PROBLEM 12.37
Draw the shear and bending–moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear,
PROBLEM 12.38
Draw the shear and bending–moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear,
PROBLEM 12.39
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
PROBLEM 12.39 (Continued)
Bending moments:
0
0 6.5089 6.5089 kN m
6.5089 0.3889 6.12 kN m
6.12 2.7 3.42 kN m
3.42 3.42 0
A
G
C
C
B
M
M
M
M
M
−
+
=
=+= ⋅
=−= ⋅
= −= ⋅
=−=
(a)
max
| | 6.75 kNV=
(b)
max
| | 6.51 kN m
M= ⋅
consent of McGraw–Hill Education.
PROBLEM 12.40
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
3
0:
0.075 (0.2)(300) (0.6)(300) 0
3.2 10 N
A
EF
EF
M
F
F
Σ=
−−=
= ×
3
0: 0 3.2 10 N
x x EF x
F AF AΣ= − = = ×
0: 300 300 0
600 N
yy
y
FA
A
Σ= − − =
=
Couple at D:
3
(0.075)(3.2 10 )
240 N m
D
M= ×
= ⋅
Shear:
A to C:
600 NV=
C to B:
600 300 300 NV=−=
Areas under shear diagram:
A to C:
(0.2)(600) 120 N mV dx = = ⋅
∫
C to D:
(0.2)(300) 60 N mV dx = = ⋅
∫
D to B:
(0.2)(300) 60 N mV dx = = ⋅
∫
Bending moments:
0
A
M=
0 120 120 N m
120 60 180 N m
180 240 60 N m
60 60 0
C
D
D
B
M
M
M
M
−
+
=+= ⋅
= += ⋅
=−=−⋅
=−+ =
Maximum 600 NV=
Maximum 180.0 N mM= ⋅
consent of McGraw–Hill Education.
PROBLEM 12.41
Using the method of Sec. 12.2, solve Prob. 12.13.
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 12.42
Using the method of Sec. 12.2, solve Prob. 12.14.
PROBLEM 12.14 For the beam and loading shown, determine
the maximum normal stress due to bending on a transverse
section at C.
SOLUTION
0:
3 (1.5)(10) (1.1)(2.2)(3) 0
2.58 kN
C
M
A
A
Σ=
−+ − =
=
0:
(1.5)(10) 3 (4.1)(2.2)(3) 0
14.02 kN
A
M
C
C
Σ=
− +− =
=
Shear:
A to D
−
:
2.58 kNV=
D+ to C
−
:
2.58 10 7.42 kNV= −=−
C+:
7.42 14.02 6.60 kNV=−+ =
B:
6.60 (2.2)(3) 0V=−=
Areas under shear diagram:
A to D:
(1.5)(2.58) 3.87 kN mV dx = = ⋅
∫
D to C:
(1.5)( 7.42) 11.13 kN mV dx =−=− ⋅
∫
C to B:
1(2.2)(6.60) 7.26 kN m
2
V dx
= = ⋅
∫
Bending moments:
0
A
M=
0 3.87 3.87 kN m
3.87 11.13 7.26 kN m
7.26 7.26 0
D
C
B
M
M
M
=+= ⋅
=−=− ⋅
=−=
3
7.26 kN m 7.26 10 N m
C
M= ⋅= × ⋅
consent of McGraw–Hill Education.
PROBLEM 12.42 (Continued)
For rectangular cross section,
22
11
(100)(200)
66
S bh
= =
3 3 62
666.67 10 mm 666.67 10 m
−
=×=×
Normal stress:
36
6
7.26 10 10.89 10 Pa
666.67 10
C
M
S
σ
−
×
= = = ×
×
10.89 MPa
consent of McGraw–Hill Education.
PROBLEM 12.43
Using the method of Sec. 12.2, solve Prob. 12.15.
PROBLEM 12.15 For the beam and loading shown, determine
the maximum normal stress due to bending on section a-a.
SOLUTION
Reactions: By symmetry, A = B.
0: 80 kN
y
F∑= = = ↑
AB
Shear diagram:
A to C:
80 kNV=
C to D:
80 30 50 kNV=−=
D to E:
50 50 0V=−=
Areas of shear diagram:
A to C:
(80)(0.8) 64 kN mV dx∫= = ⋅
C to D:
(50)(0.8) 40 kN mV dx∫= = ⋅
D to E:
0V dx∫=
Bending moments:
0
A
M=
0 64 64 kN m
C
M=+= ⋅
64 40 104 kN m
D
M=+= ⋅
104 0 104 kN m
E
M= += ⋅
3
max
104 kN m 104 10 N mM= ⋅= × ⋅
For
3 3 63
W250 67, 805 10 mm 805 10 mS−
×=× =×
Normal stress:
36
6
104 10 129.2 10 Pa
805 10
M
S
σ
−
×
= = = ×
×
129.2 MPa
σ
=
consent of McGraw–Hill Education.
