PROBLEM 9.12
Rod BD is made of steel
6
( 29 10 psi)E= ×
and is used to brace the axially
compressed member ABC. The maximum force that can be developed in member
BD is 0.02P. If the stress must not exceed 18 ksi and the maximum change in
length of BD must not exceed 0.001 times the length of ABC, determine the
smallest–diameter rod that can be used for member BD.
SOLUTION
3
0.02 (0.02)(130) 2.6 kips 2.6 10 lb= = = = ×
BD
FP
Considering stress,
3
18 ksi 18 10 psi
s
= = ×
2
2.6 0.14444 in
18
= ∴= = =
BD BD
FF
A
A
ss
Considering deformation,
(0.001)(144) 0.144 in.
δ
= =
32
6
(2.6 10 )(54) 0.03362 in
(29 10 )(0.144)
BD BD BD BD
FL FL
A
AE E
δδ
×
= ∴= = =
×
Larger area governs.
2
0.14444 in=A
2
4 (4)(0.14444)
4
π
ππ
= ∴= =
A
Ad d
0.429 in.d=
consent of McGraw–Hill Education.
PROBLEM 9.13
The specimen shown is made from a 1-in.–diameter cylindrical steel rod
with two 1.5–in.-outer-diameter sleeves bonded to the rod as shown.
Knowing that
6
29 10 psi,
= ×E
determine (a) the load P so that the
total deformation is 0.002 in., (b) the corresponding deformation of the
central portion BC.
SOLUTION
(a)
ii i
ii i
PL L
P
AE E A
δ
=Σ=Σ
1
2
4
iii
i
L
PE A d
A
π
δ
−

=Σ=


L, in. d, in. A, in2 L/A, in−1
AB 2 1.5 1.7671 1.1318
BC 3 1.0 0.7854 3.8197
CD 2 1.5 1.7671 1.1318
6.083 ← sum
6 13
(29 10 )(0.002)(6.083) 9.353 10 lbP
−
=×=×
9.53 kipsP=
(b)
3
6
9.535 10 (3.8197)
29 10
BC BC
BC BC BC
PL L
P
A E EA
δ
×
= = = ×
3
1.254 10 in.
δ
−
= ×
consent of McGraw–Hill Education.
PROBLEM 9.14
Both portions of the rod ABC are made of an aluminum for which
70 GPa.E=
Knowing that the magnitude of P is 4 kN, determine (a) the value of Q so that
the deflection at A is zero, (b) the corresponding deflection of B.
SOLUTION
(a)
2 2 62
(0.020) 314.16 10 m
44
AB AB
Ad
ππ
−
= = = ×
2 2 32
(0.060) 2.8274 10 m
44
BC BC
Ad
ππ
−
= = = ×
Force in member AB is P tension.
Elongation:
36
96
(4 10 )(0.4) 72.756 10 m
(70 10 )(314.16 10 )
δ
−
−
×
= = = ×
××
AB
AB AB
PL
EA
Force in member BC is Q − P compression.
Shortening:
9
93
() ( )(0.5) 2.5263 10 ( )
(70 10 )(2.8274 10 )
δ
−
−
−−
= = =×−
××
BC
BC BC
Q PL QP QP
EA
For zero deflection at A,
BC AB
δδ
=
96 3
2.5263 10 ( ) 72.756 10 28.8 10 NQP QP
−−
× − = × ∴ −= ×
33 3
28.3 10 4 10 32.8 10 NQ= × +× = ×
32.8 kN=Q
(b)
6
72.756 10 m
AB BC B
δδδ
−
= = = ×
0.0728 mm
δ
= ↓
AB
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PROBLEM 9.15
The rod ABC is made of an aluminum for which
70 GPa.E=
Knowing that
6 kN=P
and
42 kN,=Q
determine the deflection of (a) point A, (b) point B.
