# HW01-MAE314-002 S12

Chapter 1

PROBLEM 1.1
Two solid cylindrical rods AB and BC are welded together at B and
loaded as shown. Knowing that the average normal stress must not
exceed 175 MPa in rod AB and 150 MPa in rod BC, determine the
smallest allowable values of d
1
and d
2
.

PROBLEM 1.1
Two solid cylindrical rods AB and BC are welded together at B and
loaded as shown. Knowing that the average normal stress must not
exceed 175 MPa in rod AB and 150 MPa in rod BC, determine the
smallest allowable values of d
1
and d
2
.

SOLUTION 1.1

(a) Rod AB

3
2 2
1 1
4
3
3
1
6
40 30 70 kN 70 10 N
4
4 (4)(70 10 )
22.6 10 m
(175 10 )
t
o
t
to t
÷
= + = = ×
= = =
×
= = = ×
×
AB
AB
AB
P
P P P
A d d
P
d
1
22.6 mm = d 
(b) Rod BC

3
2 2
2 2
4
3
3
2
6
30 kN 30 10 N
4
4 (4)(30 10 )
15.96 10 m
(150 10 )
t
o
t
to t
÷
= = ×
= = =
×
= = = ×
×
BC
BC
BC
P
P P P
A d d
P
d
2
15.96 mm = d 

HW01-MAE314-002 S12

PROBLEM 1.11
The frame shown consists of four wooden members, ABC,
DEF, BE, and CF. Knowing that each member has a 2 × 4-in.
rectangular cross section and that each pin has a 1/2-in.
diameter, determine the maximum value of the average
normal stress (a) in member BE, (b) in member CF.

SOLUTION 1.11

Stress in tension member CF

Add support reactions to figure as shown.
Using entire frame as free body,
0: 40 (45 30)(480) 0
900 lb.
A x
x
M D
D
E = ÷ + =
=

Use member DEF as free body.
Reaction at D must be parallel to
BE
F and .
CF
F

4
1200 lb.
3
y x
D D = =

4
0: (30) (30 15) 0
5
2250 lb.
4
0: (30) (15) 0
5
750 lb.
| |
E = ÷ ÷ + =
|
\ .
= ÷
| |
E = ÷ =
|
\ .
=
F BE Y
BE
E CE Y
CE
M F D
F
M F D
F

Stress in compression member BE
Area:
2
2 in 4 in 8 in = × = A
(a)
2250
8
o
÷
= =
BE
BE
F
A
281 psi o = ÷
BE

Minimum section area occurs at pin.

2
min
(2)(4.0 0.5) 7.0 in = ÷ = A
(b)
min
750
7.0
o = =
CF
CF
F
A
107.1 psi o =
CF


HW01-MAE314-002 S12

PROBLEM 1.17
A load P is applied to a steel rod supported as shown by an
aluminum plate into which a 0.6-in.-diameter hole has been
drilled. Knowing that the shearing stress must not exceed
18 ksi in the steel rod and 10 ksi in the aluminum plate,
determine the largest load P that can be applied to the rod.

For steel:
1
2
(0.6)(0.4)
0.7540 in
t t = =
=
A dt

1 1 1
(0.7540)(18)
13.57 kips
t t = = =
=
P
P A
A
For aluminum:
2
2
(1.6)(0.25) 1.2566 in t t = = = A dt

2 2 2
2
(1.2566)(10) 12.57 kips t t = = = =
P
P A
A

Limiting value of P is the smaller value, so 12.57 kips = P 

PROBLEM 1.26
Link AB, of width b = 50 mm and thickness t = 6 mm, is used to support
the end of a horizontal beam. Knowing that the average normal stress in
the link is -140 MPa, and that the average shearing stress in each of the
two pins is 80 MPa, determine (a) the diameter d of the pins, (b) the
average bearing stress in the link.

HW01-MAE314-002 S12

Rod AB is in compression.
= A bt where 50 mm = b and 6 mm = t

6 2
6 6
3
(0.050)(0.006) 300 10 m
( 140 10 )(300 10 )
42 10 N
o
÷
÷
= = ×
= ÷ = ÷ ÷ × ×
= ×
A
P A

For the pin,

2
4
p
A d
t
= and t =
p
P
A

3
6 2
6
42 10
525 10 m
80 10 t
÷
×
= = = ×
×
p
P
A
(a) Diameter d

6
3
4
(4)(525 10 )
2.585 10 m
p
A
d
t t
÷
÷
×
= = = × 25.9 mm = d 
(b) Bearing stress
3
6
3
42 10
271 10 Pa
(25.85 10 )(0.006)
o
÷
×
= = = ×
×
b
P
dt
271 MPa o =
b

diameter.1 psi  .0  C F  107.0 in 2 Stress in tension member CF (b)  CF  FC F Am in  750 7. SOLUTION 1.5)  7.  Use member DEF as free body. determine the maximum value of the average normal stress (a) in member BE.11 The frame shown consists of four wooden members. Dy  4 3 D x  1200 lb.0  0. M E  0: 4  (3 0)  FC E   (1 5) D Y  0 5  FC E  7 5 0 lb.HW01-MAE314-002 S12 PROBLEM 1.11   Add support reactions to figure as shown. 4   (3 0)  F B E   (3 0  1 5) D Y  0 5   M F  0: F B E   2 2 5 0 lb. ABC. and CF. M A  0: 40 D x  (45  30)(480)  0 D x  900 lb. rectangular cross section and that each pin has a 1/2-in. Using entire frame as free body. Reaction at D must be parallel to F B E and FC F . Stress in compression member BE Area: (a) A  2 in  4 in  8 in 2  BE  FB E A   2250 8  B E   281 psi  Minimum section area occurs at pin. BE. DEF. Knowing that each member has a 2  4-in. Amin  (2)(4. (b) in member CF.

57 kips Limiting value of P is the smaller value.6-in. of width b  50 mm and thickness t  6 mm. (b) the average bearing stress in the link.-diameter hole has been drilled.57 kips  PROBLEM 1. is used to support the end of a horizontal beam.2566)(10)  12.17 A load P is applied to a steel rod supported as shown by an aluminum plate into which a 0.7540 in 2 1  P A  P  A1 1  (0.HW01-MAE314-002 S12 PROBLEM 1. determine (a) the diameter d of the pins.6)(0.26 Link AB. For steel: A1   dt   (0. .2566 in 2  P  A2 2  (1.25)  1.7540)(18)  13. determine the largest load P that can be applied to the rod. Knowing that the shearing stress must not exceed 18 ksi in the steel rod and 10 ksi in the aluminum plate.4)  0. and that the average shearing stress in each of the two pins is 80 MPa.6)(0.57 kips For aluminum: 2  P A2 A2   dt   (1. Knowing that the average normal stress in the link is -140 MPa. so P  12.

050)(0.9 mm  (b) Bearing stress b  (25.HW01-MAE314-002 S12 Rod AB is in compression. Ap   4 d 2 and   42  10 80  10 3 6 P Ap 6 2 Ap  P    525  10 m (a) Diameter d d  4 Ap  (4)(525  10 6 )  P dt   42  10 3  2.585  10 3 m d  25.006)  300  10 6 m 2 6 P   A   (  140  10 )(300  10  42  10 N 3 ) For the pin. A  bt where b  50 mm and t  6 mm 6 A  (0.006)  b  271 M P a  .85  10 3  271  10 P a 6 )(0.
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