22-Mec-A7 Advanced Strength of Materials · May 2016
Question 8 of 8: Shear flow and shear centre of a channel section
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams — May 2016, 07-Mec-A7 Advanced Strength of Materials. Open-book, 3 hours, eight problems of equal value; any five constitute a complete paper. All eight are solved as a study resource.
Reference texts. A.P. Boresi & R.J. Schmidt, Advanced Mechanics of Materials, 6th ed. (Wiley); A.C. Ugural & S.K. Fenster, Advanced Strength and Applied Elasticity, 4th ed. (Prentice Hall); R.C. Hibbeler, Mechanics of Materials, 10th ed. (Pearson); J.M. Gere & B.J. Goodno, Mechanics of Materials, 9th ed. All symbols follow the paper’s notation.
Question 8: Shear flow and shear centre of a channel section (equal value)
Given. Channel opening to the right: web height h = 40 mm, web thickness tw = 3 mm; two equal flanges width b = 30 mm, flange thickness tf = 2 mm; vertical shear V = 2500 N (up) through the shear centre; bending about the horizontal (x) axis.
Find. (a) shear-flow distribution q in flanges and web; (b) the shear-centre offset e; (c) location and value of maximum shear stress.
Channel section (left) and the shear-flow distribution (right, N/mm): linear in each flange from 0 at the tip to 46.9 at the web, parabolic in the web peaking at 70.3 on the neutral axis. The shear centre lies 11.25 mm left of the web.
Approach. Compute Ix about the horizontal axis, build q = VQ/Ix starting from a free flange tip and continuing through the web, then equate the moment of the flange shear-flow couple to V·e to locate the shear centre. Divide q by thickness for the shear stresses.
Second moment of area about the x-axis. Web plus the two flanges at ±h/2:
$$I_x=\frac{t_w h^3}{12}+2\big(b\,t_f\big)\!\left(\frac{h}{2}\right)^2=16\,000+48\,000=64\,000\ \text{mm}^4$$
Flange shear flow (part a). Measuring s from the free tip, Q = tfs(h/2), so q grows linearly:
$$q_f(s)=\frac{V\,t_f (h/2)\,s}{I_x}=1.5625\,s\quad\text{(N/mm)},\qquad q_f(b)=46.9\ \text{N/mm at the web}$$
Web shear flow. Adding the web’s first moment, q is parabolic and peaks at the neutral axis:
$$q_{\text{web,NA}}=\frac{V}{I_x}\!\left[b\,t_f\frac{h}{2}+t_w\frac{h}{2}\frac{h}{4}\right]=\frac{2500(1800)}{64\,000}=\boxed{70.3\ \text{N/mm}}$$
The web value at the flange junction, 46.9 N/mm, matches the flange result — shear flow is continuous.
Shear centre (part b). The horizontal force in each flange is
$$H=\int_0^b q_f\,ds=\frac{V\,t_f h\,b^2}{4I_x}=703\ \text{N}$$
The two equal, opposite flange forces form a couple H·h that the applied V must balance about the web:
$$V\,e=H\,h \;\Rightarrow\; e=\frac{b^2 h^2 t_f}{4 I_x}=\boxed{11.25\ \text{mm}}$$
measured from the web centreline on the side away from the flanges (i.e. to the left of the web).
Maximum shear stress (part c). Divide q by the local thickness:
$$\tau_{\text{web,NA}}=\frac{70.3}{3}=23.4\ \text{MPa},\qquad \tau_{\text{flange,junction}}=\frac{46.9}{2}=23.4\ \text{MPa}$$
$$\boxed{\tau_{\max}=23.4\ \text{MPa at the neutral axis of the web}}$$
For these particular thicknesses the flange–web corners reach the same value, but the conventional maximum sits at the mid-web neutral axis.
