978-0073398167 Chapter 8 Solution Manual Part 4

subject Type Homework Help
subject Pages 17
subject Words 1122
subject Authors David Mazurek, E. Johnston, Ferdinand Beer, John DeWolf

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PROBLEM 8.28
Two wooden members of uniform cross section are joined by the simple scarf splice
shown. Knowing that the maximum allowable tensile stress in the glued splice is 75 psi,
determine (a) the largest load P that can be safely supported, (b) the corresponding
shearing stress in the splice.
SOLUTION
2
0
2
0
(5.0)(3.0) 15 in
90 60 30
cos
A
P
A
θ
θ
s
= =
= °− °= °
=
(a)
0
22
(75)(15) 1500 lb
cos cos 30
s
θ
= = =
°
A
P
1.500 kips=P
(b)
0
sin 2 (1500)sin60
2 (2)(15)
θ
τ
°
= =
P
A
page-pf3
PROBLEM 8.29
A 240-kip load P is applied to the granite block shown. Determine the resulting
maximum value of (a) the normal stress, (b) the shearing stress. Specify the
orientation of the plane on which each of these maximum values occurs.
SOLUTION
2
0
222
0
(6)(6) 36 in
240
cos cos 6.67cos
36
A
P
A
sθ θ θ
= =
= = = −
(a) max tensile stress = 0 at
90.0
θ
= °
max. compressive stress =
6.67 ksi
at
0
θ
= °
(b)
max 0
240
2 (2)(36)
P
A
τ
= =
max
3.33 ksi
τ
=
at
45
θ
= °
consent of McGraw-Hill Education.
page-pf4
PROBLEM 8.30
A centric load P is applied to the granite block shown. Knowing that the
resulting maximum value of the shearing stress in the block is 2.5 ksi, determine
(a) the magnitude of P, (b) the orientation of the surface on which the maximum
shearing stress occurs, (c) the normal stress exerted on that surface, (d ) the
maximum value of the normal stress in the block.
SOLUTION
2
0
max
max
(6)(6) 36 in
2.5 ksi
45 for plane of
A
τ
θτ
= =
=
= °
(a)
max 0 max
0
|| | | 2 (2)(36)(2.5)
2
PPA
A
ττ
= ∴= =
180.0 kipsP=
(b)
sin 2 1 2 90
θθ
= = °
45.0
θ
= °
(c)
2
45 00
180
cos 45 2 (2)(36)
PP
AA
s
= °= =−
45 2.50 ksi
s
= −
(d )
max 0
180
36
P
A
s
= =
max
5.00 ksi
s
= −
consent of McGraw-Hill Education.
page-pf5
PROBLEM 8.31
A steel pipe of 400-mm outer diameter is fabricated from 10-mm thick
plate by welding along a helix that forms an angle of 20° with a plane
perpendicular to the axis of the pipe. Knowing that a 300-kN axial
force P is applied to the pipe, determine the normal and shearing
stresses in directions respectively normal and tangential to the weld.
SOLUTION
22 2 2
32
32
26
3
0.400 m
10.200 m
2
0.200 0.010 0.190 m
( ) (0.200 0.190 )
12.2522 10 m
20
300 10 cos 20
cos 21.621 10 Pa
12.2522 10
o
oo
io
o oi
o
d
rd
rrt
A rr
P
A
ππ
θ
=
= =
= −= =
= −=
= ×
= °
−× °
= = = = ×
×
21.6 MPa
s
= −
36
3
0
300 10 sin 40
sin 2 7.8695 10 Pa
2(2)(12.2522 10 )
τθ
−× °
= = = = ×
×
P
A
7.87 MPa
τ
=
page-pf6
PROBLEM 8.32
A steel pipe of 400-mm outer diameter is fabricated from 10-mm
thick plate by welding along a helix that forms an angle of 20° with a
plane perpendicular to the axis of the pipe. Knowing that the
maximum allowable normal and shearing stresses in the directions
respectively normal and tangential to the weld are
60 MPa
σ
=
and
36 MPa,
τ
=
determine the magnitude P of the largest axial force that
can be applied to the pipe.
