Rigid Beam on a Pin and Two Cables: 04-BS-6, December 2015, Q4

Question 4: Rigid Beam on a Pin and Two Cables (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.

Quantity	Value
Beam stations from A	B at 2 m, C at 3 m, D at 4 m
Cable at B	vertical, length 4 m (to the main ceiling)
Cable at D	vertical, length 1.6 m (to its own lower anchor)
Load at C	45 kN, downward
Cable	12 mm dia., E = 200 GPa, F_y = 350 MPa
Pin at A	30 mm dia., double shear

Rigid beam ABCD, pinned at A, with vertical cables at B (to the main ceiling) and D (to its own lower anchor).

Find. F_B, F_D; the vertical deflection at C; the shear stress in pin A.

Approach. The beam is rigid and pinned at A, so it rotates as one body through small angle θ; each cable's vertical stretch equals θ times its distance from A (compatibility), and its force follows from Hooke's law. Combine with moment equilibrium about A to solve for θ, then back out both forces.

Compatibility. The vertical drop of the rigid beam at B and D is $\theta x_B$ and $\theta x_D$; since each is also the cable's elastic stretch $FL/(AE)$, $$\theta = \frac{F_B L_B}{AE\,x_B} = \frac{F_D L_D}{AE\,x_D} \;\Rightarrow\; F_D = F_B\cdot\frac{L_B x_D}{L_D x_B} = F_B\cdot\frac{4(4)}{1.6(2)} = 5\,F_B$$
Moment equilibrium about A (cable tensions resist the load at C): $$F_B x_B + F_D x_D = P x_C \;\Rightarrow\; F_B(2)+5F_B(4) = 45(3) = 135\ \text{kN}\cdot\text{m}$$ $$\boxed{F_B = \frac{135}{22} = 6.14\ \text{kN}\qquad F_D = 5F_B = 30.68\ \text{kN}}$$ (Both stresses, 54.3 and 271.4 MPa, stay under the 350 MPa cable yield.)
Rotation and deflection at C. With $A=\pi(6)^2=113.1\ \text{mm}^2$, $$\theta = \frac{F_B L_B}{A E\,x_B} = \frac{(6136)(4000)}{113.1(200000)(2000)} = 5.426\times10^{-4}\ \text{rad}$$ $$\boxed{\delta_C = \theta\,x_C = 5.426\times10^{-4}(3000) = 1.628\ \text{mm, downward}}$$
Pin shear at A. Vertical equilibrium: $A_y = P - F_B - F_D = 45-6.14-30.68=8.18\ \text{kN}$ (no horizontal loads, so this is the full pin reaction). In double shear, $A_{pin}=\pi(15)^2=706.9\ \text{mm}^2$: $$\boxed{\tau_{pin} = \frac{A_y/2}{A_{pin}} = \frac{8182/2}{706.9} = 5.79\ \text{MPa}}$$

Quantity	Value
F_B	6.14 kN
F_D	30.68 kN
δ_C	1.628 mm, downward
τ_pin,A	5.79 MPa

Topic: Rigid-body compatibility with elastic cable supports (statically indeterminate to the first degree).

Key relations: small-rotation compatibility $\delta = \theta x$; cable stretch $\delta = FL/(AE)$; double-shear stress $\tau = V/(2A_{pin})$.

Why this works: A rigid beam pinned at one end and elastically restrained by two cables elsewhere has three unknown reactions (the two cable forces plus the pin) but only two independent equilibrium equations (moment about A, and vertical force), so it is statically indeterminate. The missing equation comes from geometry: because the beam is rigid, every point's vertical displacement is proportional to its distance from the pin, which directly relates the two cable stretches (and hence forces) to each other.

Common pitfall: Assuming the two cable forces split the load in proportion to distance from A (as if simply "sharing" the load) — that ignores that the STIFFER, SHORTER cable (D, only 1.6 m) attracts far more force per unit stretch than the longer cable at B, which is exactly what the compatibility equation captures ($F_D=5F_B$, not the 2:1 ratio distance alone would suggest).

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 4 (statically indeterminate axial members).

Question 4 of 8: Rigid Beam on a Pin and Two Cables

Question 4: Rigid Beam on a Pin and Two Cables (20 marks)