SOLUTION
2 2 62
2 2 32
(0.020) 314.16 10 m
44
(0.060) 2.8274 10 m
44
AB AB
BC BC
Ad
Ad
ππ
ππ
−
−
= = = ×
= = = ×
3
33 3
6 10 N
6 10 42 10 36 10 N
0.4 m 0.5 m
AB
BC
AB BC
PP
P PQ
LL
= = ×
=−=× − × =−×
= =
36
69
36
39
(6 10 )(0.4) 109.135 10 m
(314.16 10 )(70 10 )
( 36 10 )(0.5) 90.947 10 m
(2.8274 10 )(70 10 )
AB AB
AB AB A
BC BC
BC BC
PL
AE
PL
AE
δ
δ
−
−
−
−
×
= = = ×
××
−×
= = =−×
××
(a)
66 6
109.135 10 90.947 10 m 18.19 10 m
A AB BC
δδ δ
−− −
=+= ×− × = ×
0.01819 mm
A
δ
= ↑
(b)
6
90.9 10 m 0.0909 mm
B BC
δδ
−
==−× =−
or
0.0909 mm
δ
= ↓
B
consent of McGraw–Hill Education.
consent of McGraw-Hill Education.
PROBLEM 9.16
Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is
made of steel 6
(2910E psi), and rod BC of brass 6
(1510 psi).E
Determine (a) the total deformation of the composite rod ABC, (b) the
deflection of point B.
SOLUTION
B
PROBLEM 9.17
A 4–ft section of aluminum pipe of cross–sectional area 1.75 in2 rests on a
fixed support at A. The
5
8
–in.–diameter steel rod BC hangs from a rigid
bar that rests on the top of the pipe at B. Knowing that the modulus of
elasticity is
6
29 10×
psi for steel, and
6
10.4 10
×
psi for aluminum,
determine the deflection of point C when a 15-kip force is applied at C.
SOLUTION
Rod BC:
6
7 ft 84 in. 29 10 psi
BC BC
LE= = = ×
22 2
3
/6
(0.625) 0.30680 in
44
(15 10 )(84) 0.141618 in.
(29 10 )(0.30680)
ππ
δ
= = =
×
= = =
×
BC
BC
CB BC BC
Ad
PL
EA
Pipe AB:
6
4 ft 48 in. 10.4 10 psi
AB AB
LE= = = ×
Total:
3
//
39.560 10 0.141618 0.181178 in.
δδ δ
−
=+= ×+ =
C BA CB
0.1812 in.
C
δ
=
consent of McGraw–Hill Education.
PROBLEM 9.18
Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is
made of steel
( 200 GPa)E=
and rod BC of brass
( 105 GPa).E=
Determine
(a) the total deformation of the composite rod ABC, (b) the deflection of
point B.
SOLUTION
Rod AB:
3
30 10 N
AB
FP=−=− ×
9
2 2 62
36
96
0.250 m
200 10 GPa
(30) 706.85 mm 706.85 10 m
4
(30 10 )(0.250) 53.052 10 m
(200 10 )(706.85 10 )
π
δ
−
−
−
=
= ×
= = = ×
×
= =− =−×
××
AB
AB
AB
AB AB
AB AB AB
L
E
A
FL
EA
Rod BC:
3
30 40 70 kN 70 10 N
BC
F=+= =×
9
2 3 2 32
36
93
0.300 m
105 10 Pa
(50) 1.9635 10 mm 1.9635 10 m
4
(70 10 )(0.300) 101.859 10 m
(105 10 )(1.9635 10 )
π
δ
−
−
−
=
= ×
==×=×
×
= =− =−×
××
BC
BC
BC
BC BC
BC BC BC
L
E
A
FL
EA
(a) Total deformation:
6
tot 154.9 10 m
δδδ
−
=+=− ×
AB BC
0.1549 mm= −
(b) Deflection of Point B.
B BC
δδ
=
0.1019 mm
B
δ
= ↓
consent of McGraw–Hill Education.
