I-Beam Cantilever – Bending and Shear Stress: 04-BS-6, December 2013, Q7

Question 7: I-Beam Cantilever – Bending and Shear 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.

Cantilever I-beam, 35 kN at the 4 m tip; section 200×300 mm overall, 10 mm flanges, 15 mm web.

Given.

Quantity	Value
Length, tip load	4 m, 35 kN
Flanges	200 mm × 10 mm (top and bottom)
Web	15 mm thick, 280 mm clear height
Overall depth	300 mm
Allowable σ / τ	240 MPa / 60 MPa

Find. (a) Maximum absolute normal and shear stress in the beam. (b) The shear stress at point E (outer tip of the bottom flange), with justification.

Approach. Compute I by parallel-axis composition of the web and two flanges; the maximum moment and constant shear both occur at the fixed base; use $\sigma=Mc/I$ and $\tau=VQ/(It)$ at the neutral axis; argue point E from the free-surface condition.

Section properties (about the strong axis). $$I_{web}=\frac{15(280)^3}{12}=27.44\times10^6\ \text{mm}^4$$ $$I_{flange}= \frac{200(10)^3}{12}+200(10)(145)^2 = 42.07\times10^6\ \text{mm}^4\ \text{(each, about the NA, }d=145\text{ mm)}$$ $$I = I_{web}+2I_{flange} = \boxed{111.6\times10^6\ \text{mm}^4}, \qquad c=150\ \text{mm}$$
Maximum normal stress. Fixed-end moment $M=35(4)=140$ kN·m (hogging, maximum at the wall since it is a single tip load with no other loading): $$\sigma_{max}=\frac{Mc}{I}=\frac{140\times10^6(150)}{111.6\times10^6} =\boxed{188.2\ \text{MPa}}\quad(<240\ \text{MPa allowable, OK})$$
Maximum shear stress (at the neutral axis). $V=35$ kN throughout (single tip load, no distributed load). First moment of the area above the NA: $$Q=A_{flange}\,d_{flange} + \left(t_w\cdot\frac{h_w}{2}\right)\frac{h_w}{4} =2000(145)+(15\times140)(70)=437{,}000\ \text{mm}^3$$ $$\tau_{max}=\frac{VQ}{It_w}=\frac{35{,}000(437{,}000)}{111.6\times10^6(15)} =\boxed{9.14\ \text{MPa}}\quad(<60\ \text{MPa allowable, OK})$$
Part (b): shear stress at point E. Point E sits at the very outer tip (free edge) of the bottom flange — the first moment of area "beyond" that point, in the direction shear flow would need to develop, is $Q=0$ (there is no material further out to transfer force to). A free surface cannot sustain a complementary shear stress, so $$\boxed{\tau_E = 0}$$ Shear stress in a thin-walled open section always falls to zero at every unconnected free edge; it only builds up moving inward along the flange (feeding the shear flow into the web).

Final Results.

Quantity	Value
I	111.6 × 10⁶ mm⁴
(a) Maximum normal stress	188.2 MPa (< 240 MPa allowable)
(a) Maximum shear stress	9.14 MPa, at the NA (< 60 MPa allowable)
(b) Shear stress at E	0 (free edge)

Topic: Bending and transverse-shear stress in a built-up I-section.

Key relations: $\sigma=Mc/I$; $\tau=VQ/(It)$, with $t$ the section width AT the level being examined (the web thickness at the NA, and effectively zero "beyond" a free tip).

Why this works: The parallel-axis theorem lets the I-section's moment of inertia be built from simple rectangular pieces; the shear-flow argument that gives $\tau_E=0$ follows directly from the fact that shear flow $q=VQ/I$ must vanish wherever $Q=0$, which is true at any free (unconnected) edge by definition.

Common pitfall: Trying to compute a non-zero $\tau$ at point E by plugging numbers into $VQ/(It)$ using the flange thickness as $t$ without first recognizing $Q=0$ there — the formula still gives zero once $Q$ is correctly taken as zero, but skipping the free-edge reasoning costs the "justify your answer" marks.

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 6 (composite/built-up beams) & Ch. 7 (transverse shear).

Question 7 of 8: I-Beam Cantilever – Bending and Shear Stress

Question 7: I-Beam Cantilever – Bending and Shear Stress (20 marks)