# Design of RC Structures Questions and Answers – T-Beams – Nominal Shear Stress

This set of Design of RC Structures Multiple Choice Questions & Answers (MCQs) focuses on “T-Beams – Nominal Shear Stress”.

1. What is the maximum shear stress for a rectangular homogeneous section (V=shear force applied, D=depth of the particular section, y= distance from the neutral axis, I=moment of inertia of the section about the neutral axis)?
a) VD2/8I
b) VD2/6I
c) VD2/2I
d) VD2/4I

Explanation: The maximum shear stress of a homogeneous section is VD2/8I. Shear stress at a distance y from neutral axis is given by
q=V*A*y/I*b
where A=area of the section and b=breadth of the section;
we can derive this as
q=(V/2I)*{(D2/4)-y},
Now maximum shear stress is likely to occur at the neutral axis y=0
that results in the fact that: qmax=VD2/8I.

2. As per IS Code nominal shear stress is given by which of the following [V=shear force, b=breadth of the section, d=depth of the section]?
a) V/b*d
b) V*b/d
c) V*d/b
d) V*b*d

Explanation: According to the clause 40.1 of IS 456:2000 the nominal shear stress (ζv) is given by V/b*d.
If we check dimensionally, then also ( force*length/length) does not give the dimension of stress. Therefore neither V*b/d nor V*d/b are the answers. Again (force*length*length) does not give the dimension of stress. Hence V*b*d is not the correct answer.
Stress=force/area; V/b*d is also dimensionally correct.

3. According to IS 456:2000 for solid slabs (k ζc=Design shear strength) the value of k depends on _________________
a) length of the slab
b) depth of the slab
c) ratio between length and breadth of the slab

Explanation: The factor k is only dependent on the depth of the solid slab. Length, breadth or the ratio of length or breadth is not involved in determination of “k”. According to codal provision
k = 1.3 when depth D ≤ 150mm
k = 1.6-.002D when 150mm<D < 300mm
k = 1 for all D≥300mm.

4. What is the maximum shear strength (ζcmax) for M25 Grade concrete?
a) 3.8 N/mm2
b) 3.1 N/mm2
c) 3.5 N/mm2
d) 2.9 N/mm2

Explanation: Table 20 of IS 456:2000 gives the value of ζcmax as 3.1 N/mm2.
Besides ζcmax = .625(fck)1/2=.625 *(25)1/2=.625*5 = 3.1 N/mm2;
fck = characteristic strength of the concrete which depends on the grade of the concrete.

5. A 2 legged 10 mm φ stirrup is provided in a particular section. What is the area of shear reinforcement of the section?
a) 157 mm2
b) 160 mm2
c) 189 mm2
d) 190 mm2

Explanation: The area of shear reinforcement is 157 mm2.
Area of the shear reinforcement = Number of legs*Cross sectional area of one reinforcement bar
= 2*78.5= 157 mm2.

6. What is the main function of shear connectors?
a) To provide shear reinforcement
b) To connect two stirrups
c) To prevent the relative slip between the interfaces of two dissimilar materials
d) To provide torsional reinforcement

Explanation: The main function of shear connectors is to prevent slip between interfaces of two dissimilar material. If a composite section is required to act as a unit then proper provision has to be introduced to transmit the horizontal shear from one material to other so that it will prevent slip between interfaces of two dissimilar material. This is done by providing shear connectors.
Shear reinforcement is achieved by providing stirrups and connection of stirrup is not the principal function of shear connectors. There are other ways to provide torsional reinforcement. Therefore these are not the correct answers.

7. Which phenomenon is true for the lattice girders?
a) Top concrete acts as tension members
b) Bottom longitudinal reinforcement acts as compression members
c) Concrete between inclined bent up bars acts as diagonal compression members
d) Bent up bars act as diagonal compression members

Explanation: According to Indian standards in lattice girders Concrete between inclined bent up bars acts as diagonal compression members.
Actually following are the phenomenon in the lattice girder:
i) Top concrete acts as compression members.
ii) Bottom longitudinal reinforcement acts as tension members.
iii) Concrete between inclined bent up bars acts as diagonal compression members.
iv) Bent up bars act as diagonal tension members.

8. For a section of 300*600 what is the maximum spacing of shear reinforcement?
a) 450mm
b) 300mm
c) 100mm
d) 150mm

Explanation: The maximum spacing between the shear reinforcement is given as 450mm.According to the IS 456:2000 we know the spacing
Sv ≤ .75d where d=depth of the section;
The maximum spacing is =.75*600 = 450mm.

9. A reinforced concrete beam of 300mm*600mm carries a maximum shear strength of 150 kiloNewton. The beam is reinforced by 4 numbers of 16 mm dia bars for flexural resistance. What is the shear stress on the beam?
a) 1.21 N/ mm2
b) 1.36 N/ mm2
c) 1.5 N/ mm2
d) 1.10 N/ mm2

Explanation: The area of reinforcement is = 803.84 mm2
Assuming effective cover of 50mm, effective beam depth is=600-50=550mm.
Factored shear force is=1.5*150 kN=225kN
Shear stress=(225*1000)/(300*600)=1.36 N/ mm2.

10. To avoid shear compression failure which of the following is a necessary criteria?
a) ζcmax < ζc
b) ζc < ζcmax
c) ζc < ζv
d) ζcmax < ζv

Explanation: According to Indian Standard Code IS 456:2000, to avoid the shear compression failure we must ensure ζc < ζcmax. The shear compression failure is brittle in nature. Therefore in order to ensure ductile failure the criteria ζc < ζcmax is necessary. In Table 20 of IS 456:2000 the values of ζcmax are mentioned.

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