PROBLEM 9.19
The steel frame
( 200 GPa)E=
shown has a diagonal brace BD with
an area of 1920 mm2. Determine the largest allowable load P if the
change in length of member BD is not to exceed 1.6 mm.
SOLUTION
3 2 62
22 9
9 63
3
1.6 10 m, 1920 mm 1920 10 m
5 6 7.810 m, 200 10 Pa
(200 10 )(1920 10 )(1.6 10 )
7.81
78.67 10 N
δ
δ
δ
−−
−−
=×==×
= += = ×
=
× ××
= =
= ×
BD BD
BD BD
BD BD
BD BD BD
BD BD BD
BD BD
A
LE
FL
EA
EA
FL
Use joint B as a free body.
0:
x
FΣ=
50
7.810 −=
BD
FP
3
3
5 (5)(78.67 10 )
7.810 7.810
50.4 10 N
×
= =
= ×
BD
PF
50.4 kN=P
PROBLEM 9.20
For the steel truss
6
( 29 10 psi)E= ×
and loading shown, determine the
deformations of the members BD and DE, knowing that their cross-
sectional areas are 2 in2 and 3 in2, respectively.
SOLUTION
Free body: Portion ABC of truss
0: (15 ft) (30 kips)(8 ft) (30 kips)(16 ft) 0
48.0 kips
E BD
BD
MF
F
Σ= − − =
= +
Free body: Portion ABEC of truss
0: 30 kips 30 kips 0
60.0 kips
Σ= + − =
= +
x DE
DE
FF
F
3
26
( 48.0 10 lb)(8 12 in.)
(2 in )(29 10 psi)
δ
+× ×
= = ×
BD PL
AE
3
79.4 10 in.
δ
−
=+×
BD
3
26
( 60.0 10 lb)(15 12 in.)
(3 in )(29 10 psi)
DE PL
AE
δ
+× ×
= = ×
3
124.1 10 in.
δ
−
=+×
DE
consent of McGraw–Hill Education.
PROBLEM 9.12
Rod BD is made of steel
6
( 29 10 psi)E= ×
and is used to brace the axially
compressed member ABC. The maximum force that can be developed in member
BD is 0.02P. If the stress must not exceed 18 ksi and the maximum change in
length of BD must not exceed 0.001 times the length of ABC, determine the
smallest–diameter rod that can be used for member BD.
SOLUTION
3
0.02 (0.02)(130) 2.6 kips 2.6 10 lb= = = = ×
BD
FP
Considering stress,
3
18 ksi 18 10 psi
s
= = ×
2
2.6 0.14444 in
18
= ∴= = =
BD BD
FF
A
A
ss
Considering deformation,
(0.001)(144) 0.144 in.
δ
= =
32
6
(2.6 10 )(54) 0.03362 in
(29 10 )(0.144)
BD BD BD BD
FL FL
A
AE E
δδ
×
= ∴= = =
×
Larger area governs.
2
0.14444 in=A
2
4 (4)(0.14444)
4
π
ππ
= ∴= =
A
Ad d
0.429 in.d=
consent of McGraw–Hill Education.
PROBLEM 9.13
The specimen shown is made from a 1-in.–diameter cylindrical steel rod
with two 1.5–in.-outer-diameter sleeves bonded to the rod as shown.
Knowing that
6
29 10 psi,
= ×E
determine (a) the load P so that the
total deformation is 0.002 in., (b) the corresponding deformation of the
central portion BC.
SOLUTION
(a)
ii i
ii i
PL L
P
AE E A
δ
=Σ=Σ
1
2
4
iii
i
L
PE A d
A
π
δ
−

=Σ=


L, in. d, in. A, in2 L/A, in−1
AB 2 1.5 1.7671 1.1318
BC 3 1.0 0.7854 3.8197
CD 2 1.5 1.7671 1.1318
6.083 ← sum
6 13
(29 10 )(0.002)(6.083) 9.353 10 lbP
−
=×=×
9.53 kipsP=
(b)
3
6
9.535 10 (3.8197)
29 10
BC BC
BC BC BC
PL L
P
A E EA
δ
×
= = = ×
3
1.254 10 in.
