Combined Axial, Bending and Shear on a Cantilever: 04-BS-6, December 2013, Q4

Question 4: Combined Axial, Bending and Shear on a Cantilever (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.

Cantilever with eccentric axial load (400 kN, e = 140 mm) and transverse tip load (25 kN); 75 × 250 mm section.

Given.

Quantity	Value
Length L	3 m
Transverse tip load	25 kN (down)
Axial tip load	400 kN (horizontal, tension), eccentricity e = 140 mm above centroid
Section	75 mm × 250 mm rectangle

Find. The normal stress distribution (max/min) and shear stress distribution (max/min) at the base cross-section.

Approach. Reduce the eccentric axial force to an axial force through the centroid plus a couple; superpose the resulting axial and two bending-moment contributions for σ, and use the parabolic shear formula (with the transverse force only) for τ.

Section properties. $$A = bh = 75(250) = 18{,}750\ \text{mm}^2, \qquad I = \frac{bh^3}{12} = \frac{75(250)^3}{12} = 97.66\times10^6\ \text{mm}^4, \qquad c = 125\ \text{mm}$$
Internal actions at the base. $N = 400$ kN (tension, from the horizontal force); the 25 kN tip load gives bending moment $M_1 = 25(3) = 75$ kN·m; the eccentric 400 kN force adds a second moment $M_2 = 400(0.140) = 56$ kN·m. Both moments put the TOP fibre in tension (a rightward pull applied above the centroid, like the downward tip load, rotates the free end so the base's top fibre stretches) — they add directly.
(a) Normal stress: superpose N/A and M c/I. $$\sigma_N = \frac{N}{A}=\frac{400{,}000}{18{,}750}=21.33\ \text{MPa (tension, uniform)}$$ $$\sigma_{M1}=\frac{M_1 c}{I}=\frac{75\times10^6(125)}{97.66\times10^6}=96.0\ \text{MPa}, \qquad \sigma_{M2}=\frac{M_2 c}{I}=\frac{56\times10^6(125)}{97.66\times10^6}=71.68\ \text{MPa}$$ $$\sigma_{top}=\sigma_N+\sigma_{M1}+\sigma_{M2}=\boxed{189.0\ \text{MPa (tension)}}$$ $$\sigma_{bot}=\sigma_N-\sigma_{M1}-\sigma_{M2}=\boxed{-146.3\ \text{MPa (compression)}}$$ Stress varies linearly between these two extremes, crossing zero at $y_0 = 125(21.33)/(96.0+71.68)=15.9$ mm below the centroid.
(b) Shear stress distribution. Only the 25 kN transverse load produces transverse shear (the 400 kN force is axial); for a solid rectangle the parabolic shear formula peaks at the neutral axis: $$\tau_{max} = \frac{3}{2}\frac{V}{A} = 1.5\left(\frac{25{,}000}{18{,}750}\right) = \boxed{2.00\ \text{MPa}}\ \text{(at the NA)}, \qquad \tau=0\ \text{at top and bottom}$$

Final Results.

Quantity	Value
σ_top (max tension)	189.0 MPa
σ_bottom (max compression)	−146.3 MPa
Neutral-stress location	15.9 mm below centroid
τ_max (at NA)	2.00 MPa
τ at top/bottom fibres	0

Topic: Combined axial + biaxial-moment (eccentric load) stress and transverse shear on a rectangular section.

Key relations: $\sigma = N/A \pm Mc/I$ (superposition of axial and bending stress); $\tau_{max}=1.5V/A$ for a solid rectangular section.

Why this works: Linear-elastic beam theory allows independent load effects to be superposed for stress (unlike the deflection method-of-integration rule in Q1/Q6, which bars superposing *tabulated solutions* — superposing stress CONTRIBUTIONS at a single section is always valid). Moving the eccentric axial force to the centroid requires adding the compensating couple $M_2=N\!\cdot\!e$.

Common pitfall: Missing the eccentricity moment entirely (treating the 400 kN force as purely axial), or adding it with the wrong sign relative to the transverse load's moment — always re-derive the physical sense (which fibre is in tension) from the loading picture rather than pattern-matching a sign convention.

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 8 (combined loadings).

Question 4 of 8: Combined Axial, Bending and Shear on a Cantilever

Question 4: Combined Axial, Bending and Shear on a Cantilever (20 marks)