SOLUTION
22 2 2
32
0.400 m
10.200 m
2
0.200 0.010 0.190 m
( ) (0.200 0.190 )
12.2522 10 m
20
o
oo
io
o oi
d
rd
rrt
A rr
ππ
θ
=
= =
= −= =
= −=
= ×
= °
Based on
2
0
| | 60 MPa: cos
P
A
σ σθ
= =
36 3
22
(12.2522 10 )(60 10 ) 832.52 10 N
cos cos 20
σ
θ
××
= = = ×
°
o
A
P
Based on
| | 30 MPa: sin 2
2
τ τθ
= =
o
P
A
36 3
2 (2)(12.2522 10 )(36 10 ) 1372.39 10 N
sin 2 sin 40
τ
θ
××
= = = ×
°
o
A
P
Smaller value is the allowable value of P.
833 kNP=
page-pf7
PROBLEM 8.33
Link AB is to be made of a steel for which the ultimate normal stress is
450 MPa. Determine the cross-sectional area for AB for which the
factor of safety will be 3.50. Assume that the link will be adequately
reinforced around the pins at A and B.
SOLUTION
(1.2)(8) 9.6 kN= =P
0 : (0.8)( sin35 )
(0.2)(9.6) (0.4)(20) 0
D AB
MFΣ= − °
++=
3
ult
3
6
ult
62
21.619 kN 21.619 10 N
..
( . .) (3.50)(21.619 10 )
450 10
168.1 10 m
AB
AB
AB AB
AB
AB
F
F
A FS
FS F
A
σ
σ
σ
= = ×
= =
×
= = ×
= ×
2
168.1 mm=
AB
A
consent of McGraw-Hill Education.
page-pf8
PROBLEM 8.34
A
3
4
-in.-diameter rod made of the same material as rods AC and AD in
the truss shown was tested to failure and an ultimate load of 29 kips was
recorded. Using a factor of safety of 3.0, determine the required diameter
(a) of rod AC, (b) of rod AD.
SOLUTION
page-pf9
PROBLEM 8.35
In the truss shown, members AC and AD consist of rods made of the
same metal alloy. Knowing that AC is of 1-in. diameter and that the
ultimate load for that rod is 75 kips, determine (a) the factor of safety
for AC, (b) the required diameter of AD if it is desired that both rods
have the same factor of safety.
SOLUTION
Forces in AC and AD.
Joint C:
1
0: 10 kips 0
5
22.36 kips
y AC
AC
FF
FT
Σ= − =
=
Joint D:
1
0: 10 kips 0
17
41.23 kips
y AD
AD
FF
FT
Σ= − =
=
(a) Factor of safety for AC.
F.S. =
U
AC
P
F
75 kips
F.S. 22.36 kips
=
F.S. 3.35=
(b) For the same factor of safety in AC and AD,
.
σσ
=
AD AC
22
41.23 (1) 1.4482 in
22.36 4
π
=
= = =
AD AC
AD AC
AD
AD AC
AC
FF
AA
F
AA
F
Required diameter:
4 (4)(1.4482)
ππ
= =
AD
AD
A
d
1.358 in.=
AD
d
consent of McGraw-Hill Education.
page-pfa
PROBLEM 8.36
Members AB and BC of the truss shown are made of the same alloy. It is known
that a 20-mm-square bar of the same alloy was tested to failure and that an
ultimate load of 120 kN was recorded. If a factor of safety of 3.2 is to be
achieved for both bars, determine the required cross-sectional area of (a) bar
AB, (b) bar AC.
SOLUTION
Length of member AB:
22
0.75 0.4 0.85 m= +=AB
Use entire truss as a free body.
0: 1.4 (0.75)(28) 0
15 kN
Σ= − =
=
cx
x
MA
A
0: 28 0
28 kN
yy
y
FA
A
Σ= − =
=
Use Joint A as free body.
0.75
0: 0
0.85
(0.85)(15) 17 kN
0.75
x AB x
AB
F FA
F
Σ= − =
= =
0.4
0: 0
0.85
(0.4)(17)
28 20 kN
0.85
y y AC AB
AC
F AF F
F
Σ= − − =
=−=
For the test bar,
2 62 3
(0.020) 400 10 m 120 10 N
U
AP
==×=×
For the material,
36
6
120 10 300 10 Pa
400 10
U
U
P
A
σ
×
= = = ×
×
consent of McGraw-Hill Education.
