T-Beam – Bending and Shear Stress: 04-BS-6, May 2014, Q7

Question 7: T-Beam – 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.

Simply supported T-beam, 20 kN at mid-span (8 m span); T-section: 150×25 mm flange over a 150×30 mm web.

Given.

Quantity	Value
Span, load	8 m, 20 kN at mid-span
Flange	150 mm wide × 25 mm thick (top)
Web	150 mm deep × 30 mm wide (below the flange)
Allowable σ / τ	260 MPa / 60 MPa

Find. (a) Maximum absolute normal and shear stress. (b) The shear stress at point E (tip of the flange) at the left support, with justification.

Approach. Locate the centroid and I by composing the web and flange rectangles, apply $\sigma=Mc/I$ at the larger of the two centroidal distances and $\tau=VQ/(It)$ at the neutral axis; argue point E from the free-edge condition on shear flow.

Reactions and maximum moment. Symmetric point load at mid-span: $$R_A=R_B=\frac{20}{2}=10\ \text{kN}, \qquad M_{max}=R_A(4)=40\ \text{kN}\cdot\text{m}$$ Shear is constant at 10 kN along each half-span, including at the left support.
Centroid (y measured from the bottom of the web). $$A_{web}=30(150)=4500\ \text{mm}^2\ (y=75), \qquad A_{fl}=150(25)=3750\ \text{mm}^2\ (y=162.5)$$ $$\bar y = \frac{4500(75)+3750(162.5)}{8250} = \boxed{114.8\ \text{mm from the bottom}}$$
Moment of inertia (parallel-axis theorem). $$I_{web}=\frac{30(150)^3}{12}+4500(75-114.8)^2 = 15.56\times10^6\ \text{mm}^4$$ $$I_{fl}=\frac{150(25)^3}{12}+3750(162.5-114.8)^2 = 8.74\times10^6\ \text{mm}^4$$ $$\boxed{I = 24.29\times10^6\ \text{mm}^4}, \qquad c_{top}=60.2\ \text{mm},\ c_{bot}=114.8\ \text{mm}$$
(a) Maximum absolute normal stress. Since $c_{bot}>c_{top}$, the bottom (tension) fibre governs: $$\sigma_{bot}=\frac{Mc_{bot}}{I}=\frac{40\times10^6(114.8)}{24.29\times10^6} = \boxed{189.0\ \text{MPa (tension)}}\quad(<260\ \text{MPa allowable, OK})$$ ($\sigma_{top}=99.2$ MPa compression, smaller in magnitude.)
(a) Maximum shear stress (at the neutral axis, within the web). $$Q_{NA} = (30\times114.8)\!\left(\frac{114.8}{2}\right) = 197{,}600\ \text{mm}^3$$ $$\tau_{max}=\frac{VQ_{NA}}{I\,t_w}=\frac{10{,}000(197{,}600)}{24.29\times10^6(30)} = \boxed{2.71\ \text{MPa}}\quad(<60\ \text{MPa allowable, OK})$$
(b) Shear stress at point E. Point E sits at the very tip (free edge) of the top flange, where two boundaries meet: the top surface (no traction above it) and the flange's own end (no material beyond it sideways). The shear flow needed to develop $\tau$ at a point is $q=VQ/I$, where Q is the first moment of area BEYOND that point toward the free surface — at the outer tip itself there is no material beyond it in either direction, so $Q=0$ there. A free (unloaded) surface cannot sustain a complementary shear stress either, confirming the same conclusion independently: $$\boxed{\tau_E = 0}$$

Final Results.

Quantity	Value
ȳ, I	114.8 mm from bottom, 24.29 × 10⁶ mm⁴
(a) Maximum absolute normal stress	189.0 MPa (tension, bottom fibre)
(a) Maximum shear stress	2.71 MPa, at the neutral axis
(b) Shear stress at E	0 (free edge — no material/traction beyond it)

Topic: Bending and transverse shear in an unsymmetric (T-shaped) built-up section.

Key relations: $\sigma=Mc/I$, evaluated at BOTH extreme fibres since an unsymmetric section has $c_{top}\ne c_{bot}$; $\tau=VQ/(It)$, which is identically zero at any unconnected free edge because $Q=0$ there by definition.

Why this works: The parallel-axis theorem lets an unsymmetric built-up section be composed from simple rectangles about a single centroidal axis; the free-edge shear argument follows directly from the physical requirement that shear flow can only exist where there is material further out to transfer force to.

Common pitfall: Assuming the LARGER geometric dimension (here the flange width) automatically locates the maximum stress fibre — for a T-section it is the SMALLER centroidal distance (top) that is in compression and the LARGER one (bottom, deep into the web) that governs the absolute maximum, because the centroid sits well up inside the flange-heavy top.

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

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

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