δ
−
= ×
consent of McGraw–Hill Education.
PROBLEM 9.14
Both portions of the rod ABC are made of an aluminum for which
70 GPa.E=
Knowing that the magnitude of P is 4 kN, determine (a) the value of Q so that
the deflection at A is zero, (b) the corresponding deflection of B.
SOLUTION
(a)
2 2 62
(0.020) 314.16 10 m
44
AB AB
Ad
ππ
−
= = = ×
2 2 32
(0.060) 2.8274 10 m
44
BC BC
Ad
ππ
−
= = = ×
Force in member AB is P tension.
Elongation:
36
96
(4 10 )(0.4) 72.756 10 m
(70 10 )(314.16 10 )
δ
−
−
×
= = = ×
××
AB
AB AB
PL
EA
Force in member BC is Q − P compression.
Shortening:
9
93
() ( )(0.5) 2.5263 10 ( )
(70 10 )(2.8274 10 )
δ
−
−
−−
= = =×−
××
BC
BC BC
Q PL QP QP
EA
For zero deflection at A,
BC AB
δδ
=
96 3
2.5263 10 ( ) 72.756 10 28.8 10 NQP QP
−−
× − = × ∴ −= ×
33 3
28.3 10 4 10 32.8 10 NQ= × +× = ×
32.8 kN=Q
(b)
6
72.756 10 m
AB BC B
δδδ
−
= = = ×
0.0728 mm
δ
= ↓
AB
consent of McGraw–Hill Education.
PROBLEM 9.15
The rod ABC is made of an aluminum for which
70 GPa.E=
Knowing that
6 kN=P
and
42 kN,=Q
determine the deflection of (a) point A, (b) point B.
SOLUTION
2 2 62
2 2 32
(0.020) 314.16 10 m
44
(0.060) 2.8274 10 m
44
AB AB
BC BC
Ad
Ad
ππ
ππ
−
−
= = = ×
= = = ×
3
33 3
6 10 N
6 10 42 10 36 10 N
0.4 m 0.5 m
AB
BC
AB BC
PP
P PQ
LL
= = ×
=−=× − × =−×
= =
36
69
36
39
(6 10 )(0.4) 109.135 10 m
(314.16 10 )(70 10 )
( 36 10 )(0.5) 90.947 10 m
(2.8274 10 )(70 10 )
AB AB
AB AB A
BC BC
BC BC
PL
AE
PL
AE
δ
δ
−
−
−
−
×
= = = ×
××
−×
= = =−×
××
(a)
66 6
109.135 10 90.947 10 m 18.19 10 m
A AB BC
δδ δ
−− −
=+= ×− × = ×
0.01819 mm
A
δ
= ↑
(b)
6
90.9 10 m 0.0909 mm
B BC
δδ
−
==−× =−
or
0.0909 mm
δ
= ↓
B
consent of McGraw–Hill Education.
consent of McGraw-Hill Education.
PROBLEM 9.16
Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is
made of steel 6
(2910E psi), and rod BC of brass 6
(1510 psi).E
Determine (a) the total deformation of the composite rod ABC, (b) the
deflection of point B.
SOLUTION
B
PROBLEM 9.17
A 4–ft section of aluminum pipe of cross–sectional area 1.75 in2 rests on a
fixed support at A. The
5
8
–in.–diameter steel rod BC hangs from a rigid
bar that rests on the top of the pipe at B. Knowing that the modulus of
elasticity is
6
29 10×
psi for steel, and
6
10.4 10
×
psi for aluminum,
determine the deflection of point C when a 15-kip force is applied at C.
SOLUTION
Rod BC:
6
7 ft 84 in. 29 10 psi
BC BC
LE= = = ×
22 2
3
/6
(0.625) 0.30680 in
44
(15 10 )(84) 0.141618 in.
