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PROBLEM 12.9
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear,
(b) of the bending moment.
SOLUTION
0: (700)(3) 450 (300)(3) 1000 0
B
MAΣ= − + − =
2.55 kNA=
0: (300)(3) 450 (700)(3) 1000 0
A
MBΣ= − − − + =
3.45 kNB=
At A:
2.55 kN 0VM= =
A to C:
2.55 kNV=
At C:
0:
C
MΣ=
(300)(2.55) 0M− +=
765 N m
M= ⋅
C to E:
0.45 N mV=−⋅
At D:
0:
D
M
Σ=
(500)(2.55) (200)(3) 0M− + +=
675 N mM= ⋅
At D:
0:
D
MΣ=
(500)(2.55) (200)(3) 450 0M− + − +=
1125 N mM= ⋅
E to B:
3.45 kNV= −
At E:
0:
E
MΣ=
(300)(3.45) 0M−+ =
1035 N mM= ⋅
At B:
3.45 kN, 0VM= =
(a)
max
3.45 kNV=
(b)
max
1125 N mM= ⋅
consent of McGraw-Hill Education.
PROBLEM 12.10
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of
the shear, (b) of the bending moment.
SOLUTION
0: 16 (36)(400) (12)(1600)
G
MC
Σ=− + +
(12)(400) 0−=
1800 lbC=
0: 0
xx
F CGΣ = −+ =
1800 lb
x
G=
0: 400 1600 400 0
yy
FGΣ= − − + − =
2400 lb
y
G=
A to E:
400 lbV= −
E to F:
2000 lbV= −
F to B:
400 lbV=
At A and B,
0M=
At
,
D
−
0: (12)(400) 0
D
MMΣ = +=
4800 lb in.M=−⋅
At
+
,
D
0:
D
MΣ=
(12)(400) (8)(1800) 0M− +=
9600 lb in.M= ⋅
At E,
0: (24)(400) (8)(1800) 0
E
MM
Σ = − +=
4800 lb in.
M= ⋅
At
,
F
−
0: (8)(1800) (12)(400) 0
F
MMΣ = −− − =
19,200 lb in.M=−⋅
At
+
,
F
0: (12)(400) 0
F
MM
Σ = −− =
4800 lb in.
M=−⋅
(a) Maximum
| | 2000 lbV=
(b) Maximum
| | 19,200 lb in.M= ⋅
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
PROBLEM 12.11
Assuming that the reaction of the ground is uniformly distributed, draw
the shear and bending-moment diagrams for the beam AB and determine
the maximum absolute value (a) of the shear, (b) of the bending moment.
SOLUTION
consent of McGraw-Hill Education.
PROBLEM 12.12
Assuming that the reaction of the ground is uniformly distributed, draw
the shear and bending-moment diagrams for the beam AB and determine
the maximum absolute value (a) of the shear, (b) of the bending
moment.
SOLUTION
Over the whole beam,
0: 1.5 1.5 1.5 0
y
FwΣ= − − =
2 kN/mw=
A to C:
0 0.3 mx≤<
0: 2 0
y
F xVΣ= −=
(2 ) kNVx=
0: (2 ) 0
2
J
x
M xM
Σ= − +=
2
( ) kN mMx= ⋅
At
,
C
−
0.3 mx=
0.6 kN, 0.090 kN m
90 N m
VM= = ⋅
= ⋅
C to D:
0.3 m 1.2 mx<<
0: 2 1.5 0
y
F xVΣ= − −=
(2 1.5) kNVx= −
0: (2 ) (1.5)( 0.3) 0
2
J
x
M x xM
Σ= − + − +=
2
( 1.5 0.45) kN mMx x=−+ ⋅
At the center of the beam,
0.75 mx=
0 0.1125 kN m
112.5 N m
VM
= =−⋅
=−⋅
At
+
,
C
0.3 m, 0.9 kNxV= = −
(a) Maximum
| | 0.9 kN 900 NV= =
(b) Maximum
| | 112.5 N mM= ⋅
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
PROBLEM 12.13
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
SOLUTION
PROBLEM 12.14
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
SOLUTION
PROBLEM 12.15
For the beam and loading shown, determine the maximum normal
stress due to bending on section a-a.
SOLUTION
PROBLEM 12.16
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
SOLUTION
Use entire beam as free body.
0:
B
MΣ=
4.8 (3.6)(216) (1.6)(150) (0.8)(150) 0A−+ + + =
237 kN=A
Use portion AC as free body.
0:
C
MΣ=
(2.4)(237) (1.2)(216) 0
309.6 kN m
M
M
−+=
= ⋅
For
63
W460 113, 2390 10 mmS×=×
Normal stress:
3
63
6
309.6 10 N m
2390 10 m
129.5 10 Pa
M
S
σ
−
×⋅
= = ×
= ×
129.5
σ
=
MPa
consent of McGraw-Hill Education.
PROBLEM 12.17
For the beam and loading shown, determine the maximum normal
stress due to bending on a transverse section at C.
SOLUTION
Use entire beam as free body.
0:
B
M∑=
90 (75)(5) (60)(5) (45)(2) (30)(2) (15)(2) 0A−+ + + + + =
9.5 kipsA=
Use portion AC as free body.
0: (15)(9.5) 0
142.5 kip in.
C
MM
M
Σ= − =
= ⋅
For
3
8 18.4, 14.4 inSS×=
Normal stress:
142.5
14.4
M
S
σ
= =
9.90 ksi
σ
=
consent of McGraw-Hill Education.
