/mm per

oC, and that the building is in New Delhi. Assume that the building has 6 floors and each part has a

length of 40 m and a storey of height 3.5 m.

Solution:

(a) Gap for expansion joint:

Required gap for expansion joint = 2Δ

Where tLc

From Fig. 1 & 2 of IS 875 (Part 5):1987, For New Delhi, the maximum and minimum temperature are 47.5oC and -

2.5 oC

Temperature differential (Tmax-Tmin) = 50oC

Required gap = 2 x 12 x 10-6

x 50 x 40 = 0.048 m = 48 mm

(b) Gap required for seismic requirements:

The permissible storey drift as per IS 1893 (Clause 7.11.1) is = 0.004 H

Permissible drift per storey = 0.004 3.5 1000 = 14 mm per storey.

As per Table 7 of IS 1893(Part-1), for special moment resisting frame,

Response reduction factor, R = 5.0

Since the two units will be at the same elevation levels,

Seismic separation gap = (14+14) x 5/2 = 70 mm per storey.

For the 6 storey building the separation required at top = 70 x 6 = 420 mm

Hence gap required to prevent pounding (420 mm) governs, as it is much higher than the gap required for

contraction/expansion (48 mm).

EXERCISE 3.9

Calculate the maximum axial shortening of column at top of a 60 storey building, assuming that the variation in the

cross-sectional area of column is a linear function, with the following data: Grade of concrete is M50, P t = 250 kN,

Pb = 15,000 kN, column of size 230 mm x 300 mm at top & 800 mm x1000 mm at bottom, and height of the

building 210 m.

Solution:

For 50 MPa concrete, E = 5000 √fck = 5000√50 = 35,355 MPa

At = 230 x 300 = 69,000 mm2, Ab = 800 x 1000 = 800, 000 mm

2

Pt = 250 kN, Pb = 15,000 kN

}1ln

11ln

12

b

b

b

bz

A

zAz

EA

z

E

P



48.3000,210

000,69000,800)(

L

AA tb mm2/mm

0702.0000,210

250000,15)(

L

PP tb kN/mm = 70.2 N/mm

3248.2)0978.0ln(000,800

000,21048.31ln1ln

bA

L

)}3248.2(000,800000,21048.3

48.3

1

355,35

2.70

48.3

)3248.2(

35,355

10000,152

3

top

= 283.43 - 185.11 =98.32 mm

Note: The reinforcements in columns are not considered in the above example. They may also be

included by converting their area to equivalent area of concrete, using the modular ratio. In

addition, creep and shrinkage effect have to be considered (see the example in Samara 1995).

EXERCISE 3.10

Roof design loads include a dead load of 1.75 kN/m2, a live (or snow) load of 1.25 kN/m

2 and a wind pressure of

0.75 kN/m2 (upward or downward). Determine the governing loading.

Solution

The load combinations are:

(a) 1.5 (DL+LL) = 1.5(1.75 + 1.25) = 4.5 kN/m2

(b) 1.2 (DL+LL+WL) = 1.2(1.75 + 1.25+ 0.75) = 4.50 kN/m2

(c) 1.2 (DL+LL–WL) = 1.2(1.75 + 1.25– 0.75) = 2.7 kN/m2

(d) 0.9DL+1.5WL = 0.9 x 1.75 + 1.5 x 0.75 = 2.7 kN/m2

(e) 0.9DL–1.5WL = 0.9 x 1.75 – 1.5 x 0.75 = 0.45 kN/m2

(f) 1.5 (DL+WL) = 1.5(1.75 + 0.75) = 3.75 kN/m2

(g) 1.5 (DL–WL) = 1.5 (1.75 - 0.75) = 1.5 kN/m2

The first or second load combination governs. Hence the roof has to be designed for a total factored load of 4.5

kN/m2. It may be noted that the fifth load case produces the minimum load.

507 39 Solutions Instructor Manual Chapter 3

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