MECH 2010 Midterm Exam
Spring 2012
1. (40 pts) Simple questions
a. A circular bar with its cross sectional area of A is compressed by an axial force P. The
maximum normal stress inside the bar is P⁄A and there is no shear stress.
(i)
True
(ii)
False
Answer: False. The maximum normal stress is P/A, which exist on the cross-section plane.
But there is shear stress existing on inclined planes in this case. ………………………………..(10)
b. For most brittle materials, there is no noticeable difference between the ultimate stress
and the breaking stress (stress at rupture).
(i)
True
(ii)
False
Answer: True. For brittle materials, the typical stress-strain diagram is as follows. The
ultimate stress appears at point B and the rupture also happens at point B. So there is no
noticeable difference between the ultimate stress and the breaking stress. …………….(10)
c. The connection shown in the figure consists of five steel plates, each 3/16 in. thick,
jointed by a single ¼-in. diameter bolt. The total load transferred between the plates is
1200 lb, distributed among the plates as shown. Calculate the largest shear stress in the
bolt, disregarding friction between the plates.
Figure 1
Answer: For cross-section (a), the shear force is 360lb;
For cross-section (b), the shear force is 240lb;
For cross-section (c), the shear force is 240lb;
For cross-section (d), the shear force is 360lb; …………………………………………….(5)
Since the cross-sectional area of the bolt is uniform, so the maximum shear stress exists
on the plane (a) and (d), which equals
 max 
360
 7334(lb/in 2 ) ………………………………………………………………………….….(5)
1 1 2
( )
4 4
d. A composite bar is subject to a loading system as illustrated in Figure 2. The diameter of
the three identical cylindrical holes in the left section is d. This bar has a uniform width
(in the direction perpendicular to the paper). Mark clearly in the figure two vertical cross
sections with maximum and minimum normal stresses respectively.
(b) (a)
(c)
Figure 2
Answer: Denote the width of the bar is w. In the thick segment, the axial load is uniform
at different cross-sections, which equals to 3P. But the cross-sectional area is different at
different position. The maximum normal stress of this segment exists on the smallest
cross-section, which is at plane (a). The minimum normal stress of this segment exists on
the largest cross-section, which is at plane (b).
a 
3P
3P
3P

, b 
………………………………………………………………………….(5)
(2dw)  (dw) dw
2dw
In the thin segment, the axial load is 2P and the normal stress is uniform.
c 
2P
………………………………………………………………………………………………………………….….(2)
dw
Comparing the stresses at these three positions, the maximum stress appears at (a), the
minimum stresses appear at (b). ………….………………………………………………………………….….(3)
2.
(60pts). A clamped-clamped composite bar made of steel and aluminum is subject to a
concentrated force, 𝑃 = 1 kN , at its interface, which is located x distance away from the left
clamped end. The total length of the bar is 1 m and its cross sectional area is 10−4 m2.
(a) Find the distance x such that the reaction forces coming from the walls are equal.
(b) With the distance x determined from problem (a), find the reaction forces at both ends
when the temperature increases about 20𝑜 C.
The Young’s modulus for steel and aluminum are 𝐸𝑠 = 210 GPa and 𝐸𝑎𝑙 = 70 GPa. Their
thermal expansion coefficients are 𝛼𝑠 = 10 × 10−6 ⁄ 𝑜C and 𝛼𝑎𝑙 = 23 × 10−6 ⁄ 𝑜C respectively.
Figure 3
Answer: (a)
FA
FA
Fs
FA
Fs
Fs
………….….(10)
The free body diagrams of the whole bar and each segment are as above.
Force balance equation: FA  Fs  P
(1)
Additional equation:  A   s  0
(2) ………….………………………………………………………….….(10)
A 
F (1  x) (2) FA x Fs (1  x)
FE A
FA x
x
,s   s


0
 s A
EA A
Es A
EA A
Es A
(1  x) FA Es A
FA  Fs 
x
70GPa 1
1

  x  (m) ………….…………………………………………………….….(10)
1  x 210GPa 3
4
(b).
If temperature increases by 20𝑜 C, force balance equation (1) and additional equation (2) are still
valid. The only difference is that, the deformation of each segment is induced by the axial force
and the temperature change. ………….……………………………………………………………..……………….….(10)
A 
F (1  x)
FA x
  A xT ,  s   s
  s (1  x)T
EA A
Es A
(2)


F (1  x)
FA x
  A xT  s
  s (1  x)T  0 …………………………..…………………….….(10)
EA A
Es A
x 1/4

 Fs  FA  7420
Combine with (1), we can solve
FA  3210( N )
Fs  4210( N )
. Both two segments are compressed. ….(10)