Draw the shear stress distribution diagram for the following fig.

Note

1.For commonly used section like Rectangle. Circle, T-Section, Channel Section, I-Section, Shear Stress is Maximum at Neutral Axis

2.If the plane or section eg: 1-1, at which Shear Stress is required, is above N.A., we have to consider the area above the plan for calculating

3.If the plane, at which shear stress is required, is below the N.A, we have to consider the area below the plane calculating

Note

1.For commonly used section like Rectangle. Circle, T-Section, Channel Section, I-Section, Shear Stress is Maximum at Neutral Axis

2.If the plane or section eg: 1-1, at which Shear Stress is required, is above N.A., we have to consider the area above the plan for calculating $A\bar y$

$q_{avg}=\frac{total SF}{total area}$

1. If the plane, at which shear stress is required, is below the N.A, we have to consider the area below the plane calculating $A\bar y$

Let $V_A$ and $V_B$ be the support reaction

$\sum F_y=0$ (+ve)

$V_A$ + $V_B$=1800*8=14400N (1)

$\sum M_A=0$ (+ve)

$V_B*8-1800*8*\frac{6}{2}=0$

$V_B=7200N$ (2)

and $V_A=7200N$ (from (1) and (2))

OR

Max shear force $V_A=V_B=\frac{wL}{2}=\frac{1800*8}{2}=7200N$

i.e F=7200N (for bream ULD simply supported)

Calculation of moment of inertia:

$I_{base}=\frac{20*120^3}{3}+\frac{100*20^3}{12}+100*20*(120+\frac{20}{2})^2=45.39*10^6 mm^4$

$y_{base}=\frac{20*120*\frac{120}{2}+20*100*(120+\frac{20}{2})}{20*120+20*100}=91.81 mm$

$\sum A=20*120+20*100=4400mm$

Now, moment of inertia about N.A (neutral axis)

$I_{NA}=I_{base}-\sum A*(\bar y)^2$

$=45.39*10^6-4400*91.81^2$

$I_{NA}=8.30*10^6 mm^4$

Shear stress: (q)

1) $q_{NA}=\frac{F}{IB}*A\bar y$

$=\frac{7200*(20*91.81)*91.81}{8.30*10^6*20}=7.31 N/mm^2$

For $q_{NA}$ B-width at NA axis

A=20*91.81 because we considered area about NA

2)$q_{1-1}=\frac{7200*(100*20)*\frac{20}{2}+28.19}{8.30*10^6*b}$

When b=20mm

$q_{1-1}=3.31 N/mm^2$

When b=100mm

$q_{1-1}=0.66 N/mm^2$