Question 3 of 8: Normal and Shear Stress in the Beam

Question 3: Normal and Shear Stress in the 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. Same beam, loading and I-section as Questions 1–2. From Q1: $M_{max}=133.3$ kN·m at x=2 m, and $V_{max}=80$ kN at the B end (x=6 m). Point E is the outer tip of the bottom flange (the free edge of the cross-section).

Find. $\sigma_{max}$, $\tau_{max}$ in the beam, and $\tau_E$ at x = 2 m.

Approach. Maximum normal stress uses the flexure formula with the GLOBAL maximum moment; maximum shear stress uses the shear formula with the global maximum shear, evaluated at the neutral axis; the flange-tip shear follows from the definition of Q at a free edge.

1. Maximum normal stress. The flexure formula is $\sigma=Mc/I$ with $c=D/2=150$ mm and $I=93.78\times10^{6}\ \text{mm}^4$, evaluated at the governing moment $M_{max}=133.3$ kN·m (from Q1): $$\sigma_{max}=\frac{(133.3\times10^{6}\ \text{N}\!\cdot\!\text{mm})(150\ \text{mm})}{93.78\times10^{6}\ \text{mm}^4}=\boxed{213.3\ \text{MPa}}$$ well under the 350 MPa yield — tension on the bottom fibre, compression on the top fibre (M is positive/sagging everywhere on this span, see Q1).
2. Maximum shear stress. The shear formula is $\tau=VQ/(It)$, evaluated at the neutral axis (where $Q$ is largest) using the web thickness $t=20$ mm and the governing shear $V_{max}=80$ kN (at x=6 m): $$Q_{NA}=396\times10^{3}\ \text{mm}^3, \qquad \tau_{max}=\frac{(80\times10^{3})(396\times10^{3})}{(93.78\times10^{6})(20)}=\boxed{16.89\ \text{MPa}}$$ well under the 75 MPa shear yield.
3. Shear stress at the flange tip, point E. The shear formula's $Q$ is the first moment of the area BEYOND the point in question, taken about the neutral axis. At the very outer tip of the flange (point E) there is no material further out — the tip is a free surface with no applied traction. Formally, $Q_E=0$ because zero area lies beyond that point, so $$\boxed{\tau_E=\frac{V\,Q_E}{It}=\frac{V(0)}{It}=0}$$ This holds regardless of $x$ (so in particular at x=2 m): the shear stress must vanish at any free edge of the cross-section, by the complementary-shear argument (there is no adjoining material across that surface to supply a reactive shear traction). This is a general property of the shear-flow distribution, not a coincidence of this particular section.