PROBLEM 12.9
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear,
(b) of the bending moment.
SOLUTION
0: (700)(3) 450 (300)(3) 1000 0
B
MAΣ= − + − =
2.55 kNA=
0: (300)(3) 450 (700)(3) 1000 0
A
MBΣ= − − − + =
3.45 kNB=
At A:
2.55 kN 0VM= =
A to C:
2.55 kNV=
At C:
0:
C
MΣ=
(300)(2.55) 0M− +=
765 N m
M= ⋅
C to E:
0.45 N mV=−⋅
At D:
0:
D
M
Σ=
(500)(2.55) (200)(3) 0M− + +=
675 N mM= ⋅
At D:
0:
D
MΣ=
(500)(2.55) (200)(3) 450 0M− + − +=
1125 N mM= ⋅
E to B:
3.45 kNV= −
At E:
0:
E
MΣ=
(300)(3.45) 0M−+ =
1035 N mM= ⋅
At B:
3.45 kN, 0VM= =
(a)
max
3.45 kNV=
(b)
max
1125 N mM= ⋅
consent of McGraw-Hill Education.
PROBLEM 12.10
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of
the shear, (b) of the bending moment.
SOLUTION
0: 16 (36)(400) (12)(1600)
G
MC
Σ=− + +
(12)(400) 0−=
1800 lbC=
0: 0
xx
F CGΣ = −+ =
1800 lb
x
G=
0: 400 1600 400 0
yy
FGΣ= − − + − =
2400 lb
y
G=
A to E:
400 lbV= −
E to F:
2000 lbV= −
F to B:
400 lbV=
At A and B,
0M=
At
,
D
−
0: (12)(400) 0
D
MMΣ = +=
4800 lb in.M=−⋅
At
+
,
D
0:
D
MΣ=
(12)(400) (8)(1800) 0M− +=
9600 lb in.M= ⋅
At E,
0: (24)(400) (8)(1800) 0
E
MM
Σ = − +=
4800 lb in.
M= ⋅
At
,
F
−
0: (8)(1800) (12)(400) 0
F
MMΣ = −− − =
19,200 lb in.M=−⋅
At
+
,
F
0: (12)(400) 0
F
MM
Σ = −− =
4800 lb in.
M=−⋅
(a) Maximum
| | 2000 lbV=
(b) Maximum
| | 19,200 lb in.M= ⋅
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
PROBLEM 12.11
Assuming that the reaction of the ground is uniformly distributed, draw
the shear and bending-moment diagrams for the beam AB and determine
the maximum absolute value (a) of the shear, (b) of the bending moment.
SOLUTION
consent of McGraw-Hill Education.
PROBLEM 12.12
Assuming that the reaction of the ground is uniformly distributed, draw
the shear and bending-moment diagrams for the beam AB and determine
the maximum absolute value (a) of the shear, (b) of the bending
moment.
SOLUTION
Over the whole beam,
0: 1.5 1.5 1.5 0
y
FwΣ= − − =
2 kN/mw=
A to C:
0 0.3 mx≤<
0: 2 0
y
F xVΣ= −=
(2 ) kNVx=
0: (2 ) 0
2
J
x
M xM
Σ= − +=
2
( ) kN mMx= ⋅
At
,
C
−
0.3 mx=
0.6 kN, 0.090 kN m
90 N m
VM= = ⋅
= ⋅
C to D:
0.3 m 1.2 mx<<
0: 2 1.5 0
y
F xVΣ= − −=
(2 1.5) kNVx= −
0: (2 ) (1.5)( 0.3) 0
2
J
x
M x xM
Σ= − + − +=
2
( 1.5 0.45) kN mMx x=−+ ⋅
At the center of the beam,
0.75 mx=
0 0.1125 kN m
112.5 N m
VM
= =−⋅
=−⋅
At
+
,
C
0.3 m, 0.9 kNxV= = −
(a) Maximum
| | 0.9 kN 900 NV= =
(b) Maximum
| | 112.5 N mM= ⋅
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
PROBLEM 12.13
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
SOLUTION
PROBLEM 12.14
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
SOLUTION
PROBLEM 12.15
For the beam and loading shown, determine the maximum normal
stress due to bending on section a-a.
SOLUTION
PROBLEM 12.16
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
SOLUTION
Use entire beam as free body.
0:
B
MΣ=
4.8 (3.6)(216) (1.6)(150) (0.8)(150) 0A−+ + + =
237 kN=A
Use portion AC as free body.
0:
C
MΣ=
(2.4)(237) (1.2)(216) 0
309.6 kN m
M
M
−+=
= ⋅
For
63
W460 113, 2390 10 mmS×=×
Normal stress:
3
63
6
309.6 10 N m
2390 10 m
129.5 10 Pa
M
S
σ
−
×⋅
= = ×
= ×
129.5
σ
=
MPa
consent of McGraw-Hill Education.
PROBLEM 12.17
For the beam and loading shown, determine the maximum normal
stress due to bending on a transverse section at C.
SOLUTION
Use entire beam as free body.
0:
B
M∑=
90 (75)(5) (60)(5) (45)(2) (30)(2) (15)(2) 0A−+ + + + + =
9.5 kipsA=
Use portion AC as free body.
0: (15)(9.5) 0
142.5 kip in.
C
MM
M
Σ= − =
= ⋅
For
3
8 18.4, 14.4 inSS×=
Normal stress:
142.5
14.4
M
S
σ
= =
9.90 ksi
σ
=
consent of McGraw-Hill Education.