PROBLEM 8.28
Two wooden members of uniform cross section are joined by the simple scarf splice
shown. Knowing that the maximum allowable tensile stress in the glued splice is 75 psi,
determine (a) the largest load P that can be safely supported, (b) the corresponding
shearing stress in the splice.
SOLUTION
2
0
2
0
(5.0)(3.0) 15 in
90 60 30
cos
A
P
A
θ
θ
s
= =
= °− °= °
=
(a)
0
22
(75)(15) 1500 lb
cos cos 30
s
θ
= = =
°
A
P
1.500 kips=P
(b)
0
sin 2 (1500)sin60
2 (2)(15)
θ
τ
°
= =
P
A
PROBLEM 8.29
A 240-kip load P is applied to the granite block shown. Determine the resulting
maximum value of (a) the normal stress, (b) the shearing stress. Specify the
orientation of the plane on which each of these maximum values occurs.
SOLUTION
2
0
222
0
(6)(6) 36 in
240
cos cos 6.67cos
36
A
P
A
sθ θ θ
= =
= = = −
(a) max tensile stress = 0 at
90.0
θ
= °
max. compressive stress =
6.67 ksi
at
0
θ
= °
(b)
max 0
240
2 (2)(36)
P
A
τ
= =
max
3.33 ksi
τ
=
at
45
θ
= °
consent of McGraw-Hill Education.
PROBLEM 8.30
A centric load P is applied to the granite block shown. Knowing that the
resulting maximum value of the shearing stress in the block is 2.5 ksi, determine
(a) the magnitude of P, (b) the orientation of the surface on which the maximum
shearing stress occurs, (c) the normal stress exerted on that surface, (d ) the
maximum value of the normal stress in the block.
SOLUTION
2
0
max
max
(6)(6) 36 in
2.5 ksi
45 for plane of
A
τ
θτ
= =
=
= °
(a)
max 0 max
0
|| | | 2 (2)(36)(2.5)
2
PPA
A
ττ
= ∴= =
180.0 kipsP=
(b)
sin 2 1 2 90
θθ
= = °
45.0
θ
= °
(c)
2
45 00
180
cos 45 2 (2)(36)
PP
AA
s
= °= =−
45 2.50 ksi
s
= −
(d )
max 0
180
36
P
A
s
= =
max
5.00 ksi
s
= −
consent of McGraw-Hill Education.
PROBLEM 8.31
A steel pipe of 400-mm outer diameter is fabricated from 10-mm thick
plate by welding along a helix that forms an angle of 20° with a plane
perpendicular to the axis of the pipe. Knowing that a 300-kN axial
force P is applied to the pipe, determine the normal and shearing
stresses in directions respectively normal and tangential to the weld.
SOLUTION
22 2 2
32
32
26
3
0.400 m
10.200 m
2
0.200 0.010 0.190 m
( ) (0.200 0.190 )
12.2522 10 m
20
300 10 cos 20
cos 21.621 10 Pa
12.2522 10
o
oo
io
o oi
o
d
rd
rrt
A rr
P
A
ππ
θ
=
= =
= −= =
= −=
= ×
= °
−× °
= = = = ×
×
21.6 MPa
s
= −
36
3
0
300 10 sin 40
sin 2 7.8695 10 Pa
2(2)(12.2522 10 )
τθ
−× °
= = = = ×
×
P
A
7.87 MPa
τ
=
PROBLEM 8.32
A steel pipe of 400-mm outer diameter is fabricated from 10-mm
thick plate by welding along a helix that forms an angle of 20° with a
plane perpendicular to the axis of the pipe. Knowing that the
maximum allowable normal and shearing stresses in the directions
respectively normal and tangential to the weld are
60 MPa
σ
=
and
36 MPa,
τ
=
determine the magnitude P of the largest axial force that
can be applied to the pipe.
