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04-BS-6 · December 2017

Question 5 of 8: Combined Axial, Shear & Bending in an L-Shaped Frame

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

04-BS-6: Mechanics of Materials — National Exams, December 2017. Closed-book, 3 hours; any FIVE of the eight questions constitute a complete paper (all eight are answered below as a complete study resource). Aid sheet permitted; geometric properties of W-shape sections were supplied as an appendix (not required for any of the eight questions solved here).

Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (axial loading, shear/moment diagrams, beam deflection by integration, stress transformation & Mohr's circle, combined loadings, torsion, column buckling, unsymmetric-section flexure).

Question 5: Combined Axial, Shear & Bending in an L-Shaped Frame (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

QuantityValue
Applied load, P200 kN, downward at (the top-left joint at) A
A to vertical line through C (horizontal)800 mm
B to C (vertical)400 mm
Section a-a location500 mm from A, on member AB
Cross-section at a-a80 mm (width) × 200 mm (height)

Find. The normal and shear stress distribution across section a-a.

aaA (roller,horiz. reaction)C (fixed pin)P = 200 kN500 mm800 mm (A→ line thru C)400 mm80×200 mm rect.section at a-a
L-shaped frame with the roller (horizontal reaction) at A, fixed pin at C, and section a-a on member AB.

Approach. The roller at A bears against the vertical wall behind the frame and therefore supplies a purely horizontal reaction; the pin at C supplies both components. Find all three reactions from statics, then cut the frame at a-a and use the left-hand free body to get the internal axial force N, shear V and bending moment M there; convert to stresses with the combined-loading (axial + flexure) formula plus the shear formula for a rectangle.

  1. Reactions. Taking moments about C (which eliminates the two unknowns at C): $$\sum M_C=0:\ R_{Ax}(0.4)-200(0.8)=0 \ \Rightarrow\ R_{Ax}=400\text{ kN}$$ $$\sum F_x=0:\ C_x=-R_{Ax}=-400\text{ kN}, \qquad \sum F_y=0:\ C_y=200\text{ kN}$$
  2. Internal forces at section a-a (0.5 m from A). Cutting through AB and taking the free body from A to the cut (which contains only $R_{Ax}=400$ kN horizontal and the 200 kN downward load, both acting at A): $$N=-400\text{ kN (compression)}, \qquad V=200\text{ kN}, \qquad M = -100\text{ kN}\cdot\text{m}$$ An independent free body taken from the cut through B to C (containing only the pin reaction at C) reproduces the identical $N,V,M$, confirming the reactions.
  3. Section properties. Rectangle $b=80$ mm, $h=200$ mm: $$A=80(200)=16{,}000\text{ mm}^2,\qquad I=\frac{80(200)^3}{12}=5.333\times10^{7}\text{ mm}^4,\qquad c=100\text{ mm}$$
  4. Uniform axial stress. $$\sigma_N=\frac{N}{A}=\frac{-400{,}000}{16{,}000}=-25.0\text{ MPa (compression, uniform)}$$
  5. Bending stress at the extreme fibres (M is hogging, so the top fibre is in tension): $$\sigma_M=\frac{Mc}{I}=\frac{(100\times10^6)(100)}{5.333\times10^{7}}=187.5\text{ MPa}$$ Top fibre: $+187.5$ MPa (tension); bottom fibre: $-187.5$ MPa (compression).
  6. Combined normal stress (superposing axial + bending). $$\boxed{\sigma_{top}=187.5-25.0=+162.5\text{ MPa (net tension)}}$$ $$\boxed{\sigma_{bot}=-187.5-25.0=-212.5\text{ MPa (compression)}}$$
  7. Maximum transverse shear stress (parabolic distribution, peak at the neutral axis, zero at the outer fibres, per $\tau=VQ/(Ib)$ reduced to $\tfrac32 V/A$ for a rectangle): $$\boxed{\tau_{max}=\frac{3}{2}\frac{V}{A}=\frac{3}{2}\cdot\frac{200{,}000}{16{,}000}=18.75\text{ MPa}}$$
QuantityValue
N, V, M at section a-a−400 kN, 200 kN, −100 kN·m
σ at top fibre+162.5 MPa (tension)
σ at bottom fibre−212.5 MPa (compression)
τmax (at neutral axis)18.75 MPa