Question 3 of 8: Spreader Beam — Combined Axial and Bending Stress

Question 3: Spreader Beam — Combined Axial and Bending Stress (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.

Find. The normal-stress distribution (max & min) and the maximum shear stress at the midspan cross-section.

Approach. Recognize that the two inclined cables do double duty: their VERTICAL components act like simple end reactions (giving ordinary sagging bending), while their EQUAL and opposite HORIZONTAL components squeeze the beam axially in compression. Because the cables attach 0.12 m above the beam's top flange (i.e. above the neutral axis), that axial force is applied ECCENTRICALLY, adding a constant bending moment on top of the ordinary one. Superpose $\sigma = -N/A \pm Mc/I$ at midspan, and use $\tau=VQ/(Ib)$ for the shear check.

1. Cable tension and its components. By symmetry each cable carries half the 120 kN vertical load. With rise 1.6 m and run 2.0 m (cable length $L_c=\sqrt{1.6^2+2.0^2}=2.561$ m): $$V_{cable}=\frac{P}{2}=60\ \text{kN}\qquad T=\frac{V_{cable}}{1.6/L_c}=96.05\ \text{kN}$$ $$H = T\left(\frac{2.0}{L_c}\right) = 75.0\ \text{kN}\quad\text{(constant axial compression, both ends squeeze inward)}$$
2. Bending moment from the vertical components alone. The two 60 kN "reactions" act exactly like a simply supported beam (span 4 m between the connection plates) carrying a 120 kN centre load: $$M_{vert}=V_{cable}\left(\frac{4}{2}\right)=60(2)=120\ \text{kN}\cdot\text{m}$$
3. Extra moment from the eccentric axial force. Section centroid is $c=130$ mm below the top flange; the connection plate stands a further 0.12 m above the flange, so the cable line of action sits $e=0.12+0.130=0.25$ m above the neutral axis, at BOTH ends. An axial force applied at the same eccentricity at both ends of a straight member is equivalent to the same force through the centroid plus a CONSTANT moment $H\!\cdot\!e$, and here it adds in the SAME sense as the sagging moment (both increase top compression, bottom tension): $$M_{ecc}=He=75.0(0.25)=18.75\ \text{kN}\cdot\text{m}\qquad M_{tot}=M_{vert}+M_{ecc}=138.75\ \text{kN}\cdot\text{m}$$
4. Section properties. $$A=2(140)(30)+20(200)=12\,400\ \text{mm}^2\qquad I=\frac{140(260)^3-120(200)^3}{12}=1.2505\times10^{8}\ \text{mm}^4\qquad c=130\ \text{mm}$$
5. (a) Combined normal stress at midspan (compression negative): $$\sigma_{axial}=\frac{H}{A}=\frac{75\,000}{12\,400}=6.05\ \text{MPa (compression)}\qquad \sigma_{bend}=\frac{M_{tot}\,c}{I}=\frac{(1.3875\times10^{8})(130)}{1.2505\times10^{8}}=144.24\ \text{MPa}$$ $$\sigma_{top}=-\sigma_{axial}-\sigma_{bend}=\boxed{-150.3\ \text{MPa (compression)}}\qquad \sigma_{bot}=-\sigma_{axial}+\sigma_{bend}=\boxed{+138.2\ \text{MPa (tension)}}$$ Both are comfortably under the 350 MPa yield — the section is safe, and note the top/bottom magnitudes DIFFER (150.3 vs 138.2 MPa) precisely because of the added axial term, which is exactly why the question asks for max AND min separately rather than a single symmetric bending value.
6. (b) Maximum shear stress at the neutral axis, using the shear carried on this side of midspan ($V=P/2=60$ kN) and the first moment of the half-section above the NA: $$Q_{NA}=140(30)(115)+20(100)(50)=5.83\times10^{5}\ \text{mm}^3$$ $$\tau_{max}=\frac{VQ_{NA}}{I\,t_w}=\frac{(60\,000)(5.83\times10^{5})}{(1.2505\times10^{8})(20)}=14.0\ \text{MPa} < 60\ \text{MPa}\quad\boxed{\text{OK}}$$