Question

The shear stress at the neutral axis in a beam of triangular section with a base of 40 mm and height 20 mm, subjected to a shear force of 3 kN is

• A. 3 MPa
• B. 6 MPa
• C. 10 MPa
• D. 20 MPa

Answer 1

The correct option is C 10 MPa


Neutral Axis passes through the centroid of the section.
Shear stress at neutral axis,

$\tau =\frac{VA\overline{Y}}{Ib}$

V = shear force at the section where shear stress is to be found out.
$A\overline{Y}=$ moment of area of portion above level at which shear stress is to be found out.
I = moment of inertia about NA of complete section
b = width of section where shear stress to be found out

$b^{\prime }=\frac{2}{3}\times 40$

$V=3kN$

$h^{\prime }=\frac{2}{3}\times 20$

$\tau =\frac{VA\overline{Y}}{Ib}$

$d=\frac{3\times {10}^3\left(\frac{1}{2}{b}^{\prime }{h}^{\prime }\right)\times \left(\frac{h^{\prime }}{3}\right)}{\frac{b{h}^3}{36}\times b^{\prime }}$

$=\frac{3\times {10}^3\times (\frac{1}{2}\times \frac{2}{3}\times 40\times \frac{2}{3}\times 20)\frac{2}{3}\times 20\times \frac{1}{3}}{\frac{40\times {20}^3}{36}\times \frac{2}{3}\times 40}$

$=10MPa$

Alternatively

${\tau }_{na}=\frac{4}{3}{\tau }_{avg}$

$=\frac{4}{3}\times \frac{3000}{\frac{1}{2}\times 40\times 20}=10MPa$