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(CIVL112)[2011](s)midterm~2985^_70243.pdf
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CIVL 112 Mechanics of Materials Midterm Apr. 13, 2011
Solutions
1 (5%). A circular steel bar (E = 207 GPa) has length L = 1.2 m and diameter d = 20 mm. If it is under uniaxial tension and has acquired an elastic strain energy of U = 42 J, what is the axial stress in this bar (to nearest MPa)?
A. 200 MPa B. 205 MPa C. 210 MPa
E. None of the above
2 (10%). Consider two circular shafts of the same length and material. Shaft A has a solid cross section, while B is hollow with inner diameter din = 0.8 dout (outer diameter). The two shafts have equal cross-sectional areas and hence the same weight. If the same (constant) torque T is found in each shaft, the
C. .A = 8.31 .B D. .B = 8.31 .A
E. None of the above
3 (10%). Consider a hollow circular steel shaft (shear modulus G = 82 GPa) in uniform torsion. It has outer diameter 100 mm and inner diameter 50 mm, and is required to twist no more than 1.5 degrees over a 2 m length. What is the maximum permissible shear stress (to nearest MPa) in the shaft?
A. 12 MPa B. 29 MPa D. 68 MPa
4. For the system shown below, if a 0.5 mm initial gap exists when the temperature is 24C, find the
temperature (to 0.1C) at which the normal stress in the aluminum bar will become -75 MPa, and also
the corresponding new length (to 0.001 mm) of the aluminum bar.
(35%)
Solution: Let subscripts A, B denote aluminum and bronze, respectively. After expansion, the system is stopped by the wall and stressed. Hence, with .A = -75 MPa in aluminum bar, the force in it is
PA = .A AA = (-75106)(N/m2)180010-6 m2 = 135,000 N
By FBD analysis, the same force exists in both the aluminum and bronze bars . PA = PB = P. The compression in the whole assembly is therefore
..
..
PBLB PALA 0.35m 0.45m
.P ... 135000N . ..
9 .62 9 .62
EBAB EAAA N N .
.
105 .10 2 .1500 .10 m 73.10 2 .1800 .10 m .
. mm .
= 0.3 mm + 0.4623 mm = 0.762329 mm = 0.000762 m (or 0.762329 mm) IF the assembly were free to expand, the thermal elongation (in meters) would have been .T = LB..B .T... LA..A .T = .T (0.35 m 21.610-6 + 0.45 m 23.210-6) .T = 1810-6 .T This .T is partly taken up by the 0.5 mm gap, while the rest of it is resisted by the force in the bars.
Hence, .T = 0.0005 m + .P = 0.5 mm + 0.762329 mm = 1.262329 10-3 m Thus 0.0005m ..P 1.262329 .10.3
.T .. . 70.1294.C
.6 .6
18.10 18.10 . new temperature = (24 + 70.1) = 94.1C True elongation of the aluminum bar is therefore
PALA
.. (. ) . (. ) . L ..T .
A TA PA AA
EAAA
= 0.732151 C 0.462329 = 0.269822 mm . The new length of the aluminum bar is 450 + .A = 450.270 mm
3.(40%) As shown, a steel bar ABC of constant
circular cross section is rigidly fixed to a wall at its left end A. The right end C is attached to two vertical steel rods MN and QP (thus elastically restrained against free rotation). These thin rods are attached to the bar at points N and Q, with NQ being a 80 mm horizontal line (= diameter of bar ABC). A 6000 Nm t