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
• B. increases parabolically to the neutral axis and then remains constant over the remaining depth
• C. increases linearly to the neutral axis and then remains constant up to the tensioned steel
• D. increases parabolically to the neutral axis and then remains constant up to the tensioned steel.

Answer 1

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,
$q=\frac{V\left(A\overline{y}\right)}{lb}=\frac{V}{2l}(x_u^2-y^2)$

Above the neutral axis variation of shear stress is parabolic

Figure shows two sections mn and $m_1{n}_1$ at distance dx apart of a R.C. beam.

Due to variation of bending moment over dsitance dx compressive force C and C+ $\delta C$ and tensile force in steel T and T + $\delta$ 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 $(T+\delta T-T)=\delta T$
Hence, qbdx $=\delta T$
$q=\frac{\delta T}{bdx}$ ...(i)
For equilibrium, $\Sigma M=0$
$(T+\delta T).jd=T.jd+V.dx$
$\delta T=\frac{V}{bjd}dx$
Putting in equation (i)
$q=\frac{V}{bjd}$

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.