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PROBLEM 12.45
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
SOLUTION
0
01
2
012
2
2
1
1
sin
cos
sin
0 at 0 0
0 at 0 0 0
0
dV x
ww
dx L
wL x dM
VC
L dx
wL x
M Cx C
L
MxC
M x L CL
C
π
π
π
π
π
=−=−
= +=
= ++
= = =
= = =++
=
(a)
0cos
wL x
VL
π
π
=
2
02
sin
wL x
ML
π
π
=
0 at 2
dM L
Vx
dx = = =
(b)
2
0
max 2
sin 2
wL
M
π
π
=
2
0
max 2
wL
M
π
=
PROBLEM 12.46
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
SOLUTION
0
01
2012
2
1
20
22
cos 2
2sin 2
4cos 2
0 at 0. Hence, 0.
4
0 at 0. Hence, .
dV x
ww
dx L
Lw x dM
VC
L dx
Lw x
M Cx C
L
Vx C
Lw
Mx C
π
π
π
π
π
π
=−=−
=− +=
= ++
= = =
= = = −
(a)
0
(2 / sin( /2 )V Lw x L
ππ
=−)
20
(4 / )[1 cos( /2 )]M Lw x L
ππ
2
=−−
(b)
22
0
max 4/M wL
π
=
PROBLEM 12.47
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
SOLUTION
0
2
01
3
0 12
1
2
1
6
dV x
ww
dx L
x dM
VwC
L dx
x
M w Cx C
L
=−=−
=− +=
=− ++
2
0 at 0 0M xC= = =
2
01 10
11
0 at 0 66
M x L wL CL C wL= ==−+ =
(a)
22
00
11
26
x
V w wL
L
=−+
22
0
1( 3 )/
6
V wL x L= −
3
00
11
66
x
M w w Lx
L
=−+
3
0
1( /)
6
M w Lx x L= −
(b)
max
M
occurs when
22
0. 3 0
m
dM V Lx
dx == −=
22
max 0
1
6
3 3 33
m
L LL
x Mw
= = −
2
max 0
0.0642M wL=
PROBLEM 12.48
For the beam and loading shown, determine the equations of the shear
and bending-moment curves, and the maximum absolute value of the
bending moment in the beam, knowing that (a) k = 1, (b) k = 0.5.
SOLUTION
00 0
0
0
2
0
01
1
2
0
0
()
(1 )
(1 )
(1 ) 2
0 at 0 0
(1 ) 2
wx kw L x wx
w k kw
LL L
wx
dV w kw k
dx L
wx
V kw x k C
L
V xC
wx
dM V kw x k
dx L
−
=− =+−
=−= −+
= −+ +
= = =
== −+
23
00
2
2
23
00
(1 )
26
0 at 0 0
(1 )
26
kwx wx
M kC
L
M xC
kwx kwx
ML
= −+ +
= = =
+
= −
(a)
1.k=
2
0
0wx
V wx L
= −
23
00
23
wx wx
ML
= −
Maximum M occurs at
.xL=
2
0
max
6
wL
M=
(b)
1.
2
k=
2
00
3
24
wx wx
VL
= −
23
00
44
wx wx
ML
= −
2
0 at 3
V xL= =
At
( ) ( )
23
22
2
00
33 2
00
2, 0.03704
3 4 4 27
wL wL wL
x L M wL
L
= =−==
At
,0xL M= =
PROBLEM 12.49
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal
stress due to bending.
SOLUTION
0: (0.9)(9) (1.5)(3)(12) 3 0
C
MBΣ= − +=
15.3 kNB=
0: (3.9)(9) 3 (1.5)(3)(12) 0
B
MCΣ= −+ =
29.7 kNC=
Shear:
to : 9 kNAC V= −
: 9 29.7 20.7 kNCV
+
=−+ =
: 20.7 (3)(12) 15.3 kNBV=−=−
max
20.7 kNV=
Locate point E where
0.V=
336 (20.7)(3)
20.7 15.3
ee
e
−
= =
1.725 ft 3 1.275 ftee= −=
Areas under shear diagram:
A to C:
(0.9)(9) 8.1kN mV dx = = ⋅
∫
C to E:
1(1.725)(20.7) 17.8538 kN m
2
V dx
= = ⋅
∫
E to B:
1( 1.275)(15.3) 9.7538 kN m
2
V dx
=− =−⋅
∫
consent of McGraw-Hill Education.
