This set of Design of RC Structures Multiple Choice Questions & Answers (MCQs) focuses on “Slabs – Rankine Grashoff Theory”.
1. According to Grashoff – Rankine method which of the following is correct statement?
a) B.M. (bending moment) is determined on the basis of maximum deflection at the centre of slab
b) B.M. is determined on the basis of (B / L) ratio
c) B.M. is determined on the basis of udl in both the directions
d) B.M. is determined on the basis of length of longer span
View Answer
Explanation: In the Grashoff – Rankine method, the bending moments carried in the long and short directions are determined on the basis of maximum deflection at the centre of the slab. For the calculation of bending moments consider two middle strips of unit width, parallel to long and short span.
2. Which of the following is the equation for load carried by slab in long direction (L)?
a) wL = w / (1 + (L / B)2)
b) wL = w / (1 + (L / B)4)
c) wL = w / (1 + L4)
d) wL = w / (1 + B4)
View Answer
Explanation: The load carried by slab in the longer direction (L) is given by equation:
wL = w / (1 + (L / B)4)
This is reduced to:
wL = w / (1 + r4) = rLw
Here (L / B) = r and rL = 1 / (1 + r4),
this depends on the value of (L / B) ratio.
3. Which of the following is the equation for load carried by slab in short direction (B)?
a) wB = w / (1 + (B / L)2)
b) wB = w / (B + L4)
c) wB = w / (1 + (B / L)4)
d) wB = w / (L + B4)
View Answer
Explanation: The load that the slab carries in the shorter direction (B) is given by equation:
wB = w / (1 + (B / L)4)
This can be reduced to:
wB = wr4 / (1 + r4) = rBw.
Here (L / B) = r and rB = 1 / (1 + r4),
this depends on the value of (L / B) ratio.
4. Which of the following is the equation for maximum bending moment in longer direction (L)?
a) ML = wrL (L4 / 8)
b) ML = wrL (L / 8)2
c) ML = wrL (L / 8)
d) ML = wrL (L2 / 8)
View Answer
Explanation: The maximum bending moment in the longer direction is:
ML = w (L2 / 8) = [w / (1 + r4)] (L2 / 8) = wrL (L2 / 8)
Here r = (L / B),
L is long span of a slab and B is the short span of a slab.
5. Which of the following is the equation for maximum bending moment in shorter direction (L)?
a) MB = wrB (B4 / 8)
b) MB = wrB (B / 8)2
c) MB = wrB (B / 8)
d) MB = wrB (B2 / 8)
View Answer
Explanation: The value of maximum bending moment in the longer direction (B) is given by:
MB = w (B2 / 8) = [wr4 / (1 + r4)] (B2 / 8) = wrB (L2 / 8).
Here r = (L / B),
L is long span of a slab and B is the short span of a slab.
6. Slab thickness is determined on the basis of MB. Because the moment in short span will be more than that in the long span.
a) True
b) False
View Answer
Explanation: Normally, the bending moment in the short span will be more than the bending moment on the long span. Hence the slab thickness will be determined on the basis of the moment in short span that is MB.
7. Which of the following is used to calculate Ast in the short direction (B)?
a) MB / (σst jc d)
b) 2MB / (σst jc d)
c) MB / 2(σst jc d)
d) MB / 4 (σst jc d)
View Answer
Explanation: The area of steel in the short direction is calculated by:
AstB = MB / (σst jc d)
Here MB is moment in the short span,
d is effective depth of slab determined using MB.
σst is the permissible tensile stress in steel.
8. Which of the following is used to calculate Ast in the long direction (L)?
a) ML / 2(σst jc (d – ΦB))
b) 2ML / (σst jc (d – ΦB))
c) ML / (σst jc (d – ΦB))
d) ML / 4 (σst jc (d – ΦB))
View Answer
Explanation: The area of steel in the short direction is calculated by:
AstL = ML / (σst jc (d – ΦB)).
Here ML is moment in the long span,
d is effective depth of slab determined using MB,
σst is the permissible tensile stress in steel and
ΦB is diameter of bars in short span.
9. Which of the following is used to calculate S.F. (shear force) along the long span (L)?
a) w.B.(w / (2 + r))
b) 2.w.B.(w / (2 + r))
c) w.B.(2w / (2 + r))
d) w.B.(0.5w / (2 + r))
View Answer
Explanation: The force acting in a direction parallel to a surface or a planar cross section of a body is shear force. S.F. along the long span, per unit width is taken as:
w.B. (w / (2 + r)),
subject to a maximum of (0.5wB) when r exceeds 2.
10. Design a R.C. slab for a room of size 4 m × 5 m from inside. The total load on the slab 5480 N. Take
c = σcbc = 5 N/mm2,
t = σst = 140 N/mm2
m = 19,
kc = 0.404,
jc = 0.865,
Rc = 0.874.
If the slab is simply supported at all the four edges, with corners free to lift, what is the areas of steel in short and long directions?
a) 600 mm2 and 400 mm2
b) 671.30 mm2 and 481 mm2
c) 700 mm2 and 500 mm2
d) 670 mm2 and 480 mm2
View Answer
Explanation: Let us take effective depth = 100 mm, we have
L = 5 + 0.1 = 5.1 m and B = 4 + 0.1 = 4.1 m
r = L / B = 5.1 / 4.1 = 1.245; r4 = 2.4
Hence wB = w(r4 / (1 + r4)) = 5480 (2.4 / (1 + 2.4)) = 3868 N;
wL = w(1 / (1 + r4)) = 5480 (1 / (1 + 2.4)) = 1612 N;
MB = wB (B2 / 8) = 3868 ((4.1)2 / 8) . 1000 = 8.13 × 106 N – mm;
ML = wL (L2 / 8) = 1612 ((5.1)2 / 8) . 1000 = 5.24 × 106 N – mm
The effective depth of slab is found on the basis of greater moment MB.
d = √(MB / Rcb) = 97 mm. Adopt total thickness as 120 mm
Using 10 mm diameter Bars and a nominal cover of 15 mm, we get d = 120 – 15 – 5 = 100 mm for short span. The effective depth for long span = 100 – 10 = 90 mm, since the reinforcement in long span is placed above the reinforcement in short span.
AstB = MB / (σst jc d) = (8.13 × 106) / (140 × 0.865 × 100) = 671.3 mm2. Using 10 mm diameter bars.
AΦ = π / 4(102) = 78.5 mm2.
Spacing sB = 1000 AΦ / Ast = 1000(78.5 / 671.3) = 117 mm.
Hence provide 10 mm diameter bars @ 100 mm c/c along the short span. Bend – up alternate bars at B / 7 = 4.1 / 7 = 60 cm approx. from the centre of each support.
AstL = (5.24 × 106) / (140 × 0.865 × 90) = 481 mm2.
Hence spacing of 10 mm diameter bars is given by sL = 1000 (78.5 / 481) = 163 mm.
However, provide 10 mm diameter bars @ 150 mm c/c.
Bend – up alternate bars at L / 7 = 5100 / 7 = 700 mm approx. from the centre of each support.
11. When the L / B ratio is greater than 2, one may design it as a slab supported on two edges.
a) False
b) True
View Answer
Explanation: The slabs in which the L / B ratio is greater than 2 (L / B > 2), one may design it as a slab supported on two edges. This is without any serious errors. Here, (L) is longer span and (B) is the shorter span.
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