Question 5 of 8: Combined Axial and Bending in a Propped Beam

Question 5: Combined Axial and Bending in a Propped Beam (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. Horizontal beam AB, span 3 m, pinned at A to a wall; a tension member runs from a second wall pin C (directly above A, 1.5 m higher) down to B at the far end of the beam. A 40 kN/m UDL covers the full span. The T-section (150 mm flange at the bottom, 30×200 mm stem on top) is given in the figure.

Find. Normal-stress distribution and maximum shear stress at the section 1 m from A; and the effect of an off-centroid pin connection.

Approach. The tension member CB is a two-force member; find its tension and the pin reaction at A from equilibrium of the whole beam, then cut at x=1 m (a section that does not include B or the cable) to get the internal N, V, M there, and finally combine axial and bending stress (uniform + linear, since the pins sit at the centroid so there is no added eccentricity).

1. Cable tension and reactions. The cable direction is B(3000,0) to C(0,1500), unit vector $(-0.894,\,0.447)$. Summing moments about A for the whole beam (UDL resultant $=40(3)=120$ kN at midspan): $$3000(0.447\,T) - 120000(1500) = 0 \;\Rightarrow\; T=134.2\text{ kN}$$ and then $A_x=0.894\,T=120.0$ kN, $A_y=120-0.447\,T=60.0$ kN.
2. Internal forces at x=1 m. Cutting between A and the cable attachment (no cable force acts on this segment), the segment $[0,1000\text{ mm}]$ carries only $A_x$, $A_y$, and 40 kN/m of UDL over 1 m: $$N=-A_x=\boxed{-120000\text{ N (compression)}}, \qquad V=A_y-40(1)=\boxed{20000\text{ N}}$$ $$M=A_y(1000)-\tfrac12(40)(1000)^2/1000 = \boxed{4.000e+07\ \text{N}\!\cdot\!\text{mm}}$$ The beam is pushed into compression because the cable's horizontal pull toward the wall must be balanced by the pin at A pushing the beam outward.
3. Section properties. With the flange at the bottom and pins at the centroid, $\bar y=81.73$ mm above the bottom, $I=49.4\times10^{6}\ \text{mm}^4$; the neutral axis falls inside the stem ($\bar y>t_f=25$ mm) so the stem's constant 30 mm width governs the shear formula.
4. Combined normal stress. Uniform axial ($N/A$) plus linear bending ($M(y-\bar y)/I$), since the pins are at the centroid (no eccentric axial moment): $$\sigma_{top}=\frac{N}{A}-\frac{M\,y_{top}}{I}=\boxed{-128.3\text{ MPa}}\ (\text{compression}), \qquad \sigma_{bot}=\frac{N}{A}+\frac{M\,y_{bot}}{I}=\boxed{53.9\text{ MPa}}\ (\text{tension})$$ both comfortably under the 350 MPa normal yield.
5. Maximum shear stress. The N.A. sits inside the stem, so $\tau_{max}$ occurs there using the stem width $t=30$ mm: $$\tau_{max}=\frac{VQ_{NA}}{It}=\boxed{4.155\text{ MPa}}$$ well under the 60 MPa shear yield.

Part (c). If the pins at A and B were NOT at the section centroid, the axial force N (from the pin reaction / cable component) would act with an eccentricity $e$ relative to the centroidal axis, introducing an ADDITIONAL bending moment $M_{ecc}=N\!\cdot\!e$ on top of the moment already caused by the transverse loading. The stress distribution would no longer be simply "uniform axial + bending about the applied transverse loads" — every section along the beam would carry this extra constant moment from the eccentric axial force, shifting the true neutral axis and increasing the peak stress on whichever fibre the eccentricity moment reinforces. (This is exactly the mechanism exploited deliberately in Question 3's spreader-beam sibling problems, where an off-axis cable pin is used to counteract bending — here it would be an unwanted side effect.)