Question 8 of 8: Combined Axial + Bending Stress in a Cable-Supported T-Beam
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam, May 2019 — 04-BS-06: Mechanics of Materials. Closed book, one 8.5"×11" handwritten aid sheet permitted, 3 hours, 8 questions of equal value (any 5 of 8 constitute a complete paper — all 8 are answered here as a complete study resource).
Reference texts: R.C. Hibbeler, Mechanics of Materials, 10th ed. (Pearson) — axial deformation of statically indeterminate assemblies, beam deflection by integration, shear/moment diagrams, plane-stress transformation and Mohr's circle, column buckling, torsion of circular shafts, composite (transformed-section) beams, combined axial-flexural loading, transverse shear.
Question 8: Combined Axial + Bending Stress in a Cable-Supported T-Beam
Eccentricity of cable attachment at B (above beam's neutral axis)
0.2 m
Tip load
80 kN, downward, at B/tip (x = 8 m)
Section of interest
$x=4$ m from C
T-section
flange 100 mm × 15 mm (top); web 40 mm × 100 mm
Find. The normal stress distribution (sketch) and the maximum shear stress at the section $4$ m from C.
T-beam C-B: pin at C, cable A-B (0.2 m eccentric), 80 kN tip load at B; section 4 m from C.
Approach. Use global equilibrium (moment about C, using the cable's TRUE eccentric attachment point) to find the cable tension and pin reaction; the section 4 m from C lies entirely within the unloaded span C–B, so its internal axial force, shear, and moment come directly from that pin reaction alone.
Cable geometry and tension. With C at the origin and the beam's neutral axis along $y=0$, the cable runs from $A=(0,6.2)$ to $B=(8,0.2)$ (0.2 m above the beam line) — length $\sqrt{8^2+6^2}=10$ m exactly (a 6-8-10 triangle), direction cosines $(-0.8,\,0.6)$. Taking moments about C for the whole beam (tip load $80$ kN at $(8,0)$; cable force $T(-0.8,0.6)$ at $B=(8,0.2)$):
$$\sum M_C=0:\ \ T\big[8(0.6)-0.2(-0.8)\big]-8(80)=0\ \Rightarrow\ T(24)=640\ \Rightarrow\ T=129.03\text{ kN}$$
Pin reaction at C.
$$C_x=-T(-0.8)=103.23\text{ kN},\qquad C_y=80-T(0.6)=80-77.42=2.58\text{ kN}$$
Internal forces at $x=4$ m. This section lies between C and B — only $C_x,C_y$ act on the free body from C to the cut (the cable and tip load are both further out, at $x=8$):
$$N=-C_x=-103.23\text{ kN (i.e. 103.2 kN COMPRESSION)}$$
$$V=C_y=2.58\text{ kN}$$
$$M=C_y(4)=10.32\text{ kN}\!\cdot\!\text{m (sagging)}$$
(The eccentricity enters only through $T$ and the reactions — it does not add a separate local moment term here, since the section is well clear of the actual cable attachment point.)
T-section properties. Flange $100\times15$, web $40\times100$ (total depth 115 mm):
$$\bar y=\frac{(100)(15)(7.5)+(40)(100)(65)}{(100)(15)+(40)(100)}=49.32\text{ mm from the top}$$
$$I=\underbrace{\tfrac{100(15)^3}{12}+1500(49.32-7.5)^2}_{\text{flange}}+\underbrace{\tfrac{40(100)^3}{12}+4000(65-49.32)^2}_{\text{web}}=6.968\times10^{6}\text{ mm}^4$$
Combined axial + bending stress ($\sigma=N/A-My/I$, $y$ up from centroid).
$$A=5500\text{ mm}^2,\quad \sigma_{axial}=\frac{-103{,}230}{5500}=-18.77\text{ MPa}$$
$$\boxed{\sigma_{top}=-18.77-\frac{(10.32\times10^6)(49.32)}{6.968\times10^6}=-91.8\text{ MPa (compression)}}$$
$$\boxed{\sigma_{bottom}=-18.77+\frac{(10.32\times10^6)(65.68)}{6.968\times10^6}=+78.5\text{ MPa (tension)}}$$
Stress crosses zero between the flange/web junction ($-69.6$ MPa) and the bottom fibre — a linear ramp from $-91.8$ MPa at the top to $+78.5$ MPa at the bottom.
Maximum shear stress ($\tau=VQ/(Ib)$, at the neutral axis, $b=$ web width).
$$Q_{NA}=A_{above}\bar y_{above}=2872.7(30.0)=86{,}282\text{ mm}^3$$
$$\boxed{\tau_{max}=\frac{(2.58\times10^3)(86{,}282)}{(6.968\times10^6)(40)}=0.80\text{ MPa}}$$
