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Question

# Assuming concrete below the neutral axis to be cracked, the shear stress across the depth of a singly-reinforced rectangular beam section

A
increases parabolically to the neutral axis and then drops suddenly to zero value
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B
increases parabolically to the neutral axis and then remains constant over the remaining depth
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C
increases linearly to the neutral axis and then remains constant up to the tensioned steel
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D
increases parabolically to the neutral axis and then remains constant up to the tensioned steel.
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Solution

## The correct option is D increases parabolically to the neutral axis and then remains constant up to the tensioned steel.Above the neutral axis concrete is assumed homogenous for rectangular homogeneous section Shear stress, q=V(A¯¯¯y)lb=V2l(x2u−y2) Above the neutral axis variation of shear stress is parabolic Figure shows two sections mn and m1n1 at distance dx apart of a R.C. beam. Due to variation of bending moment over dsitance dx compressive force C and C+ δC and tensile force in steel T and T + δ T are acting as shown in figure. Below N.A. concrete does not take any tension. Consider any plane between N.A. and centre of steel reinforcement, let q be the intensity of shear at that plane, hence total horizontal shear = q.bdx at the layer, also the total horizontal force that tends to slide this layer past the adjacent one is equal to (T+δT−T)=δT Hence, qbdx =δT q=δTbdx ...(i) For equilibrium, ΣM=0 (T+δT).jd=T.jd+V.dx δT=Vbjddx Putting in equation (i) q=Vbjd This value of q is evidently the same for any layer between the N.A. and the centre of steel thus shear stress distribution below the N.A. is rectangular Conclusion : Shear stress across the depth of singly reinforced rectangular beam section increases parabolically to the N.A. and then remains constant upto the tensioned steel.

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