Question 6 of 8: Combined Axial and Bending Stress in an I-Beam
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2015 – 04-BS-6: Mechanics of Materials
3 hours duration, closed book (one hand-written aid sheet permitted). Any five of the eight
questions constitute a complete paper; all eight are solved below as a complete study resource.
Reference texts: R.C. Hibbeler, Mechanics of Materials, 10th ed.
(Ch. 5 torsion, Ch. 6 shear/moment diagrams & composite beams, Ch. 7 transverse shear,
Ch. 8 combined loadings, Ch. 9 stress transformation, Ch. 12 deflection by integration,
Ch. 13 buckling); F. Beer & E.R. Johnston, Mechanics of Materials; W. Gere &
B. Goodno, Mechanics of Materials (alternates). A wide-flange (W-shape) properties
table printed on the exam's last page supplies the section used in Question 2.
Question 6: Combined Axial and Bending Stress in an I-Beam (20 marks, 4 bonus)
Beam AB (pinned at A) propped by shallow strut BC (0.6 m horizontal, 0.2 m vertical), two 80 kN loads; I-beam cross-section and the resulting normal-stress distribution at the 0.7 m cut.
Given.
Quantity
Value
Beam AB
3 m, pinned at A; 80 kN at x=1 m and x=2 m (downward)
Strut BC offsets
0.6 m horizontal, 0.2 m vertical (pin-pin, withstands applied loads)
Section
I-beam: flanges 100×15 mm, overall depth 300 mm, web 40 mm thick
fy (normal / shear)
350 MPa / 60 MPa
Cut location
x = 0.7 m from A
Find. (a) The normal-stress distribution at x=0.7 m (max/min values). (b) The
maximum shear stress at the same section. (c, bonus) The plastic moment capacity of the section.
Approach. Strut BC is a shallow two-force member, so its force at B has both a
large horizontal and a smaller vertical component; find that force from moment equilibrium of AB
about A, then take a free body from A to the 0.7 m cut (before either 80 kN load) to get N, V and
M there, and combine $\sigma=N/A\pm Mc/I$ with $\tau_{max}=VQ/(It)$.
Strut force (moment equilibrium of AB about A). Strut unit vector
B→C $=(0.9487,-0.3162)$; compression pushes the beam at B in the OPPOSITE sense,
$(-0.9487,+0.3162)F$:
$$3(0.3162)F = 80(1)+80(2)=240 \ \Rightarrow\ F=253.0\ \text{kN}$$
$$F_x=-0.9487(253.0)=-240.0\ \text{kN}\ (\text{leftward on the beam}), \qquad
F_y=+0.3162(253.0)=80.0\ \text{kN (upward)}$$
Reactions at A.
$$A_x=+240.0\ \text{kN (rightward)}, \qquad A_y=160-80=80.0\ \text{kN (upward)}$$
Internal N, V, M at the cut (0.7 m from A, before either load). The only
loads on the free body from A to the cut are the pin reactions themselves:
$$N=-A_x=\boxed{-240.0\ \text{kN (compression, constant over the whole beam — no
horizontal loads anywhere)}}$$
$$V=A_y=\boxed{80.0\ \text{kN}}, \qquad M=A_y(0.7)=\boxed{56.0\ \text{kN}\cdot\text{m
(sagging)}}$$
(a) Combine axial and bending stress.
$$\sigma_N=\frac{N}{A}=\frac{-240{,}000}{13{,}800}=-17.4\ \text{MPa (uniform compression)}$$
$$\sigma_M=\frac{Mc}{I}=\frac{56\times10^6(150)}{126.6\times10^6}=66.4\ \text{MPa}$$
Top fibre (sagging → bending compression, adding to the axial compression):
$$\sigma_{top}=-17.4-66.4=\boxed{-83.7\ \text{MPa (compression, maximum magnitude)}}$$
Bottom fibre (bending tension, partly offset by axial compression):
$$\sigma_{bot}=-17.4+66.4=\boxed{+49.0\ \text{MPa (tension, minimum/net-tension value)}}$$
Both are well below the 350 MPa yield.
(b) Maximum shear stress ($\tau=VQ/(It)$, at the neutral axis).
$$Q_{NA}=(100\times15)(142.5)+(40\times135)(67.5)=213{,}750+364{,}500=578{,}250\
\text{mm}^3$$
$$\boxed{\tau_{max}=\frac{VQ_{NA}}{I\,t_w}=\frac{80{,}000(578{,}250)}{126.6\times10^6(40)}=
9.14\ \text{MPa}}\quad(<60\ \text{MPa allowable, OK})$$
(c, bonus) Plastic moment capacity. The plastic section modulus is twice the
first moment of the half-section about the NA (already computed as $Q_{NA}$ above):
$$Z=2Q_{NA}=1{,}156{,}500\ \text{mm}^3$$
$$\boxed{M_p=f_y\,Z=350(1{,}156{,}500)=404.8\ \text{kN}\cdot\text{m}}$$
