Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
PROBLEM 12.19
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
SOLUTION
0:
B
M∑=
(11)(25) 10 (8)(25) (2)(25) 0 52.5 kipsC−+ + = =C
0:
C
M∑=
(1)(25) (2)(25) (8)(25) 10 0 22.5 kipsB− − += =B
Shear:
A to
C−
:
25 kipsV= −
C+ to D−:
27.5 kipsV=
D+ to E−:
2.5 kipsV=
E+ to B:
22.5 kipsV= −
Bending moments:
At C,
0: (1)(25) 0
C
MM∑ = +=
25 kip ftM=−⋅
At D,
0: (3)(25) (2)(52.5) 0
D
MM∑ = − +=
30 kip ftM= ⋅
At E,
0: (2)(22.5) 0 45 kip ft
E
MM M∑ = −+ = = ⋅
max 45 kip ft 540 kip in.M= ⋅= ⋅
For
12 35
S×
rolled steel section,
3
38.1 in
S=
Normal stress:
540 14.17 ksi
38.1
M
S
s
= = =
14.17 ksi
s
=
consent of McGraw-Hill Education.
PROBLEM 12.20
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
SOLUTION
0:
M∑=
consent of McGraw-Hill Education.
PROBLEM 12.20 (Continued)
PROBLEM 12.21
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal stress
due to bending.
SOLUTION
Reaction at C:
0: (18)(5) 13 +(5)(10) 0
10.769 kips
B
MC
C
Σ= − =
=
Reaction at B:
0: (5)(5) (8)(10) 13 0
4.231 kips
C
MB
B
= − +=
=
Shear diagram:
to : 5 kips
to : 5 10.769 5.769 kips
to : 5.769 10 4.231 kips
AC V
CD V
DB V
−
+−
+
= −
=−+ =
= −=−
At A and B,
0M=
At C,
0: (5)(5) 0
25 kip ft
CC
C
MM
M
= +=
=−⋅
At D,
0: (5)(4.231)
21.155 kip ft
DD
D
MM
M
Σ=−+
= ⋅
max
5.77 kipsV=
max
||M
occurs at C.
max
| | 25 kip ft 300 kip in.M= ⋅= ⋅
For
W14 22×
rolled-steel section,
3
29.0 inS=
Normal stress:
300
29.0
M
S
σ
= =
10.34 ksi
σ
=
consent of McGraw-Hill Education.
PROBLEM 12.22
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
SOLUTION
Reactions:
0: 4 64 (24)(2)(1) 0 28 kN
D
MA∑ = −− = =A
0: 28 (24)(2) 0 76 kN
y
FD∑ = − +− = =D
A to C:
0 2mx<<
0: 28 0
28 kN
y
FV
V
∑ = −− =
= −
0: 28 0
( 28 ) kN m
J
M Mx
Mx
∑= + =
=−⋅
C to D:
2m 4mx<<
0: 28 0
28 kN
y
FV
V
∑ = −− =
= −
0: 28 64 0
( 28 64) kN m
J
M Mx
Mx
∑ = + −=
=−+ ⋅
D to B:
4m 6mx<<
0:
24(6 ) 0
( 24 144) kN
y
F
Vx
Vx
∑=
− −=
=−+
2
0:
6
24(6 ) 0
2
12(6 ) kN m
J
M
x
Mx
Mx
∑=
−
−− − =
=−− ⋅
3
max 56 kN m 56 10 N mM= ⋅=× ⋅
For
S250 52×
section,
33
482 10 mmS= ×
Normal stress:
36
63
56 10 N m 116.2 10 Pa
482 10 m
M
S
σ
−
×⋅
= = = ×
×
116.2 MPa
σ
=
consent of McGraw-Hill Education.
PROBLEM 12.23
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal
PROBLEM 12.23 (Continued)
Bending Moment at D:
( )
0: 11.20 5 0
56.0 kip ft
DD
D
MM
M
Σ= − + =
= ⋅
From the diagram,
max
56 kip ft 672 kip in.M= ⋅= ⋅
For
W12 40
×
rolled-steel shape,
3
51.5 in
x
S=
Stress:
max
m
M
S
σ
=
672 13.05 ksi
51.5
m
σ
= =
13.05 ksi
m
σ
=
consent of McGraw-Hill Education.
PROBLEM 12.24
Knowing that
3 kipsW=
, draw the shear and bending-moment
diagrams for beam AB and determine the maximum normal stress
due to bending.
