Transcript Home Work #1 - Hong Kong University of Science and Technology

Home Work #3
Due Date: 11 Mar, 2010
(Turn in your assignment at the mail box
of S581 outside the ME general office)
The solutions must be written on single-side A4 papers only.
HW 3-Problem #1
A steel rod 110 ft long hangs inside a tall tower and holds a 200-pound
weight at its lower end (see figure).
If the diameter of the circular rod is ¼ inch, calculate the maximum
normal stress σmax in the rod, taking into account the weight of the rod
itself.
Weight density of steel: 490lb/ft3
The maximum force Fmax in the rod occurs at the upper end
1 1 2 2
Wo   AL  (490lb / ft )( (  ) ft )(110 ft )
8 12
 18.37lb
W  200lb
F
W  WO (200  18.37)lb
 max  max 

 4448.67 psi
1
A
A
 ( )2
8
3
HW 3-Problem #2
A loading crane consisting of a steel girder ABC supported by a cable BD
is subjected to a load P. The cable has an effective cross-sectional area
A=0.471 in2. The dimensions of the crane are H=9 ft, L1=12 ft, and L2=4
ft.
a.If the load P=9000 lb, what is the average tensile stress in the cable?
b.If the cable stretches by 0.382 in, what is the average strain?
Ray
TBy
Rax
TBx
A
(a)
 Ma  0
C
B
P
( L1  L 2) P  L1TBy
TBy  12000lb
15
TB  TBy ( )  20000lb
9
TB 20000lb
 average  
 424628.5 psi
2
A 0.471in

0.382in

 2.122 103
(b)  average 
L0 (15 12)
HW 3-Problem #3
The data shown in the accompanying table were obtained from a tensile
test of high-strength steel. The test specimen had a diameter of 0.505 in
and a gage length of 2 in. At fracture, the elongation between the gage
marks was 0.12 in and the minimum diameter was 0.42 in.
Plot the conventional stress-strain curve for the steel and determine the
proportional limit, modulus of elasticity (i.e., the slope of the initial part
of the stress-strain curve), yield stress at 0.1% offset, ultimate stress,
percent elongation in 2 in, and percent reduction in area.
HW 3-Problem #3
Data for the conventional stress-strain curve
Ultimate stress
112832.9lb/in2
Yield stress
69000lb/in2
65000
lb/in2
Proportinal limit
0.1%
As shown in the plot,
Proportinal limit≈ 65000 lb/in2
(Any two point in the proportinal
59911.28139  4992.606
 29686ksi line is OK)
Young’s Modulus =
0.00`95  0.0001
Yield stress of 0.1% offset ≈ 69000 lb/in2
Ultimate stress ≈ 112832.9 lb/in2
Percentage of elongation =0.12/2 = 6% 2
1  (0.525  0.422 )
 30.8 %
Percentage of reduction in area = 4 1
2
4
 0.525
HW 3-Problem #4
A bar of monel metal as in the figure (length L=8 in, diameter d=0.25 in) is
loaded axially by a tensile force P=1500 lb. Determine the increase in length
of the bar and the percent decrease in this cross-sectional area.
Monel metal material property:
Modulus of elasticity E=25000 ksi
Poisson’s Ratio ν=0.32
PL
E
X A
1500  8
3
25000 10 



9.7785

10
X
 X (0.125)2 
3
 d   X (0.32)  3.9104 104
d  3.9104 104  0.25  9.776  105
Percentage of reduction in area
A 1/ 4 (0.252  (0.25  9.776 105 ) 2 )

 0.078
2
A
1/ 4 (0.25)
HW 3-Problem #5
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.
a.Calculate the largest shear stress in the bolt, disregarding friction
between the plates.
b.Calculate the largest bearing stress acting against the bolt.
360
A
B
C
480
600
600
D
360
360
A
①
360
②
A
600
240
360
③
B
A
B
C
480
360
④
360
A
B
C
480
360
D
600
240
600
600
As shown in the figure above,
the shear force of A, D= 360 lb
The shear force of B,C= 240lb
The largest shear force is:
Vmax
360
 max 

 7.334 103 lbin2
A
 ( 1 )2
4 4
The largest bearing force is:
 max
Fmax
600


 1.28 104 lbin 2
1 3
A

4 16
HW 3-Problem #6
An elastomeric bearing pad consisting of two steel plates bonded to a
chloroprene elastomer (an artificial rubber) is subjected to a shear force
V during a static loading test. The pad has dimensions a=150 mm and
b=250mm, and the elastomer has thickness t=50 mm. When the force V
equals 12 kN, the top plate is found to have displaced laterally 8.0mm
with respect to the bottom plate.
What is the shear modulus of elasticity G of the chloroprene.
 aver 
V
ab
Shear strain :

 aver
Ge

V
abGe
d
   0.16
t
Ge 
V
ab
120000
6
2


2

10
Nm
150  103  250  103  0.16
HW 3-Problem #7
A steel bar AD has a cross-sectional area of 0.4 in2 and is loaded by forces
P1=2700 lb, P2=1800 lb, and P3=1300 lb. The lengths of the segments of
the bar are a=60 in, b=24 in, and c=36 in.
a.Assuming that the modulus of elasticity E=30*106 psi, calculate the
change in length δ of the bar. Does the bar elongate or shorten?
b.By what amount P should the load P3 be increased so that the bar does
not change in length when the three loads are applied?
F3
P3
C
D
F2
P2-P3
B
C
F1
A
P3  LCD
1300  36
3
 CD 



3.9

10
in
6
E A
30 10  0.4
( P2  P3 )  LBC (1800  1300)  24
3
 BC 


1

10
in
6
E A
30 10  0.4
( P1  P2  P3 )  LAB (1800  1300)  60
 AB 

 0.016in
6
E A
30 10  0.4
 total  0.0131in(elongate)
b) CD   BC   AB  0
36 P3  (1800  P3 )24  (2700  1800  P3 )60  0
P3  2610lb
P3 has to be increased by 2610-1310=1310 lb
P1+P2-P3
B
HW 3-Problem #8
A bar ABC of length L consists of two parts of equal lengths but different
diameters. Segment AB has diameter d1=100 mm, and segment BC has
diameter d2=60 mm. Both segments have length L/2=0.6 m. A
longitudinal hole of diameter d is drilled through segment AB for one-half
of its length (distance L/4=0.3 m). The bar is made of plastic having
modulus of elasticity E=4.0 Gpa. Compressive loads P=110 kN act at the
ends of the bar.
If the shortening of the bar is limited to 8.0mm, what is the maximum
allowable diameter dmax of the hole?
A’
 AA '   A ' B   BC  0.008m
110 103  0.6
110 103  0.3
110 103  0.3


9
3 2
9
3 2
4 10 (3 10 )  4 10 (50 10 )  4 109 ((50 103 )2  ( d )2 )
2
d
 (50 103 )2  ( )2  2.357 103
2
d  23.87mm