Question 6 of 8: Combined Axial, Bending and Shear in an L-Shaped Frame

Question 6: Combined Axial, Bending and Shear 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.

[Figure not reproduced: L-shaped element: 150 kN at A, roller at A bearing against a VERTICAL wall (horizontal reaction only, confirmed on the source drawing), pin at C; section a-a at 400 mm. See the official exam paper.]

Given.

Find. The normal and shear stress distribution at section a-a, with maximum and minimum values.

Approach. The roller at A bears against a VERTICAL wall (visible on the source drawing as rollers stacked against vertical hatching), so it supplies a HORIZONTAL reaction only — the full 150 kN vertical load must therefore be carried entirely by the pin at C through the rigid L-shaped frame. Solve the frame's three reactions from statics, then take a free body from A to the cut at a-a to get the axial force, shear and moment there, and combine $\sigma=N/A\pm Mc/I$ with $\tau_{max}=1.5V/A$.

1. Reactions. With $A_x$ the only reaction at A, and $C_x, C_y$ at C (800 mm horizontal, 300 mm vertical from A): $$\Sigma F_y=0:\ C_y = P = 150\ \text{kN}$$ $$\Sigma M_A=0:\ (800)C_y + (300)C_x = 0 \ \Rightarrow\ C_x = -\frac{800(150)}{300} = -400\ \text{kN}$$ $$\Sigma F_x=0:\ \boxed{A_x = -C_x = 400\ \text{kN}}$$ (A push of 400 kN from the wall through the roller, needed to balance the couple that the offset 150 kN load and the pin's vertical reaction create about the corner.)
2. Internal actions at section a-a (400 mm from A). Cutting the horizontal arm and taking the free body from A to the cut (only $A_x$ and $P$ act on it, both applied right at A): $$N = A_x = \boxed{400\ \text{kN (compression, uniform along the whole arm)}}$$ $$V = P = \boxed{150\ \text{kN}}$$ $$M = P(0.4) = \boxed{60.0\ \text{kN}\cdot\text{m (hogging – tension on top, like a tip-loaded cantilever)}}$$
3. Section properties. $$A = 60(180) = 10{,}800\ \text{mm}^2, \qquad I = \frac{60(180)^3}{12} = 29.16\times10^6\ \text{mm}^4, \qquad c=90\ \text{mm}$$
4. Combine axial and bending stress. $$\sigma_{axial} = \frac{N}{A} = \frac{400{,}000}{10{,}800} = 37.0\ \text{MPa (compression, uniform)}$$ $$\sigma_{bend} = \frac{Mc}{I} = \frac{60\times10^6(90)}{29.16\times10^6} = 185.2\ \text{MPa}$$ Top fibre (bending tension combines with the uniform compression): $$\sigma_{top} = -37.0+185.2 = \boxed{148.1\ \text{MPa (net tension)}}$$ Bottom fibre (bending compression adds to the axial compression): $$\sigma_{bot} = -37.0-185.2 = \boxed{-222.2\ \text{MPa (net compression, GOVERNS)}}$$
5. Shear stress. Rectangular section, parabolic distribution peaking at the neutral axis: $$\tau_{max} = \frac{3}{2}\frac{V}{A} = 1.5\!\left(\frac{150{,}000}{10{,}800}\right) = \boxed{20.8\ \text{MPa, at the NA}}, \qquad \tau=0\ \text{at top and bottom fibres}$$

Final Results.