22-Mec-B9 Advanced Engineering Structures · December 2014
Question 4 of 8: Thermal stress in a restrained two-material bar
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2014 — 07-Mec-B9 Advanced Engineering Structures. Three hours, open book, any non-communicating calculator permitted. Eight questions of equal total value (20 marks each); the rubric states that any five constitute a complete paper. All eight are solved here, because the complete set is the study resource.
Reference texts.
T. H. G. Megson, Aircraft Structures for Engineering Students, 6th ed. — open and closed thin-walled sections, shear flow, shear centre, structural idealisation (Ch. 16, 17, 18, 20, 21, 23).
A. C. Ugural and S. K. Fenster, Advanced Strength and Applied Elasticity, 5th ed. — yield criteria, torsion of thin-walled members, statically indeterminate axial members (Ch. 2, 4, 6).
A. P. Boresi and R. J. Schmidt, Advanced Mechanics of Materials, 6th ed. — torsion of open and closed sections, multiply connected cells (Ch. 6).
N. E. Dowling, Mechanical Behavior of Materials, 4th ed. — linear elastic fracture mechanics, the Paris law, strain-life and Miner’s rule (Ch. 8, 11, 14).
R. C. Hibbeler, Mechanics of Materials, 10th ed. — thermal stress in restrained members and combined loading (Ch. 4, 8).
Axes and sign convention used throughout. A right-handed set is used for every thin-walled question: the span axis runs along the member, $Y$ is vertical (upward) in the cross-section and $Z$ is horizontal. Section constants are written $I_{ZZ}=\int Z^{2}\,dA$, $I_{YY}=\int Y^{2}\,dA$ and $I_{YZ}=\int ZY\,dA$ about the centroid, and the bending stress is carried as the linear field $\sigma = aZ + bY$. Shear flows are counted positive in the direction of the walk stated with each figure; a negative answer simply means the flow runs the other way round the circuit.
Question 4: Thermal stress in a restrained two-material bar (20 marks)
Given. Two dissimilar rods welded end to end at B and built in at both ends, heated uniformly. Working in newtons and millimetres throughout: 10 cm2 is 1 000 mm2, 165 cm is 1 650 mm, and 165 GPa is $165\times10^{3}$ MPa.
Quantity
Symbol
Value
Rod 1 modulus, area, length
$E_1, A_1, L_1$
165 GPa, 1 000 mm2, 1 650 mm
Rod 1 expansion coefficient
$\alpha_1$
$13\times10^{-6}\ \text{K}^{-1}$
Rod 2 modulus, area, length
$E_2, A_2, L_2$
90 GPa, 1 500 mm2, 1 100 mm
Rod 2 expansion coefficient
$\alpha_2$
$25\times10^{-6}\ \text{K}^{-1}$
Uniform temperature rise
$\Delta T$
70 K
End conditions
—
A and C rigidly fixed
Find. (a) the axial stress carried by each rod after the temperature rise, and (b) how far and in which direction the welded joint B moves.
Figure Q4 — the two-segment rod between rigid supports. Rod 2 is the softer and the more expansive of the two, so it wins the tug of war and drives B toward A.
Approach. Release one support, let the assembly expand freely, then apply the single redundant force that pushes the released end back to its original position; because the bar is a series chain, that force is common to both rods and the compatibility equation has one unknown.
Part (a) — free thermal expansion of the released assembly. Cut the bar loose at C and let it grow: $$\delta_T = \alpha_1 L_1 \Delta T + \alpha_2 L_2 \Delta T$$which is $13\times10^{-6}(1650)(70) + 25\times10^{-6}(1100)(70) = 1.5015 + 1.9250$, so $\delta_T = 3.4265$ mm. Rod 2 contributes more despite being shorter, because its expansion coefficient is nearly twice as large.
Series flexibility of the chain. The same axial force $F$ passes through both rods, so their elastic extensions add: $$f = \frac{L_1}{E_1 A_1} + \frac{L_2}{E_2 A_2} = \frac{1650}{1.65\times10^{8}} + \frac{1100}{1.35\times10^{8}}$$giving $f = 1.0000\times10^{-5} + 0.8148\times10^{-5} = 1.8148\times10^{-5}$ mm/N.
Compatibility at the restored support. The rigid supports permit no net length change, so the elastic deformation must exactly cancel the thermal one: $\delta_T + fF = 0$, hence $$\boxed{F = -\frac{\delta_T}{f} = -\frac{3.4265}{1.8148\times10^{-5}} = -188.8\ \text{kN}}$$The negative sign means compression, as expected when heated material is prevented from growing.
Axial stresses. Dividing the common force by each area, $$\sigma_1 = \frac{-188\,807}{1000} = -188.8\ \text{MPa},\qquad \sigma_2 = \frac{-188\,807}{1500} = -125.9\ \text{MPa}$$both compressive. The slender rod is the more highly stressed simply because it is the smaller in section; the stresses are in inverse proportion to the areas, not to the moduli.
Part (b) — displacement of joint B, computed from A. Rod 1 experiences its own thermal growth plus its elastic shortening: $$u_B = \alpha_1 L_1 \Delta T + \frac{F L_1}{E_1 A_1} = 1.5015 - 1.8881 = -0.3866\ \text{mm}$$The negative result is a movement in the direction of A.
Check the same displacement from C. Working back from the other support, rod 2 grows $1.9250$ mm thermally and shortens $1.5385$ mm elastically, a net extension of $0.3866$ mm, which places B exactly $0.3866$ mm nearer to A. The two independent routes agree, so $$\boxed{|u_B| = 0.387\ \text{mm, directed from B toward support A}}$$Physically, rod 2 is both the more expansive and the more compliant member, so it pushes the joint back into rod 1.
