sm10 40 43

________________________________________________________________________
PROBLEMS (10.40 through 10.43) A beam is supported and loaded as shown in Figs. P10.40
through P10.43. Determine:
(a) The equation of the elastic curve.
(b) The slope at the free end.
(c) The deflection at the free end.
SOLUTION (10.40)
(a)
y
3
M A = wa 2
2
w
A
a
C
a
B
x
RA=wa
3
1
EIv '' = wax − wa 2 − w < x − a > 2
2
2
1
3
1
EIv ' = wax 2 − wa 2 x − w < x − a >3 +C1
2
2
6
1
3
1
EIv = wax3 − wa 2 x 2 − w < x − a > 4 +C1 x + C2
6
4
24
Boundary conditions:
v(0) = 0 : C2 = 0
v ' (0) = 0 : C1 = 0
Equation (a) and (b) become
1
wax
EIv ' = − w < x − a >3 −
(3a − x)
6
2
1
wax 2
EIv = − w < x − a > 4 −
(9a − 2 x)
24
12
(a)
(b)
(1)
(2)
(b) Make x=2a in Eq.(1):
7 wa 3
θB =
6 EI
(c) Make x=2a in Eq.(2):
41 wa 4
vB =
↓
24 EI
Continued on next slide
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SOLUTION (10.41)
(a)
y
P
PL
A
L/2
C
L/2
B
x
L
>
2
P
L
EIv ' = PLx − < x − > 2 +C1
2
2
1
P
L
EIv = PLx 2 − < x − >3 +C1 x + C2
2
6
2
EIv '' = PL − P < x −
(a)
(b)
Boundary conditions:
PL2
7
v '( L) = 0 : PL2 −
+ C1 ,
C1 = − PL2
8
8
2
3
3
PL PL 7 PL
19
v( L) = 0 :
−
−
+ C2 = 0,
C2 = − PL3
2
48
8
48
Equations (a) and (b) become
P
L
v' =
[8Lx − 4 P < x − > 2 −7 L2 ]
8 EI
2
P
L
v=
[24 Lx 2 − 8P < x − >3 −42 L2 + 19 L3 ]
48 EI
2
(1)
(2)
(b) Make x=2a in Eq.(1):
7 PL2
θA =
8 EI
(c) Make x=0 in Eq.(2):
19 PL3
vB =
↑
48 EI
Continued on next slide
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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
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SOLUTION (10.42)
(a)
y
w
B
MA=2wa2
A
C
2a
x
a
RA=2wa
1
1
EIv '' = 2 wax − 2 wa 2 − wx 2 + w < x − 2a > 2
2
2
1
1
EIv ' = wax 2 − 2wa 2 x − wx 3 + w < x − 2a >3 +C1
6
6
1
1
1
EIv = wax 3 − wa 2 x 2 − wx 4 + w < x − 2a > 4 +C1 x + C2
3
24
24
Boundary conditions:
v '(0) = 0 : C1 = 0
v(0) = 0 : C2 = 0
Then, Eqs. (a) and (b) become
w
v' =
[6ax 2 − 12a 2 x − x 3 + < x − 2a >3 ]
6 EI
w
v=
[8ax 3 − 24a 2 x 2 − x 4 + < x − 2a > 4 ]
24 EI
(a)
(b)
(1)
(2)
(b) Make x=3a in Eq.(1):
4 wa 3
θB =
3 EI
(c) Make x=3a in Eq.(2):
10 wa 4
vB =
↓
3 EI
Continued on next slide
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instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
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SOLUTION (10.43)
y
w
A
L
RA
RB
x
C
B
w
M0
Refer to the Fig. P10.43:
1
∑ M A = 0 : − M 0 − 2 wL2 + RB L = 0
M
1
RB = wL + 0
or
L
2
M
1
wL M
∑ Fy = 0 : RA = wL − 2 wL − L0 = 2 − L0
Thus
d 2v
1
1
EI 2 = M ( x) = RA x − wx 2 + RB < x − L > + w < x − L > 2
dx
2
2
dv 1
1
1
1
EI
= RA x 2 − wx3 + RB < x − L > 2 + w < x − L >3 +C1
dx 2
6
2
6
1
1
1
1
EIv = RA x 3 − wx 4 + RB < x − L >3 + w < x − L > 4 +C1 x + C2
6
24
6
24
Boundary conditions:
EIv(0) = 0 : C2 = 0
1
1
EIv( L) = 0 :
RA L3 − wL4 + C1 L = 0
6
24
or
R L2 wL3
wL3 M 0 L
C1 = − A +
=−
+
6
24
24
6
(1)
(2)
Continued on next slide
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Equation (1) and (2) become then
dv
1 wL 2 M 0 2 wx 3 wL M 0
(a)
=θ =
x −
x −
+(
+
[
) < x − L >2
dx
L
L
2 EI 2
3
2
w
wL3 M 0 L
]
+ < x − L >3 −
+
3
12
3
1 wL 3 M 0 x 3 wx 4
wL M 0
v=
[
x −
−
+(
+
) < x − L >3
6 EI 2
L
4
2
L
w
wL3
+ < x − L >4 −
x + M 0 Lx]
4
4
(b)
3L
1 wL3
θC = θ ( ) =
− 5M 0 L)
(
2
6 EI 4
(c) vC = v(
3L
1 9wL4 31
)=
(
− M 0 L2 )
2
24 EI 16
3
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