PROBLEM 12.37
Draw the shear and bending–moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear,
PROBLEM 12.38
Draw the shear and bending–moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear,
PROBLEM 12.39
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
PROBLEM 12.39 (Continued)
Bending moments:
0
0 6.5089 6.5089 kN m
6.5089 0.3889 6.12 kN m
6.12 2.7 3.42 kN m
3.42 3.42 0
A
G
C
C
B
M
M
M
M
M
−
+
=
=+= ⋅
=−= ⋅
= −= ⋅
=−=
(a)
max
| | 6.75 kNV=
(b)
max
| | 6.51 kN m
M= ⋅
consent of McGraw–Hill Education.
PROBLEM 12.40
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
3
0:
0.075 (0.2)(300) (0.6)(300) 0
3.2 10 N
A
EF
EF
M
F
F
Σ=
−−=
= ×
3
0: 0 3.2 10 N
x x EF x
F AF AΣ= − = = ×
0: 300 300 0
600 N
yy
y
FA
A
Σ= − − =
=
Couple at D:
3
(0.075)(3.2 10 )
240 N m
D
M= ×
= ⋅
Shear:
A to C:
600 NV=
C to B:
600 300 300 NV=−=
Areas under shear diagram:
A to C:
(0.2)(600) 120 N mV dx = = ⋅
∫
C to D:
(0.2)(300) 60 N mV dx = = ⋅
∫
D to B:
(0.2)(300) 60 N mV dx = = ⋅
∫
Bending moments:
0
A
M=
0 120 120 N m
120 60 180 N m
180 240 60 N m
60 60 0
C
D
D
B
M
M
M
M
−
+
=+= ⋅
= += ⋅
=−=−⋅
=−+ =
Maximum 600 NV=
Maximum 180.0 N mM= ⋅
consent of McGraw–Hill Education.
PROBLEM 12.41
Using the method of Sec. 12.2, solve Prob. 12.13.
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 12.42
Using the method of Sec. 12.2, solve Prob. 12.14.
PROBLEM 12.14 For the beam and loading shown, determine
the maximum normal stress due to bending on a transverse
section at C.
SOLUTION
0:
3 (1.5)(10) (1.1)(2.2)(3) 0
2.58 kN
C
M
A
A
Σ=
−+ − =
=
0:
(1.5)(10) 3 (4.1)(2.2)(3) 0
14.02 kN
A
M
C
C
Σ=
− +− =
=
Shear:
A to D
−
:
2.58 kNV=
D+ to C
−
:
2.58 10 7.42 kNV= −=−
C+:
7.42 14.02 6.60 kNV=−+ =
B:
6.60 (2.2)(3) 0V=−=
Areas under shear diagram:
A to D:
(1.5)(2.58) 3.87 kN mV dx = = ⋅
∫
D to C:
(1.5)( 7.42) 11.13 kN mV dx =−=− ⋅
∫
C to B:
1(2.2)(6.60) 7.26 kN m
2
V dx
= = ⋅
∫
Bending moments:
0
A
M=
0 3.87 3.87 kN m
3.87 11.13 7.26 kN m
7.26 7.26 0
D
C
B
M
M
M
=+= ⋅
=−=− ⋅
=−=
3
7.26 kN m 7.26 10 N m
C
M= ⋅= × ⋅
consent of McGraw–Hill Education.
PROBLEM 12.42 (Continued)
For rectangular cross section,
22
11
(100)(200)
66
S bh
= =
3 3 62
666.67 10 mm 666.67 10 m
−
=×=×
Normal stress:
36
6
7.26 10 10.89 10 Pa
666.67 10
C
M
S
σ
−
×
= = = ×
×
10.89 MPa
consent of McGraw–Hill Education.
PROBLEM 12.43
Using the method of Sec. 12.2, solve Prob. 12.15.
PROBLEM 12.15 For the beam and loading shown, determine
the maximum normal stress due to bending on section a-a.
SOLUTION
Reactions: By symmetry, A = B.
0: 80 kN
y
F∑= = = ↑
AB
Shear diagram:
A to C:
80 kNV=
C to D:
80 30 50 kNV=−=
D to E:
50 50 0V=−=
Areas of shear diagram:
A to C:
(80)(0.8) 64 kN mV dx∫= = ⋅
C to D:
(50)(0.8) 40 kN mV dx∫= = ⋅
D to E:
0V dx∫=
Bending moments:
0
A
M=
0 64 64 kN m
C
M=+= ⋅
64 40 104 kN m
D
M=+= ⋅
104 0 104 kN m
E
M= += ⋅
3
max
104 kN m 104 10 N mM= ⋅= × ⋅
For
3 3 63
W250 67, 805 10 mm 805 10 mS−
×=× =×
Normal stress:
36
6
104 10 129.2 10 Pa
805 10
M
S
σ
−
×
= = = ×
×
129.2 MPa
σ
=
consent of McGraw–Hill Education.