SOLUTION
By symmetry,
AB=
0: 2 3 2 0
0.5 kips
y
FA B
AB
Σ = −+−+ =
= =
Shear:
to : 0.5 kipsAC V
−
=
to : 1.5 kipsCD V
+−
= −
to : 1.5 kipsDE V
+− =
to : 0.5 kipsEB V
+
= −
Bending moment:
max
1.50 kips
V=
At C,
0: (3)(0.5) 0
CC
MMΣ= − =
1.5 kip ft
C
M= ⋅
At D,
0: (6)(0.5) (3)(2) 0
DD
MM+Σ = − + =
3 kip ft
D
M=−⋅
By symmetry,
1.5 kip ft at .
ME= ⋅
1.5 kip ft
E
M= ⋅
max| | 3.00 kip ft 36.0 kip in.M= ⋅= ⋅
occurs at E.
For
3
W12 16, 17.1 in
x
S×=
Normal stress:
max
max
| | 36
17.1
x
M
S
σ
= =
max
2.11 ksi
σ
=
consent of McGraw-Hill Education.
PROBLEM 12.25
Knowing that
480 N,PQ
= =
determine (a) the distance a
for which the absolute value of the bending moment in the
beam is as small as possible, (b) the corresponding maximum
normal stress due to bending. (Hint: Draw the bending-
moment diagram and equate the absolute values of the largest
positive and negative bending moments obtained.)
SOLUTION
480 N 480 NPQ= =
Reaction at A:
0: 480( 0.5)
480(1 ) 0
720
960 N
D
M Aa a
a
Aa
Σ = −+ −
− −=
= −
Bending moment at C:
0: 0.5 0
360
0.5 480 N m
CC
C
M AM
MA a
Σ=− + =
==−⋅
Bending moment at D:
0: 480(1 ) 0
480(1 ) N m
DD
D
MM a
Ma
Σ = − − −=
=−−⋅
(a) Equate:
360
480(1 ) 480
DC
MM a a
− = −= −
0.86603 m
a=
866 mm
a=
128.62 N 64.31 N m 64.31 N m
CD
AM M= = ⋅ =−⋅
(b) For rectangular section,
2
1
6
S bh=
2 3 93
1(12)(13) 648 mm 648 10 m
6
S
−
= = = ×
6
max
max 9
| | 64.31 99.2 10 Pa
6.48 10
M
S
σ
−
= = = ×
×
max
99.2 MPa
σ
=
consent of McGraw-Hill Education.
PROBLEM 12.19
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
SOLUTION
0:
B
M∑=
(11)(25) 10 (8)(25) (2)(25) 0 52.5 kipsC−+ + = =C
0:
C
M∑=
(1)(25) (2)(25) (8)(25) 10 0 22.5 kipsB− − += =B
Shear:
A to
C−
:
25 kipsV= −
C+ to D−:
27.5 kipsV=
D+ to E−:
2.5 kipsV=
E+ to B:
22.5 kipsV= −
Bending moments:
At C,
0: (1)(25) 0
C
MM∑ = +=
25 kip ftM=−⋅
At D,
0: (3)(25) (2)(52.5) 0
D
MM∑ = − +=
30 kip ftM= ⋅
At E,
0: (2)(22.5) 0 45 kip ft
E
MM M∑ = −+ = = ⋅
max 45 kip ft 540 kip in.M= ⋅= ⋅
For
12 35
S×
rolled steel section,
3
38.1 in
S=
Normal stress:
540 14.17 ksi
38.1
M
S
s
= = =
14.17 ksi
s
=
consent of McGraw-Hill Education.
PROBLEM 12.20
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
SOLUTION
0:
M∑=
consent of McGraw-Hill Education.
PROBLEM 12.20 (Continued)
PROBLEM 12.21
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal stress
due to bending.
SOLUTION
Reaction at C:
0: (18)(5) 13 +(5)(10) 0
10.769 kips
B
MC
C
Σ= − =
=
Reaction at B:
0: (5)(5) (8)(10) 13 0
4.231 kips
C
MB
B
= − +=
=
Shear diagram:
to : 5 kips
to : 5 10.769 5.769 kips
to : 5.769 10 4.231 kips
AC V
CD V
DB V
−
+−
+
= −
=−+ =
= −=−
At A and B,
0M=
At C,
0: (5)(5) 0
25 kip ft
CC
C
MM
M
= +=
=−⋅
At D,
0: (5)(4.231)
21.155 kip ft
DD
D
MM
M
Σ=−+
= ⋅
max
5.77 kipsV=
max
||M
occurs at C.
max
| | 25 kip ft 300 kip in.M= ⋅= ⋅
For
W14 22×
rolled-steel section,
3
29.0 inS=
Normal stress:
300
29.0
M
S
σ
= =
10.34 ksi
σ
=
consent of McGraw-Hill Education.