PROBLEM 12.49 (Continued)
Bending moments:
0
A
M=
0 8.1 8.1kN m
C
M=−=− ⋅
8.1 17.8538 9.7538 kN m
9.7538 9.7538 0
E
B
M
M
=−+ = ⋅
=−=
3
max
9.75 10 N m at point ME=×⋅
For
W200 19.3×
rolled-steel section,
3 3 63
162 10 mm 162 10 mS
−
=×=×
Normal stress:
36
6
9.7538 10 60.2 10 Pa 60.2 MPa
162 10
M
S
σ
−
×
== =×=
×
60.2 MPa
σ
=
consent of McGraw-Hill Education.
PROBLEM 12.50
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal
PROBLEM 12.51
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal
PROBLEM 12.52
Draw the shear and bending-moment diagrams for the beam and
loading shown and determine the maximum normal stress due to
bending.
SOLUTION
0: 12 (9)(6)(2) (2)(6) 0
8 kips
D
MA
A
=−+ − =
=
0: (3)(6) 12 (14)(6) 0
10 kips
A
MD
D
=− +− =
=
Shear:
8 kips
8 (6)(2) 4 kips
A
C
V
V
=
=−=−
to : 4 kips
to : 4 10 6 kips
CD V
DB V
= −
=−+ =
Locate point E where
0.V=
612 48
84
4 ft 6 2 ft
ee
e
ee
−
= =
= −=
Areas of the shear diagram:
1
to : (4)(8) 16 kip ft
2
1
to : (2)( 4) 4 kip ft
2
to : (6)( 4) 24 kip ft
to : (2)(6) 12 kip ft
A E Vdx
E C Vdx
C D Vdx
D B Vdx
= = ⋅
= −=− ⋅
= −=− ⋅
= = ⋅
∫
∫
∫
∫
Bending moments:
0
A
M=
0 16 16 kip ft
16 4 12 kip ft
12 24 12 kip ft
12 12 0
E
C
D
B
M
M
M
M
=+= ⋅
= −= ⋅
=−=− ⋅
=−+ =
Maximum
| | 16 kip ft 192 kip inM= ⋅= ⋅
For
W8 31×
rolled steel section,
3
27.5 inS=
Normal stress:
| | 192
27.5
M
S
σ
= =
6.98 ksi
σ
=
consent of McGraw-Hill Education.
PROBLEM 12.45
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
SOLUTION
0
01
2
012
2
2
1
1
sin
cos
sin
0 at 0 0
0 at 0 0 0
0
dV x
ww
dx L
wL x dM
VC
L dx
wL x
M Cx C
L
MxC
M x L CL
C
π
π
π
π
π
=−=−
= +=
= ++
= = =
= = =++
=
(a)
0cos
wL x
VL
π
π
=
2
02
sin
wL x
ML
π
π
=
0 at 2
dM L
Vx
dx = = =
(b)
2
0
max 2
sin 2
wL
M
π
π
=
2
0
max 2
wL
M
π
=
PROBLEM 12.46
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
SOLUTION
0
01
2012
2
1
20
22
cos 2
2sin 2
4cos 2
0 at 0. Hence, 0.
4
0 at 0. Hence, .
dV x
ww
dx L
Lw x dM
VC
L dx
Lw x
M Cx C
L
Vx C
Lw
Mx C
π
π
π
π
π
π
=−=−
=− +=
= ++
= = =
= = = −
(a)
0
(2 / sin( /2 )V Lw x L
ππ
=−)
20
(4 / )[1 cos( /2 )]M Lw x L
ππ
2
=−−
(b)
22
0
max 4/M wL
π
=
PROBLEM 12.47
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
SOLUTION
0
2
01
3
0 12
1
2
1
6
dV x
ww
dx L
x dM
VwC
L dx
x
M w Cx C
L
=−=−
=− +=
=− ++
2
0 at 0 0M xC= = =
2
01 10
11
0 at 0 66
M x L wL CL C wL= ==−+ =
(a)
22
00
11
26
x
V w wL
L
=−+
22
0
1( 3 )/
6
V wL x L= −
3
00
11
66
x
M w w Lx
L
=−+
3
0
1( /)
6
M w Lx x L= −
(b)
max
M
occurs when
22
0. 3 0
m
dM V Lx
dx == −=
22
max 0
1
6
3 3 33
m
L LL
x Mw
= = −
2
max 0
0.0642M wL=
PROBLEM 12.48
For the beam and loading shown, determine the equations of the shear
and bending-moment curves, and the maximum absolute value of the
bending moment in the beam, knowing that (a) k = 1, (b) k = 0.5.
