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PROBLEM 15.1
For the loading shown, determine (a) the equation of the elastic
curve for the cantilever beam AB, (b) the deflection at the free end,
(c) the slope at the free end.
SOLUTION
0: ( ) 0
J
M M PL xΣ = −− −=
()M PL x=−−
2
2
()
dy
EI P L x PL Px
dx =− −=−+
21
1
2
dy
EI PLx Px C
dx =−+ +
11
0, 0 : 0 0 0 0
= = =−+ + =
dy
x CC
dx
23
12
11
26
EIy PLx Px C x C=− + ++
22
[ 0, 0]: 0 000 0= = =−+++ =x y CC
(a) Elastic curve.
2
(3 )
6
Px
y Lx
EI
=−−
(2 )
2
dy Px Lx
dx EI
=−−
(b)
at .yxL=
23
(3 )
63
BPL PL
y LL
EI EI
=− −=−
3
3
B
PL
yEI
= ↓
(c)
at .
dy xL
dx =
2
(2 )
22
B
dy PL PL
LL
dx EI EI
=− −=−
2
2
BPL
EI
θ
=
consent of McGraw-Hill Education.
PROBLEM 15.2
For the loading shown, determine (a) the equation of the elastic curve
for the cantilever beam AB, (b) the deflection at the free end, (c) the
slope at the free end.
SOLUTION
0
0
0: 0
K
M MM
MM
=− +=
∑
=
2
0
01
dy
EI M M
dx
dy
EI M x C
dx
= =
= +
01 1 0
, 0: 0
===+=−
dy
x L ML C C ML
dx
2
0 12
1
2
EIy M x C x C= ++
22 2
0 022 0
11
[ , 0] 0 22
x Ly ML ML C C ML== = −+ =
(a) Elastic curve:
22
0(2 )
2
M
y x Lx L
EI
= −+
2
0
()
2
M
y Lx
EI
= −
(b)
at 0:
yx
=
2
0
( 0)
2
A
M
yL
EI
= −
2
0
2
AML
yEI
= ↑
(c)
at 0:
dy x
dx =
000
( ) ( 0)
dy M M M L
Lx L
dx EI EI EI
=− −=− −=−
0
AML
EI
θ
=
consent of McGraw-Hill Education.
PROBLEM 15.3
For the loading shown, determine (a) the equation of the elastic curve
for the cantilever beam AB, (b) the deflection at the free end, (c) the
slope at the free end.
SOLUTION
Use Free body AJ.
0:Σ=
J
M
2
0
0
23
+ ⋅=
wx x
ML
[ , 0]
,0
= =
= =
x Ly
dy
xL
dx
3
0
1
6
= − wx
ML
23
0
1
6
= −
d y wx
dx L
4
01
1
24
=−+
dy w x
EI C
dx L
5
012
1
120
=− ++
wx
EIy C x C
L
, 0 :
dy
xL
dx
= =
3
01
10
24 wL C− +=
3
10
1
24
=C wL
[ , 0]= =x Ly
44
0 02
11 0
120 24
=− + +=EIy wL wL C
4
20
1
30
C wL=
(a) Elastic curve.
54 5
0
( 5 4)
120
=− −+
w
y x Lx L
EIL
44
0()
24
= −+
dy w xL
dx EIL
(b)
@ 0:
=yx
4
0
30
= −
A
wL
yEI
4
0
30
= ↓
A
wL
yEI
◄
(c)
@ 0:=
dy x
dx
3
0
24
=
A
dy w L
dx EI
3
0
24
θ
=
AwL
EI
consent of McGraw-Hill Education.
PROBLEM 15.4
For the loading shown, determine (a) the equation of the elastic curve
for the cantilever beam AB, (b) the deflection at the free end, (c) the
slope at the free end.
SOLUTION
2
0: ( ) 0
2
1
2
J
x
M wx M
M wx
Σ= +=
= −
22
2
31
1
2
1
6
dy
EI M wx
dx
dy
EI wx C
dx
= = −
=−+
33
11
11
, 0 : 0 66
= = =−+ =
dy
x L wL C C wL
dx
33
11
66
dy
EI wx wL
dx =−+
43
2
11
24 6
EIy wx wL x C=−++
44
2
11
[ , 0] 0 0
24 6
x L y wL wL C= = =− + +=
44
211 3
24 6 24
C wL wL
=−=−
(a) Elastic curve.
43 4
( 4 3)
24
w
y x Lx L
EI
=− −+
(b)
at 0.yx=
44
3
24 8
A
wL wL
yEI EI
=−=−
4
8
A
wL
yEI
= ↓
(c)
at 0.
dy x
dx =
3
6
A
dy wL
dx EI
=
3
6
AwL
EI
θ
=
consent of McGraw-Hill Education.