PROBLEM 12.22
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
SOLUTION
Reactions:
0: 4 64 (24)(2)(1) 0 28 kN
D
MA∑ = −− = =A
0: 28 (24)(2) 0 76 kN
y
FD∑ = − +− = =D
A to C:
0 2mx<<
0: 28 0
28 kN
y
FV
V
∑ = −− =
= −
0: 28 0
( 28 ) kN m
J
M Mx
Mx
∑= + =
=−⋅
C to D:
2m 4mx<<
0: 28 0
28 kN
y
FV
V
∑ = −− =
= −
0: 28 64 0
( 28 64) kN m
J
M Mx
Mx
∑ = + −=
=−+ ⋅
D to B:
4m 6mx<<
0:
24(6 ) 0
( 24 144) kN
y
F
Vx
Vx
∑=
− −=
=−+
2
0:
6
24(6 ) 0
2
12(6 ) kN m
J
M
x
Mx
Mx
∑=
−
−− − =
=−− ⋅
3
max 56 kN m 56 10 N mM= ⋅=× ⋅
For
S250 52×
section,
33
482 10 mmS= ×
Normal stress:
36
63
56 10 N m 116.2 10 Pa
482 10 m
M
S
σ
−
×⋅
= = = ×
×
116.2 MPa
σ
=
consent of McGraw-Hill Education.
PROBLEM 12.23
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal
PROBLEM 12.23 (Continued)
Bending Moment at D:
( )
0: 11.20 5 0
56.0 kip ft
DD
D
MM
M
Σ= − + =
= ⋅
From the diagram,
max
56 kip ft 672 kip in.M= ⋅= ⋅
For
W12 40
×
rolled-steel shape,
3
51.5 in
x
S=
Stress:
max
m
M
S
σ
=
672 13.05 ksi
51.5
m
σ
= =
13.05 ksi
m
σ
=
consent of McGraw-Hill Education.
PROBLEM 12.24
Knowing that
3 kipsW=
, draw the shear and bending-moment
diagrams for beam AB and determine the maximum normal stress
due to bending.
SOLUTION
By symmetry,
AB=
0: 2 3 2 0
0.5 kips
y
FA B
AB
Σ = −+−+ =
= =
Shear:
to : 0.5 kipsAC V
−
=
to : 1.5 kipsCD V
+−
= −
to : 1.5 kipsDE V
+− =
to : 0.5 kipsEB V
+
= −
Bending moment:
max
1.50 kips
V=
At C,
0: (3)(0.5) 0
CC
MMΣ= − =
1.5 kip ft
C
M= ⋅
At D,
0: (6)(0.5) (3)(2) 0
DD
MM+Σ = − + =
3 kip ft
D
M=−⋅
By symmetry,
1.5 kip ft at .
ME= ⋅
1.5 kip ft
E
M= ⋅
max| | 3.00 kip ft 36.0 kip in.M= ⋅= ⋅
occurs at E.
For
3
W12 16, 17.1 in
x
S×=
Normal stress:
max
max
| | 36
17.1
x
M
S
σ
= =
max
2.11 ksi
σ
=
consent of McGraw-Hill Education.
PROBLEM 12.25
Knowing that
480 N,PQ
= =
determine (a) the distance a
for which the absolute value of the bending moment in the
beam is as small as possible, (b) the corresponding maximum
normal stress due to bending. (Hint: Draw the bending-
moment diagram and equate the absolute values of the largest
positive and negative bending moments obtained.)
SOLUTION
480 N 480 NPQ= =
Reaction at A:
0: 480( 0.5)
480(1 ) 0
720
960 N
D
M Aa a
a
Aa
Σ = −+ −
− −=
= −
Bending moment at C:
0: 0.5 0
360
0.5 480 N m
CC
C
M AM
MA a
Σ=− + =
==−⋅
Bending moment at D:
0: 480(1 ) 0
480(1 ) N m
DD
D
MM a
Ma
Σ = − − −=
=−−⋅
(a) Equate:
360
480(1 ) 480
DC
MM a a
− = −= −
0.86603 m
a=
866 mm
a=
128.62 N 64.31 N m 64.31 N m
CD
AM M= = ⋅ =−⋅
(b) For rectangular section,
2
1
6
S bh=
2 3 93
1(12)(13) 648 mm 648 10 m
6
S
−
= = = ×
6
max
max 9
| | 64.31 99.2 10 Pa
6.48 10
M
S
σ
−
= = = ×
×
max
99.2 MPa
σ
=
consent of McGraw-Hill Education.