SOLUTION
00 0
0
0
2
0
01
1
2
0
0
()
(1 )
(1 )
(1 ) 2
0 at 0 0
(1 ) 2
wx kw L x wx
w k kw
LL L
wx
dV w kw k
dx L
wx
V kw x k C
L
V xC
wx
dM V kw x k
dx L
−
=− =+−
=−= −+
= −+ +
= = =
== −+
23
00
2
2
23
00
(1 )
26
0 at 0 0
(1 )
26
kwx wx
M kC
L
M xC
kwx kwx
ML
= −+ +
= = =
+
= −
(a)
1.k=
2
0
0wx
V wx L
= −
23
00
23
wx wx
ML
= −
Maximum M occurs at
.xL=
2
0
max
6
wL
M=
(b)
1.
2
k=
2
00
3
24
wx wx
VL
= −
23
00
44
wx wx
ML
= −
2
0 at 3
V xL= =
At
( ) ( )
23
22
2
00
33 2
00
2, 0.03704
3 4 4 27
wL wL wL
x L M wL
L
= =−==
At
,0xL M= =
PROBLEM 12.49
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal
stress due to bending.
SOLUTION
0: (0.9)(9) (1.5)(3)(12) 3 0
C
MBΣ= − +=
15.3 kNB=
0: (3.9)(9) 3 (1.5)(3)(12) 0
B
MCΣ= −+ =
29.7 kNC=
Shear:
to : 9 kNAC V= −
: 9 29.7 20.7 kNCV
+
=−+ =
: 20.7 (3)(12) 15.3 kNBV=−=−
max
20.7 kNV=
Locate point E where
0.V=
336 (20.7)(3)
20.7 15.3
ee
e
−
= =
1.725 ft 3 1.275 ftee= −=
Areas under shear diagram:
A to C:
(0.9)(9) 8.1kN mV dx = = ⋅
∫
C to E:
1(1.725)(20.7) 17.8538 kN m
2
V dx
= = ⋅
∫
E to B:
1( 1.275)(15.3) 9.7538 kN m
2
V dx
=− =−⋅
∫
consent of McGraw-Hill Education.
PROBLEM 12.49 (Continued)
Bending moments:
0
A
M=
0 8.1 8.1kN m
C
M=−=− ⋅
8.1 17.8538 9.7538 kN m
9.7538 9.7538 0
E
B
M
M
=−+ = ⋅
=−=
3
max
9.75 10 N m at point ME=×⋅
For
W200 19.3×
rolled-steel section,
3 3 63
162 10 mm 162 10 mS
−
=×=×
Normal stress:
36
6
9.7538 10 60.2 10 Pa 60.2 MPa
162 10
M
S
σ
−
×
== =×=
×
60.2 MPa
σ
=
consent of McGraw-Hill Education.
PROBLEM 12.50
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal
PROBLEM 12.51
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal
PROBLEM 12.52
Draw the shear and bending-moment diagrams for the beam and
loading shown and determine the maximum normal stress due to
bending.
SOLUTION
0: 12 (9)(6)(2) (2)(6) 0
8 kips
D
MA
A
=−+ − =
=
0: (3)(6) 12 (14)(6) 0
10 kips
A
MD
D
=− +− =
=
Shear:
8 kips
8 (6)(2) 4 kips
A
C
V
V
=
=−=−
to : 4 kips
to : 4 10 6 kips
CD V
DB V
= −
=−+ =
Locate point E where
0.V=
612 48
84
4 ft 6 2 ft
ee
e
ee
−
= =
= −=
Areas of the shear diagram:
1
to : (4)(8) 16 kip ft
2
1
to : (2)( 4) 4 kip ft
2
to : (6)( 4) 24 kip ft
to : (2)(6) 12 kip ft
A E Vdx
E C Vdx
C D Vdx
D B Vdx
= = ⋅
= −=− ⋅
= −=− ⋅
= = ⋅
∫
∫
∫
∫
Bending moments:
0
A
M=
0 16 16 kip ft
16 4 12 kip ft
12 24 12 kip ft
12 12 0
E
C
D
B
M
M
M
M
=+= ⋅
= −= ⋅
=−=− ⋅
=−+ =
Maximum
| | 16 kip ft 192 kip inM= ⋅= ⋅
For
W8 31×
rolled steel section,
3
27.5 inS=
Normal stress:
| | 192
27.5
M
S
σ
= =
6.98 ksi
σ
=
consent of McGraw-Hill Education.