PROBLEM 15.5
For the cantilever beam and loading shown, determine (a) the equation of
the elastic curve for portion AB of the beam, (b) the deflection at B, (c) the
slope at B.
SOLUTION
Using ABC as a free body,
PROBLEM 15.6
For the cantilever beam and loading shown, determine (a) the equation of
the elastic curve for portion AB of the beam, (b) the deflection at B, (c) the
slope at B.
SOLUTION
Using ABC as a free body,
0: 0 0
22
yA A
wL wL
FR RΣ= − + = =
2
0: 0
22 4
=−+ = =
AA A
wL L wL
MM M
Using AJ as a free body (J between A and B),
2
0: ( ) 0
42
JwL x
M wx MΣ=− + +=
22
222
2
23
1
11
42
11
42
11
46
M wL wx
dy
EI wL wx
dx
dy
EI wL x wx C
dx
= −
= −
= −+
11
0, 0 : 0 0 0 0
dy
x CC
dx
= = =−+ =
22 4 12
11
8 24
EIy wL x wx C x C= − ++
22
[ 0, 0] : 0 0 0 0 0xy CC= = =−++ =
(a) Elastic curve.
22 4
11
8 24
w
y Lx x
EI
= −
23
11
46
dy w Lx x
dx EI
= −
(b)
at .
2
L
yx=
24
4
2
1 1 11
8 2 24 2 32 384
B
w L L wL
yL
EI EI
= −=−
4
11
384
wL
EI
=
4
11
384
BwL
yEI
= ↑
consent of McGraw-Hill Education.
PROBLEM 15.6 (Continued)
(c)
at .
2
dy L
x
dx =
33
2
1 1 11
4 2 6 2 8 48
B
w L L wL
L
EI EI
θ
= −=−
3
5
48
wL
EI
=
3
5
48
B
wL
EI
θ
=
consent of McGraw-Hill Education.
PROBLEM 15.7
For the beam and loading shown, determine (a) the equation of the elastic
curve for portion AB of the beam, (b) the slope at A, (c) the slope at B.
SOLUTION
0:
310
24
3
8
B
A
A
M
R L wL L
R wL
Σ=
−+ =
=
For portion AB only,
(0 )xL≤<
2
3
0: ( ) 0
82
31
82
J
x
M wLx wx M
M wLx wx
Σ= − + +=
= −
22
2
23
1
34
12
31
82
31
16 6
11
16 24
dy
EI wLx wx
dx
dy
EI wLx wx C
dx
EIy wLx wx C x C
= −
= −+
= − ++
22
34 3
11
[ 0, 0] 0 0 0 0 0
11 1
[ , 0] 0 16 24 48
xy C C
x L y wL wL C L C wL
= = =−++ =
== =−+ =−
(a) Elastic curve.
343
1 11
16 24 48
w
y Lx x L x
EI
= −−
23 3
3 11
16 6 48
dy w Lx x L
dx EI
= −−
(b)
at 0.
dy x
dx =
3
3
1
0048 48
A
dy w wL
L
dx EI EI
= −− =−
3
48
A
wL
EI
θ
=
(c)
at .
dy xL
dx =
33 3
311 0
16 6 48
B
dy w LL L
dx EI
= −− =
0
B
θ
=
PROBLEM 15.8
For the beam and loading shown, determine (a) the equation of the elastic
curve for portion AB of the beam, (b) the deflection at mid-span, (c) the
slope at B.
SOLUTION
Reactions:
00
0
1111
0: 0
2346
1
8
BA
A
M RL wL L wL L
R wL
Σ= − + − =
=
Boundary conditions:
[ 0, 0] [ , 0]x y x Ly= = = =
For portion AB only,
(0 )xL≤<
0
0
3
0
0
11
0: ( ) 0
82 3
11
86
Σ=− + +=
= −
J
wx
M w Lx x x M
L
w
M w Lx x
L
23
0
0
2
24
0
01
35
0
0 12
11
86
11
16 24
11
48 120
dy w
EI w Lx x
L
dx
dy w
EI w Lx x C
dx L
w
EIy w Lx x C x C
L
= −
=−+
= − ++
2
[ 0, 0]: 0 0 0 0
xy C= = =−++
2
0C=
44
0 01
11
[ , 0]: 0 48 120
x L y wL wL CL===−+
3
10
1
80
C wL= −
(a) Elastic curve.
23 5 4
0
1 11
48 120 80
w
y Lx x Lx
EIL
= −−
22 4 4
0
1 11
16 24 80
dy w Lx x L
dx EIL
= −−
(b)
at .
