Question 5 of 8: Rigid Beam on a Pin and Two Cables
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2015 – 04-BS-6: Mechanics of Materials
3 hours duration, closed book (one hand-written aid sheet permitted). Any five of the eight
questions constitute a complete paper; all eight are solved below as a complete study resource.
Reference texts: R.C. Hibbeler, Mechanics of Materials, 10th ed.
(Ch. 5 torsion, Ch. 6 shear/moment diagrams & composite beams, Ch. 7 transverse shear,
Ch. 8 combined loadings, Ch. 9 stress transformation, Ch. 12 deflection by integration,
Ch. 13 buckling); F. Beer & E.R. Johnston, Mechanics of Materials; W. Gere &
B. Goodno, Mechanics of Materials (alternates). A wide-flange (W-shape) properties
table printed on the exam's last page supplies the section used in Question 2.
Question 5: Rigid Beam on a Pin and Two Cables (20 marks)
Rigid beam ABCD, pin at A; cable at B (1500 mm) and cable at C (750 mm) hang from a fixed ceiling; P = 200 kN applied at D.
Given.
Quantity
Value
Spacing A–B–C–D
1 m, 2 m, 1.4 m (4.4 m from A to D)
Cable diameter
30 mm (both)
Cable lengths
LB = 1500 mm, LC = 750 mm
Cable E, fy
200 GPa, 800 MPa
Load at D
P = 200 kN, downward
Pin at A
50 mm diameter, double shear
Find. (a) Cable forces FB, FC. (b) The vertical
displacement at D. (c) The shear stress in the double-shear pin at A.
Approach. The rigid beam is held by one pin and two cables — four
reaction unknowns against three equilibrium equations, so it is indeterminate to one degree;
close it with a rigid-body rotation compatibility equation (the beam rotates by one angle
θ about the pin at A), then finish with statics.
Compatibility. The beam is rigid and pinned at A, so every point's vertical
drop is proportional to its distance from A: $\delta_B=1\theta$, $\delta_C=3\theta$. With
$\delta=FL/(AE)$ (same A, E for both cables, but DIFFERENT lengths):
$$F_B=\frac{AE}{L_B}\,\delta_B=\frac{AE}{1.5}(1\theta), \qquad
F_C=\frac{AE}{L_C}\,\delta_C=\frac{AE}{0.75}(3\theta)$$
Moment equilibrium about A.
$$F_B(1)+F_C(3)=P(4.4)$$
Substituting the compatibility expressions ($A=\tfrac{\pi}{4}(30)^2=706.9\ \text{mm}^2$):
$$\left[\frac{AE}{1.5}(1)^2+\frac{AE}{0.75}(3)^2\right]\theta = 200(4.4)=880\
\text{kN}\cdot\text{m}$$
$$\left[0.667AE+12.0AE\right]\theta=12.667AE\,\theta=880 \ \Rightarrow\
\theta=4.914\times10^{-4}\ \text{rad}$$
(a) Cable forces.
$$\boxed{F_B=\frac{AE}{1.5}(1)\theta=46.3\ \text{kN}}, \qquad
\boxed{F_C=\frac{AE}{0.75}(3)\theta=277.9\ \text{kN}}$$
Check: $F_B(1)+F_C(3)=46.3+833.7=880.0=P(4.4)$. Cable stresses
$\sigma_B=F_B/A=65.5$ MPa, $\sigma_C=F_C/A=393.1$ MPa — both well below the 800 MPa
yield, confirming the linear-elastic compatibility solution is valid.
(b) Displacement at D.
$$\boxed{\delta_D=4.4\,\theta=4.4(4.914\times10^{-4})=2.16\ \text{mm, downward}}$$
(c) Pin shear at A. Vertical equilibrium (up positive):
$$A_y+F_B+F_C-P=0 \ \Rightarrow\ A_y=200-46.3-277.9=-124.2\ \text{kN}$$
i.e. the pin must pull DOWN 124.2 kN on the beam — the two cables together (324.2 kN)
over-carry the 200 kN load, so the pin supplies the balancing downward force. With double shear
($A_{pin}=\tfrac{\pi}{4}(50)^2=1963.5\ \text{mm}^2$, two shear planes):
$$\boxed{\tau_{pin}=\frac{|A_y|}{2A_{pin}}=\frac{124{,}200}{2(1963.5)}=31.6\ \text{MPa}}$$
