sm10 108 111

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________________________________________________________________________
PROBLEM (10.108 through 10.111) A cantilever beam is loaded as shown in Figs. P10.108
through P10.111.Determine:
(a) The slope at the free end.
(b) The deflection at the free end.
SOLUTION (10.108)
We have 0
A
θ
= and 0
A
v=.
332
4432
LL
xb== =
13 bh
AEI =− 2
112 2
339
LwL
EI
=− 3
4
81
wL
EI
=−
(a) Equation (10.48) becomes
0
BA B A
θ
θ =−=, or
3
4
81
BwL
EI
θ =
(b) Equation (10.50):
3
45
() ()
3816
BBA LwLL
vt Ax EI ==−+=− 4
10
243 wL EI
=
↓
Continued on next slide
M
EI
x
B
v
2
2
9wL−
B
θ
A
y
A
2
3
A
R
wL=
x
C
C
w
L/
3
C
B
Parabolic
Spandrel
2
2
9
A
M
wL=2L/3
A
B
Tangent at A

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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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Copyright Act without the permission of the copyright owner is unlawful.
SOLUTION (10.109)
(a) 22
/12
11
24
BBA PL PL
AA EI EI
θθ ==+=− − 2
3
4PL
EI
=
(b) 11 22BBA
vt AxAx== + 22
25
() ()
2346
PL L PL L
EI EI
=− − 3
13
24 PL
EI
=
↓
SOLUTION (10.110)
23
00
1()( )
44 6 24
wL wL
bh
AL
EI EI EI
== =−
Continued on next slide
M
EI
x
A
1
1
x
C
L/
2
C
L/
2
A
B
Tangent at A
-PL
P
tC/A
A
2
P
M
EI
x
tB/A
2
x
-PL
A
2
0
A
θ
=
0
A
v
=
M
EI
Cubic
spandrel
A
x
B
L
A
B
Tangent at A
w0L2/6
w0
A
2
0
6
wL
−
tAB
w0L2/6
0
B
v
=
0
B
θ
=
14
25
n
x
bL
n
+
==
+

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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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(a) 0
/BA B A A
θθθ =−=, 3
0
24
AwL
EI
θ =
(b) 3
04
()
24 5
BA A wL L
tvAx EI
== =− 4
0
30
wL
EI
=
↓
SOLUTION (10.111)
2
11
2Pa
AEI
=
2
22Pa
AEI
=−
(a) 0
/12BA B A AA
θθθ =− =−
22 2
23
22
BPa Pa Pa
EI EI EI
θ =− =
(b) /12
34
() ()
23
BA B aa
tvA A
== +
or 22
32 2
() (2)
22 3
BPa a Pa
va
EI EI
=−⋅ 3
23
12 Pa
EI
=
↓
2
P
a
x
A
1 32a
a
C
a
A
B
Tangent at A
-
P
A
P
a
/
2
M
EI
tB/A
43a
0
0
A
A
v
θ
=
=
A
2