2
L
yx=
555 4
00
2
15
384 3840 160 3840
L
w L L L wL
yEIL EI
= −−=−
4
0
2
256
L
wL
yEI
= ↓
(c)
at .
dy xL
dx =
444 3
00
2
16 24 80 240
B
dy w L L L w L
dx EIL EI
= −− =+
3
0
120
BwL
EI
θ
=
consent of McGraw-Hill Education.
PROBLEM 15.9
Knowing that beam AB is an
S200 34×
rolled shape and that
60 kN, 2 m,PL= =
and
200 GPa,
E=
determine (a) the slope at
consent of McGraw-Hill Education.
PROBLEM 15.2
For the loading shown, determine (a) the equation of the elastic curve
for the cantilever beam AB, (b) the deflection at the free end, (c) the
slope at the free end.
SOLUTION
0
0
0: 0
K
M MM
MM
=− +=
∑
=
2
0
01
dy
EI M M
dx
dy
EI M x C
dx
= =
= +
01 1 0
, 0: 0
===+=−
dy
x L ML C C ML
dx
2
0 12
1
2
EIy M x C x C= ++
22 2
0 022 0
11
[ , 0] 0 22
x Ly ML ML C C ML== = −+ =
(a) Elastic curve:
22
0(2 )
2
M
y x Lx L
EI
= −+
2
0
()
2
M
y Lx
EI
= −
(b)
at 0:
yx
=
2
0
( 0)
2
A
M
yL
EI
= −
2
0
2
AML
yEI
= ↑
(c)
at 0:
dy x
dx =
000
( ) ( 0)
dy M M M L
Lx L
dx EI EI EI
=− −=− −=−
0
AML
EI
θ
=
consent of McGraw-Hill Education.
PROBLEM 15.3
For the loading shown, determine (a) the equation of the elastic curve
for the cantilever beam AB, (b) the deflection at the free end, (c) the
slope at the free end.
SOLUTION
Use Free body AJ.
0:Σ=
J
M
2
0
0
23
+ ⋅=
wx x
ML
[ , 0]
,0
= =
= =
x Ly
dy
xL
dx
3
0
1
6
= − wx
ML
23
0
1
6
= −
d y wx
dx L
4
01
1
24
=−+
dy w x
EI C
dx L
5
012
1
120
=− ++
wx
EIy C x C
L
, 0 :
dy
xL
dx
= =
3
01
10
24 wL C− +=
3
10
1
24
=C wL
[ , 0]= =x Ly
44
0 02
11 0
120 24
=− + +=EIy wL wL C
4
20
1
30
C wL=
(a) Elastic curve.
54 5
0
( 5 4)
120
=− −+
w
y x Lx L
EIL
44
0()
24
= −+
dy w xL
dx EIL
(b)
@ 0:
=yx
4
0
30
= −
A
wL
yEI
4
0
30
= ↓
A
wL
yEI
◄
(c)
@ 0:=
dy x
dx
3
0
24
=
A
dy w L
dx EI
3
0
24
θ
=
AwL
EI
consent of McGraw-Hill Education.
PROBLEM 15.4
For the loading shown, determine (a) the equation of the elastic curve
for the cantilever beam AB, (b) the deflection at the free end, (c) the
slope at the free end.
SOLUTION
2
0: ( ) 0
2
1
2
J
x
M wx M
M wx
Σ= +=
= −
22
2
31
1
2
1
6
dy
EI M wx
dx
dy
EI wx C
dx
= = −
=−+
33
11
11
, 0 : 0 66
= = =−+ =
dy
x L wL C C wL
dx
33
11
66
dy
EI wx wL
dx =−+
43
2
11
24 6
EIy wx wL x C=−++
44
2
11
[ , 0] 0 0
24 6
x L y wL wL C= = =− + +=
44
211 3
24 6 24
C wL wL
=−=−
(a) Elastic curve.
43 4
( 4 3)
24
w
y x Lx L
EI
=− −+
(b)
at 0.yx=
44
3
24 8
A
wL wL
yEI EI
=−=−
4
8
A
wL
yEI
= ↓
(c)
at 0.
dy x
dx =
3
6
A
dy wL
dx EI
=
3
6
AwL
EI
θ
=
consent of McGraw-Hill Education.
PROBLEM 15.5
For the cantilever beam and loading shown, determine (a) the equation of
the elastic curve for portion AB of the beam, (b) the deflection at B, (c) the
slope at B.
SOLUTION
Using ABC as a free body,
PROBLEM 15.6
For the cantilever beam and loading shown, determine (a) the equation of
the elastic curve for portion AB of the beam, (b) the deflection at B, (c) the
slope at B.