(29 10 )(0.30680)
ππ
δ
= = =
×
= = =
×
BC
BC
CB BC BC
Ad
PL
EA
Pipe AB:
6
4 ft 48 in. 10.4 10 psi
AB AB
LE= = = ×
Total:
3
//
39.560 10 0.141618 0.181178 in.
δδ δ
−
=+= ×+ =
C BA CB
0.1812 in.
C
δ
=
consent of McGraw–Hill Education.
PROBLEM 9.18
Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is
made of steel
( 200 GPa)E=
and rod BC of brass
( 105 GPa).E=
Determine
(a) the total deformation of the composite rod ABC, (b) the deflection of
point B.
SOLUTION
Rod AB:
3
30 10 N
AB
FP=−=− ×
9
2 2 62
36
96
0.250 m
200 10 GPa
(30) 706.85 mm 706.85 10 m
4
(30 10 )(0.250) 53.052 10 m
(200 10 )(706.85 10 )
π
δ
−
−
−
=
= ×
= = = ×
×
= =− =−×
××
AB
AB
AB
AB AB
AB AB AB
L
E
A
FL
EA
Rod BC:
3
30 40 70 kN 70 10 N
BC
F=+= =×
9
2 3 2 32
36
93
0.300 m
105 10 Pa
(50) 1.9635 10 mm 1.9635 10 m
4
(70 10 )(0.300) 101.859 10 m
(105 10 )(1.9635 10 )
π
δ
−
−
−
=
= ×
==×=×
×
= =− =−×
××
BC
BC
BC
BC BC
BC BC BC
L
E
A
FL
EA
(a) Total deformation:
6
tot 154.9 10 m
δδδ
−
=+=− ×
AB BC
0.1549 mm= −
(b) Deflection of Point B.
B BC
δδ
=
0.1019 mm
B
δ
= ↓
consent of McGraw–Hill Education.
PROBLEM 9.19
The steel frame
( 200 GPa)E=
shown has a diagonal brace BD with
an area of 1920 mm2. Determine the largest allowable load P if the
change in length of member BD is not to exceed 1.6 mm.
SOLUTION
3 2 62
22 9
9 63
3
1.6 10 m, 1920 mm 1920 10 m
5 6 7.810 m, 200 10 Pa
(200 10 )(1920 10 )(1.6 10 )
7.81
78.67 10 N
δ
δ
δ
−−
−−
=×==×
= += = ×
=
× ××
= =
= ×
BD BD
BD BD
BD BD
BD BD BD
BD BD BD
BD BD
A
LE
FL
EA
EA
FL
Use joint B as a free body.
0:
x
FΣ=
50
7.810 −=
BD
FP
3
3
5 (5)(78.67 10 )
7.810 7.810
50.4 10 N
×
= =
= ×
BD
PF
50.4 kN=P
PROBLEM 9.20
For the steel truss
6
( 29 10 psi)E= ×
and loading shown, determine the
deformations of the members BD and DE, knowing that their cross-
sectional areas are 2 in2 and 3 in2, respectively.
SOLUTION
Free body: Portion ABC of truss
0: (15 ft) (30 kips)(8 ft) (30 kips)(16 ft) 0
48.0 kips
E BD
BD
MF
F
Σ= − − =
= +
Free body: Portion ABEC of truss
0: 30 kips 30 kips 0
60.0 kips
Σ= + − =
= +
x DE
DE
FF
F
3
26
( 48.0 10 lb)(8 12 in.)
(2 in )(29 10 psi)
δ
+× ×
= = ×
BD PL
AE
3
79.4 10 in.
δ
−
=+×
BD
3
26
( 60.0 10 lb)(15 12 in.)
(3 in )(29 10 psi)
DE PL
AE
δ
+× ×
= = ×
3
124.1 10 in.
δ
−
=+×
DE
consent of McGraw–Hill Education.