Question 3 of 6: Structural Mechanics

Question 3: Structural Mechanics (20 marks)

(a) Section properties of an L-frame on hull plate [10]

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. A built-up T-shaped section: bottom hull plate 320×10 mm, vertical web 10×200 mm rising from it, top flange 80×10 mm at the web tip — total depth 220 mm, all plate 10 mm thick.

Find. Neutral-axis location $\bar y$, moment of inertia $I_{NA}$, elastic section moduli $Z_{top}$/$Z_{bot}$, and plastic section modulus $Z_p$.

Approach. Composite-section method: locate the centroid by first moments of area, build $I_{NA}$ piece-by-piece with the parallel-axis theorem, then locate the separate equal-area (plastic) axis for $Z_p$.

1. i. Neutral axis (centroid). $$\bar y=\frac{\sum A_iy_i}{\sum A_i}=\frac{3200(5)+2000(110)+800(215)}{6000}=\frac{408{,}000}{6000}$$ $$\boxed{\bar y = 68.0\text{ mm above the base}}$$
2. ii. Moment of inertia about the N.A. Parallel-axis theorem on each piece, $I=I_{own}+Ad^2$: bottom plate contributes $12.73\times10^6$, web contributes $10.19\times10^6$, top flange contributes $17.29\times10^6\ \text{mm}^4$. Summing: $$\boxed{I_{NA} = 40.22\times10^6\ \text{mm}^4}$$
3. iii. Elastic section modulus. Extreme-fibre distances $c_{top}=220-68.0=152.0\text{ mm}$, $c_{bot}=68.0\text{ mm}$: $$Z_{top}=\frac{I_{NA}}{c_{top}}=\frac{40.22\times10^6}{152.0}=0.2646\times10^6\ \text{mm}^3,\qquad Z_{bot}=\frac{I_{NA}}{c_{bot}}=\frac{40.22\times10^6}{68.0}=0.5914\times10^6\ \text{mm}^3$$ The top fibre governs first yield ($Z_{top}$ is smaller, so it reaches $\sigma_y$ first under a given moment): $$\boxed{Z_{elastic}=0.2646\times10^6\ \text{mm}^3\ \text{(top-governing)}}$$
4. iv. Plastic section modulus — equal-area axis. Full plastification splits the 6000 mm² total area exactly in half (3000 mm² each side). Since the bottom plate alone (3200 mm²) already exceeds this half-area, the plastic N.A. lies within the bottom plate, at $$y_p=\frac{3000}{320}=9.375\text{ mm}$$
5. iv. Plastic section modulus — value. $Z_p=\sum A_i|y_i-y_p|$ summed piece-by-piece (splitting the bottom plate at $y_p$): $$Z_p = 3000(4.6875)+200(0.3125)+2000(100.625)+800(205.625)$$ $$\boxed{Z_p = 0.3799\times10^6\ \text{mm}^3}$$ Shape factor $Z_p/Z_{elastic}=0.3799/0.2646=1.44$, a typical value for an asymmetric T-section — consistent with the elastic–perfectly plastic idealization of Q1a.v.

(b) Beam bending — compound (hinged) beam [10]

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. A 3 m beam fixed at A, with an internal pin (hinge) at B, 2 m from A, and a roller support at C, 1 m further on; a UDL $w=50\text{ N/m}$ acts over the full B–C span; $EI=1.0\text{ kN}\!\cdot\!\text{m}^2$.

Find. SFD, BMD (with values), qualitative slope/deflection diagrams, and the vertical deflection at the pin B.

Approach. The internal hinge at B carries zero moment, so it splits the structure into two independently solvable pieces: solve the statically-determinate suspended span B–C first, then apply its reaction back onto the cantilever A–B as a tip point load.

1. Suspended span B–C (simply supported, full-span UDL). By symmetry of the UDL over the 1 m span, reactions at both ends are equal: $$R_B=R_C=\frac{wb}{2}=\frac{50(1)}{2}=25\text{ N}$$ Maximum sagging moment at midspan of BC: $$M_{max,BC}=\frac{wb^2}{8}=\frac{50(1)^2}{8}=\boxed{6.25\text{ N}\!\cdot\!\text{m (sagging)}}$$
2. Cantilever A–B (point load = the hinge reaction). By Newton's third law, the suspended span pushes down on the hinge with exactly $R_B=25\text{ N}$, which the cantilever must carry as a tip point load. Shear is constant along AB: $$V_{AB}=25\text{ N (constant)}$$ Moment is maximum (hogging) at the fixed support: $$M_A=-R_B\cdot a=-25(2)=\boxed{-50\text{ N}\!\cdot\!\text{m}}$$ falling linearly to zero at the hinge (consistent with the hinge carrying no moment).
3. Assembled SFD/BMD. $V(x)=+25\text{ N}$ throughout AB, then ramps from $+25$ to $-25\text{ N}$ linearly across BC. $M(x)$ rises linearly from $-50\text{ N}\!\cdot\!\text{m}$ at A to $0$ at the hinge B, then follows the UDL parabola in BC, peaking at $+6.25\text{ N}\!\cdot\!\text{m}$ at midspan and returning to $0$ at the roller C (as required by a simple support).
4. Slope and deflection (qualitative). Slope $\theta=0$ at the fixed end A and grows in magnitude toward B (cantilever curvature only, since AB carries no distributed load, only a tip point load $\Rightarrow$ constant curvature would only appear if $M$ were constant; here $M$ varies linearly so curvature does too, giving a smoothly increasing slope). Deflection $v=0$ at A, increasing (downward) to a local extreme at the pin B, then the suspended span BC sags between the (now-displaced) hinge and the ground-fixed roller at C, returning to $v=0$ at C.
5. Deflection at the pin (part ii). The pin's physical position is exactly the tip of cantilever AB, so its deflection is the standard cantilever tip-deflection formula for an end point load (from the supplied beam table), using $P=R_B=25\text{ N}=0.025\text{ kN}$ and $a=2\text{ m}$: $$v_B=\frac{Pa^3}{3EI}=\frac{0.025(2)^3}{3(1.0)}=\frac{0.2}{3}=\boxed{0.0667\text{ m}=66.7\text{ mm (downward)}}$$