Question 6 of 8: Rigid Bar on a Pin and Two Cables — Statically Indeterminate
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2016 — 04-BS-6: Mechanics of Materials (3 hours, closed book,
one hand-written aid sheet permitted). Any FIVE of the eight questions constitute a complete
paper on the official exam; every question is answered below. A W-shape section-property table is attached at the end of the official exam; it is a red herring here — every question below supplies its own
built-up or standard section, and Question 2 cites the CISC-standard W610×125 properties
directly.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (composite
beams, deflection by integration, combined loading, transformation of stress, columns, torsion,
shear/moment diagrams); CISC Handbook of Steel Construction (W-shape section properties).
Question 6: Rigid Bar on a Pin and Two Cables — Statically Indeterminate (20 marks)
pin at A (y=0), B at y=6 m, C at y=9 m (rigid, vertical)
Distributed load
triangular, 0 at A rising to 20 kN/m at B, pushing horizontally
Concentrated load
25 kN at C, same direction
Cables (both horizontal, to fixed walls)
at B: 12 mm dia, L = 4 m; at C: 12 mm dia, L = 2 m; Fy = 400 MPa, E = 200 GPa
Pin at A
20 mm dia, double shear
Rigid bar on a pin, with triangular + point load and two horizontal cables.
Find. Cable forces TB, TC; horizontal displacement at C;
shear stress in the pin at A.
Approach. The pin provides two reactions and the two cables a third restraining
force, against only one useful equilibrium equation (moment about A — both force-equilibrium
equations either vanish trivially or just give the pin reaction), so the system is one-degree
statically indeterminate. Close it with small-rotation compatibility: since the bar is rigid, its
whole deformed shape is one rotation $\varphi$ about the pin, so each cable's elongation is
$\varphi$ times its attachment height.
Resultant of the triangular load.
$$R=\frac12(20)(6)=60\ \text{kN, acting at } y=\frac{2}{3}(6)=4.0\ \text{m from A}$$
Moment of the applied loads about A (CCW+, both loads push the same way):
$$M_{loads}=25(9)+60(4)=225+240=465\ \text{kN}\cdot\text{m}$$
Compatibility. For a small rotation $\varphi$ of the rigid bar about A, the
horizontal displacement (= cable elongation, since both cables are horizontal) at height $y$ is
$\delta=\varphi y$. With cable stiffness $k=A_cE/L$ ($A_c=\pi(6)^2=113.1\ \text{mm}^2$ for the
12 mm cables):
$$T_B=k_B(\varphi\, y_B)=\frac{A_cE}{4000}(6000\varphi)\qquad T_C=k_C(\varphi\,y_C)=\frac{A_cE}{2000}(9000\varphi)$$
Moment equilibrium about A (cable tensions resist the applied-load moment):
$$T_B\,y_B+T_C\,y_C=M_{loads}\;\Rightarrow\; 6000\!\left(1.5A_cE\varphi\right)+9000\!\left(4.5A_cE\varphi\right)=465\times10^6$$
Solving for $\varphi$ and back-substituting:
$$\varphi=4.153\times10^{-4}\ \text{rad}\qquad T_B=14.09\ \text{kN}\qquad T_C=42.27\ \text{kN}$$
(The shorter, stiffer cable at C also has the larger lever arm, so it attracts nearly three times
the force in B — both values are far below the 400 MPa yield stress on their 113.1 mm²
area.)
Horizontal displacement at C.
$$\delta_C=\varphi\,y_C=(4.153\times10^{-4})(9000)=\boxed{3.74\ \text{mm}}$$
Pin reaction and shear stress at A. All applied loads and both cable forces are
horizontal, so $A_y=0$ and:
$$A_x = 25+60-T_B-T_C = 85-14.09-42.27=28.64\ \text{kN}$$
$$\tau_{pin}=\frac{A_x/2}{A_{pin}}=\frac{14\,320}{\pi(10)^2}=\boxed{45.6\ \text{MPa (double shear)}}$$