SOLUTION
Using ABC as a free body,
0: 0 0
22
yA A
wL wL
FR RΣ= − + = =
2
0: 0
22 4
=−+ = =
AA A
wL L wL
MM M
Using AJ as a free body (J between A and B),
2
0: ( ) 0
42
JwL x
M wx MΣ=− + +=
22
222
2
23
1
11
42
11
42
11
46
M wL wx
dy
EI wL wx
dx
dy
EI wL x wx C
dx
= −
= −
= −+
11
0, 0 : 0 0 0 0
dy
x CC
dx
= = =−+ =
22 4 12
11
8 24
EIy wL x wx C x C= − ++
22
[ 0, 0] : 0 0 0 0 0xy CC= = =−++ =
(a) Elastic curve.
22 4
11
8 24
w
y Lx x
EI
= −
23
11
46
dy w Lx x
dx EI
= −
(b)
at .
2
L
yx=
24
4
2
1 1 11
8 2 24 2 32 384
B
w L L wL
yL
EI EI
= −=−
4
11
384
wL
EI
=
4
11
384
BwL
yEI
= ↑
consent of McGraw-Hill Education.
PROBLEM 15.6 (Continued)
(c)
at .
2
dy L
x
dx =
33
2
1 1 11
4 2 6 2 8 48
B
w L L wL
L
EI EI
θ
= −=−
3
5
48
wL
EI
=
3
5
48
B
wL
EI
θ
=
consent of McGraw-Hill Education.
PROBLEM 15.7
For the beam and loading shown, determine (a) the equation of the elastic
curve for portion AB of the beam, (b) the slope at A, (c) the slope at B.
SOLUTION
0:
310
24
3
8
B
A
A
M
R L wL L
R wL
Σ=
−+ =
=
For portion AB only,
(0 )xL≤<
2
3
0: ( ) 0
82
31
82
J
x
M wLx wx M
M wLx wx
Σ= − + +=
= −
22
2
23
1
34
12
31
82
31
16 6
11
16 24
dy
EI wLx wx
dx
dy
EI wLx wx C
dx
EIy wLx wx C x C
= −
= −+
= − ++
22
34 3
11
[ 0, 0] 0 0 0 0 0
11 1
[ , 0] 0 16 24 48
xy C C
x L y wL wL C L C wL
= = =−++ =
== =−+ =−
(a) Elastic curve.
343
1 11
16 24 48
w
y Lx x L x
EI
= −−
23 3
3 11
16 6 48
dy w Lx x L
dx EI
= −−
(b)
at 0.
dy x
dx =
3
3
1
0048 48
A
dy w wL
L
dx EI EI
= −− =−
3
48
A
wL
EI
θ
=
(c)
at .
dy xL
dx =
33 3
311 0
16 6 48
B
dy w LL L
dx EI
= −− =
0
B
θ
=
PROBLEM 15.8
For the beam and loading shown, determine (a) the equation of the elastic
curve for portion AB of the beam, (b) the deflection at mid-span, (c) the
slope at B.
SOLUTION
Reactions:
00
0
1111
0: 0
2346
1
8
BA
A
M RL wL L wL L
R wL
Σ= − + − =
=
Boundary conditions:
[ 0, 0] [ , 0]x y x Ly= = = =
For portion AB only,
(0 )xL≤<
0
0
3
0
0
11
0: ( ) 0
82 3
11
86
Σ=− + +=
= −
J
wx
M w Lx x x M
L
w
M w Lx x
L
23
0
0
2
24
0
01
35
0
0 12
11
86
11
16 24
11
48 120
dy w
EI w Lx x
L
dx
dy w
EI w Lx x C
dx L
w
EIy w Lx x C x C
L
= −
=−+
= − ++
2
[ 0, 0]: 0 0 0 0
xy C= = =−++
2
0C=
44
0 01
11
[ , 0]: 0 48 120
x L y wL wL CL===−+
3
10
1
80
C wL= −
(a) Elastic curve.
23 5 4
0
1 11
48 120 80
w
y Lx x Lx
EIL
= −−
22 4 4
0
1 11
16 24 80
dy w Lx x L
dx EIL
= −−
(b)
at .
2
L
yx=
555 4
00
2
15
384 3840 160 3840
L
w L L L wL
yEIL EI
= −−=−
4
0
2
256
L
wL
yEI
= ↓
(c)
at .
dy xL
dx =
444 3
00
2
16 24 80 240
B
dy w L L L w L
dx EIL EI
= −− =+
3
0
120
BwL
EI
θ
=
consent of McGraw-Hill Education.
PROBLEM 15.9
Knowing that beam AB is an
S200 34×
rolled shape and that
60 kN, 2 m,PL= =
and
200 GPa,
E=
determine (a) the slope at
consent of McGraw-Hill Education.