SOLUTION
22 2 2
32
0.400 m
10.200 m
2
0.200 0.010 0.190 m
( ) (0.200 0.190 )
12.2522 10 m
20
o
oo
io
o oi
d
rd
rrt
A rr
ππ
θ
=
= =
= −= =
= −=
= ×
= °
Based on
2
0
| | 60 MPa: cos
P
A
σ σθ
= =
36 3
22
(12.2522 10 )(60 10 ) 832.52 10 N
cos cos 20
σ
θ
××
= = = ×
°
o
A
P
Based on
| | 30 MPa: sin 2
2
τ τθ
= =
o
P
A
36 3
2 (2)(12.2522 10 )(36 10 ) 1372.39 10 N
sin 2 sin 40
τ
θ
××
= = = ×
°
o
A
P
Smaller value is the allowable value of P.
833 kNP=
PROBLEM 8.33
Link AB is to be made of a steel for which the ultimate normal stress is
450 MPa. Determine the cross-sectional area for AB for which the
factor of safety will be 3.50. Assume that the link will be adequately
reinforced around the pins at A and B.
SOLUTION
(1.2)(8) 9.6 kN= =P
0 : (0.8)( sin35 )
(0.2)(9.6) (0.4)(20) 0
D AB
MFΣ= − °
++=
3
ult
3
6
ult
62
21.619 kN 21.619 10 N
..
( . .) (3.50)(21.619 10 )
450 10
168.1 10 m
AB
AB
AB AB
AB
AB
F
F
A FS
FS F
A
σ
σ
σ
= = ×
= =
×
= = ×
= ×
2
168.1 mm=
AB
A
consent of McGraw-Hill Education.
PROBLEM 8.34
A
3
4
-in.-diameter rod made of the same material as rods AC and AD in
the truss shown was tested to failure and an ultimate load of 29 kips was
recorded. Using a factor of safety of 3.0, determine the required diameter
(a) of rod AC, (b) of rod AD.
SOLUTION
PROBLEM 8.35
In the truss shown, members AC and AD consist of rods made of the
same metal alloy. Knowing that AC is of 1-in. diameter and that the
ultimate load for that rod is 75 kips, determine (a) the factor of safety
for AC, (b) the required diameter of AD if it is desired that both rods
have the same factor of safety.
SOLUTION
Forces in AC and AD.
Joint C:
1
0: 10 kips 0
5
22.36 kips
y AC
AC
FF
FT
Σ= − =
=
Joint D:
1
0: 10 kips 0
17
41.23 kips
y AD
AD
FF
FT
Σ= − =
=
(a) Factor of safety for AC.
F.S. =
U
AC
P
F
75 kips
F.S. 22.36 kips
=
F.S. 3.35=
(b) For the same factor of safety in AC and AD,
.
σσ
=
AD AC
22
41.23 (1) 1.4482 in
22.36 4
π
=
= = =
AD AC
AD AC
AD
AD AC
AC
FF
AA
F
AA
F
Required diameter:
4 (4)(1.4482)
ππ
= =
AD
AD
A
d
1.358 in.=
AD
d
consent of McGraw-Hill Education.
PROBLEM 8.36
Members AB and BC of the truss shown are made of the same alloy. It is known
that a 20-mm-square bar of the same alloy was tested to failure and that an
ultimate load of 120 kN was recorded. If a factor of safety of 3.2 is to be
achieved for both bars, determine the required cross-sectional area of (a) bar
AB, (b) bar AC.
SOLUTION
Length of member AB:
22
0.75 0.4 0.85 m= +=AB
Use entire truss as a free body.
0: 1.4 (0.75)(28) 0
15 kN
Σ= − =
=
cx
x
MA
A
0: 28 0
28 kN
yy
y
FA
A
Σ= − =
=
Use Joint A as free body.
0.75
0: 0
0.85
(0.85)(15) 17 kN
0.75
x AB x
AB
F FA
F
Σ= − =
= =
0.4
0: 0
0.85
(0.4)(17)
28 20 kN
0.85
y y AC AB
AC
F AF F
F
Σ= − − =
=−=
For the test bar,
2 62 3
(0.020) 400 10 m 120 10 N
U
AP
==×=×
For the material,
36
6
120 10 300 10 Pa
400 10
U
U
P
A
σ
×
= = = ×
×
consent of McGraw-Hill Education.

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