Question 8 of 8: Combined Axial + Bending + Shear on an Overhang Beam

Question 8: Combined Axial + Bending + Shear on an Overhang 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. Beam pin-supported at $A$ ($x=0$), roller at $B$ ($x=6\text{ m}$), free overhang tip at $C$ ($x=9\text{ m}$). UDL $w=10\text{ kN/m}$ over the FIRST span $A$ to $B$ ($0\le x\le6\text{ m}$). Axial load $200\text{ kN}$ applied HORIZONTALLY at the free tip C. Cross-section: $100\text{ mm}\times200\text{ mm}$ rectangle; point B on the section is marked at the extreme TOP fibre.

Find. Normal and shear stress distribution over the section at B, and a volume element at the marked point.

Approach. Find the reactions, then the internal axial force, shear and moment at the section through B; the axial load produces uniform tension over the whole beam (the pin at A is the only horizontal restraint), and the extreme top fibre (point B) has zero shear by definition, so its stress state is pure axial tension once the bending moment there is evaluated.

1. Reactions. UDL resultant $=10\times6=60\text{ kN}$ at its centroid $x=3\text{ m}$ (exactly mid-span between A and B). $\sum M_A=0$: $R_B(6)-60(3)=0\Rightarrow R_B=\boxed{30\text{ kN}}$; $\sum F_y=0$: $R_A=60-30=\boxed{30\text{ kN}}$. $\sum F_x=0$: the pin at A alone resists the 200 kN axial pull, so the internal axial force is $N=\boxed{200\text{ kN tension}}$, CONSTANT along the entire beam (no other horizontal load or support anywhere).
2. Shear and moment just left of B. $V(6^-)=R_A-w(6)=30-60=\boxed{-30\text{ kN}}$. $M(6)=R_A(6)-\dfrac{w}{2}(6)^2=30(6)-5(36)=180-180=\boxed{0}$ — the UDL is centred exactly between the two supports, so the moment at the second support is exactly zero (a designed check, not a coincidence).
3. Stress at point B (extreme top fibre, $M=0$ there). With $M=0$ at this section, bending stress is zero everywhere on the cross-section, so the normal-stress DISTRIBUTION is UNIFORM tension from axial load alone: $\sigma=N/A=200{,}000/(100\times200)=\boxed{10.0\text{ MPa tension, uniform across the whole section}}$. Shear stress follows the usual parabolic profile for $V=-30\text{ kN}$: zero at the top/bottom extreme fibres, maximum at the neutral axis $\tau_{max}=1.5\,V/A=1.5(30{,}000)/20{,}000=\boxed{2.25\text{ MPa}}$; AT point B itself (the extreme top fibre) $\tau=0$.
4. Volume element at B. Since $\tau=0$ and $\sigma=10.0\text{ MPa}$ tension at the extreme top fibre, the stress element at B is simple UNIAXIAL TENSION (no shear on the element's faces) — sketched at right.