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.