# Design of RC Structures Questions and Answers – Slabs – Marcus Correction

This set of Design of RC Structures Multiple Choice Questions & Answers (MCQs) focuses on “Slabs – Marcus Correction”.

1. In context to the Marcus method the bending moment (B.M.) MB and ML calculated by Grashoff – Rankine method which of the following is correct?
a) B.M. MB and ML are added to a reduction factor (C)
b) B.M. MB and ML are divided by a reduction factor (C)
c) B.M. MB and ML are multiplied by a reduction factor (C)
d) B.M. MB and ML are subtracted from the reduction factor (C)

Explanation: Marcus method is an approximate method by which maximum B.M. in a slab with corners held down can be determined. In this method, the B.M. MB and ML calculated by the Grashoff – rankine method are multiplied by a reduction factor C.

2. What is the value of the reduction factor (C)?
a) C = 1 – (5 / 6) (r2 / (1 + r4))
b) C = 1 – (6 / 5) (r2 / (1 + r4))
c) C = 1 – (5 / 6) (r4 / (1 + r2))
d) C = 1 – (6 / 5) (r4 / (1 + r2))

Explanation: The value of reduction factor C is given by:
C = 1 – (5 / 6) (r2 / (1 + r4))
This value depends on the L / B ratio. Also (C) should be less than unity because the positive bending moment in slab is reduced as the corners are held down.

3. What is the value of mid – span bending moment per unit width in shorter direction (B)?
a) MB = CrB (wB4 / 8)
b) MB = CrB (wB2 / 8)
c) MB = CrB (wB / 8)
d) MB = CrB (wB3 / 8)

Explanation: The value of the mid – span bending moment per unit width in the shorter direction (B) is given by equation:
MB = CrB (wB2 / 8).
This value includes the reduction factor (C).
Here C = 1 – (5 / 6) (r2 / (1 + r4)) and
r = (L / B).

4. Which of the following is the equation of mid – span bending moment per unit width in the longer direction (L)?
a) ML = CrL (wL2 / 8)
b) ML = CrL (wL4 / 8)
c) ML = CrL (wL / 8)
d) ML = CrL (wL3 / 8)

Explanation: The equation for mid – span bending moment per unit width in the longer direction is:
ML = ML = CrL (wL2 / 8)
Here C is the reduction factor and
r = (L / B).
L is long span of a slab and B is the short span of a slab.

5. What is the value of resulting bending moment in the shorter direction (B) when Poisson’s ratio is taken equal to 0.15?
a) MB = CrB (1 + (0.15 / r2)) (wB2 / 8)
b) MB = CrB (0.15 / r2) (wB2 / 8)
c) MB = CrB (1 + (0.15 / r2)2) (wB4 / 8)
d) MB = CrB (1 + (0.15 / r2)) (wB4 / 8)

Explanation: The value of mid – span bending moment in the shorter direction when Poisson’s ratio is taken as 0.15:
MB = CrB (1 + (0.15 / r2)) (wB2 / 8)
Here C is the reduction factor and
r = (L / B).
Also L is long span of a slab and B is the short span of a slab.

6. What is the equation for resulting bending moment when Poisson’s ratio is taken equal to 0.15 in the longer direction (L)?
a) ML = CrL (0.15 / r2) (wL2 / 8)
b) ML = CrL (1 + (0.15 / r2)2) (wL4 / 8)
c) ML = CrL ((1 / r2) + 0.15) (wL2 / 8)
d) ML = CrL (r4 + (0.15 / r2)) (wL4 / 8)

Explanation: The equation for mid – span bending moment in the longer direction where Poisson’s ratio is taken as 0.15:
ML = CrL ((1 / r2) + 0.15) (wL2 / 8)
In this C is the reduction factor and
r = (L / B).
Here L is long span of a slab and B is the short span of a slab.

7. To resist the torsional moments induced at the held down corners, it is necessary to provide mesh reinforcement at both the faces of the slab, at each corner.
a) True
b) False

Explanation: According to Dr. Marcus to resist the torsional moments induced at the held down corners, it is necessary to provide mesh reinforcement at both the faces of the slab, at each corner. According to common practice both top and bottom reinforcements should consist of two layers bars. These should be parallel to the sides of the slab, extending in these directions for a distance of one – fifth of the shorter span.

8. What is the value of shear force along the short span (B) as per the Marcus method?
a) FB = 1 / 2 (wB)
b) FB = 1 / 4 (wB)
c) FB = 1 / 5 (wB)
d) FB = 1 / 3 (wB)

Explanation: The value of shear force, per unit width, along the short span is calculated by using:
FB = 1 / 3 (wB)
This value for shear force is similar to that derived in the Grashoff – Rankine method. This is the similarity in both the methods.

9. Which of the following equation is correct when calculating the shear force in the long span (L) as per the Marcus method?
a) w.B. (r / (2 + r))
b) 2.w.B. (w / (2 + r))
c) w.B. (2w / (2 + r))
d) w.B. (0.5w / (2 + r))

Explanation: The equation for S.F. (shear force) along the long span per unit width is taken as:
[w.B.(r / (2 + r))]
Subject to a maximum of (0.5wB).

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. The slab is simply supported at all the four edges, with all the corners held down.What is the value of reduced bending moment in the short direction calculated according to the Marcus method?
(Take c = σcbc = 5 N / mm2,
t = σst = 140 N / mm2,
m = 19,
jc = 0.865,
kc = 0.404,
Rc = 0.874).
a) 5.24 × 106 N – mm
b) 5.04 × 106 N – mm
c) 5.90 × 106 N – mm
d) 6.24 × 106 N – mm

Explanation: Assuming the overall depth = 120 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
The bending moment, according to the Marcus’s method is given by:
MB = [1 – (5 / 6) (r2 / (1 + r4))] r4 / (1 + r4) (wB2 / 8) = [1 – (5 / 6) (1.55 / (1 + 2.4))] 2.4 / (1 + 2.4) (5480(4.1)2 / 8) × 1000
= 5.04 × 106 N – mm

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