Question 5 of 8: Statically Indeterminate Composite Axial Bar
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams December 2013 – 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 W-shape properties table printed on
the exam's last page is not needed — every question supplies its own section dimensions.
Question 5: Statically Indeterminate Composite Axial Bar (20 marks)
Composite bar: Al + 2×Cu in parallel (800 mm) in series with a steel bar (400 mm), pulled by P = 500 kN.
Given.
Member
A (mm2)
E (GPa)
fy (MPa)
Length (mm)
Aluminum (1)
1200
70
400
800
Copper (2, parallel)
600 each
100
350
800
Steel
2500
200
350
400
Applied load P = 500 kN, tension, at the free end.
Find. (a) The axial force carried by each bar. (b) The total horizontal
displacement of the free end.
Approach. The steel bar is the only load path in its segment, so it carries
the full 500 kN. In the left segment, the three parallel bars share the same rigid-plate
displacement (compatibility); combine that with equilibrium (their forces must sum to 500 kN) to
solve the once-indeterminate parallel system, then add the two segments' elongations.
Parallel bundle: stiffness and compatibility. Each bar shares the same
elongation $\delta$ over the common 800 mm length, so $F_i = k_i\delta$ with
$k_i = A_iE_i/L$:
$$k_{Al}=\frac{1200(70{,}000)}{800}=105{,}000\ \text{N/mm}, \qquad
k_{Cu}=\frac{600(100{,}000)}{800}=75{,}000\ \text{N/mm (each)}$$
$$\Sigma k = 105{,}000+2(75{,}000)=255{,}000\ \text{N/mm}$$
Solve for the shared elongation and each bar force.
$$\delta = \frac{P}{\Sigma k}=\frac{500{,}000}{255{,}000}=1.961\ \text{mm}$$
$$F_{Al}=k_{Al}\delta = \boxed{205.9\ \text{kN}}, \qquad
F_{Cu}=k_{Cu}\delta = \boxed{147.1\ \text{kN (each)}}$$
Check: $205.9+2(147.1)=500.0$ kN $=P$. σAl = 171.6 MPa ($<400$, OK);
σCu = 245.1 MPa ($<350$, OK) — both bars remain elastic.
Total end displacement. The steel segment's elongation adds in series to
the (common) parallel-bundle elongation:
$$\delta_{steel}=\frac{F_{steel}L_{steel}}{A_{steel}E_{steel}}
=\frac{500{,}000(400)}{2500(200{,}000)}=0.400\ \text{mm}$$
$$\boxed{\delta_{total}=\delta_{left}+\delta_{steel}=1.961+0.400=2.361\ \text{mm}}$$
