PROBLEM 1.29

PROBLEM 1.29
P'
150 mm
P
Two wooden members of uniform rectangular cross section are joined
11 kN,
by the simple glued scarf splice shown. Knowing that P
determine the normal and shearing stresses in the glued splice.
45!!
45
75 mm
SOLUTION
90
P
A0
45
11 kN
45
11 103 N
(150)(75)
11.25 103 mm 2
11.25 10 3 m 2
P cos 2
A0
(11 103 ) cos 2 45
11.25 10 3
489 103 Pa
489 kPa
P sin 2
2 A0
(11 103 )(sin 90 )
(2)(11.25 10 3 )
489 103 Pa
489 kPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
33
A
PROBLEM 1.42
600 lb/ft
Link AB is to be made of a steel for which the ultimate normal stress is
65 ksi. Determine the cross-sectional area of AB for which the factor
of safety will be 3.20. Assume that the link will be adequately
reinforced around the pins at A and B.
35!
B
C
D
E
5 kips
1.4 ft
1.4 ft
1.4 ft
SOLUTION
P
MD
(4.2)(0.6)
0:
2.52 kips
(2.8)( FAB sin 35 )
(0.7)(2.52)
FAB
AB
AAB
(1.4)(5)
0
5.4570 kips
FAB
AAB
ult
F. S .
( F. S .) FAB
ult
(3.20)(5.4570 kips)
65 ksi
0.26854 in 2
AAB
0.268 in 2
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
48
PROBLEM 1.69
2.4 kips
The two portions of member AB are glued together along a plane
forming an angle with the horizontal. Knowing that the ultimate stress
for the glued joint is 2.5 ksi in tension and 1.3 ksi in shear, determine
(a) the value of
for which the factor of safety of the member is
maximum, (b) the corresponding value of the factor of safety. (Hint:
Equate the expressions obtained for the factors of safety with respect to the
normal and shearing stresses.)
A
!
B
1.25 in.
2.0 in.
SOLUTION
A0
At the optimum angle,
( F. S.)
( F. S.)
PU ,
U A0
2
P
cos 2
A0
Normal stress:
cos
P
sin
A0
cos
Equating,
Solving,
(b)
PU
U A0
2
P cos
sin
cos
P
P cos
U A0
PU ,
sin
cos
U A0
PU ,
( F. S.)
U A0
2
PU ,
( F. S .)
Shearing stress:
2.50 in 2
(2.0)(1.25)
P sin
P
cos
U A0
P sin
tan
U A0
2
cos
cos
U
U
(12.5)(2.50)
cos 2 27.5
F. S.
1.3
2.5
0.520
(a)
opt
27.5
7.94 kips
PU
P
7.94
2.4
F. S.
3.31
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
76

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