sm10 120 121

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________________________________________________________________________
PROBLEMS (10.120 and 10.121) A simple beam with two different moments of inertia supports
loading as shown in Figs. P10.120 and P10.121.
SOLUTION (10.120)
2
11
21
()
3224
Pa Pa
xaA a
EI EI
===
2
22
25 1 ()
33 26 12
a a Pa Pa
xa A a
EI EI
=+ = = =
2
33
3
22 6
aa Pa
xa A EI
=+ = =
(a) 0
C
θ
= from symmetry about C
(b) ///123
0, ,
CA A A CA CA AAA
θ
θθ θ θ =− =− = + +
Thus, 2
123 (3 1 2)
12
APa
AAA EI
θ =− + + =− + +
2
2
Pa
EI
=
3
/112233 (6 5 9)
36
CCA Pa
vt AxAxAx EI
==+ + = ++
3
5
9Pa
EI
=↓
Continued on next slide
x
x
a
C
v
B
1
x
A
tC/A C
v
=
a
P
/2
A
C
B
P
/2
y
M/EI
P
a
/
6EI
P
a
/
2E
a
P
a
/
3EI
a
P
A
θ
2
x
3E
I
EI
3
x
A
1
A
2
A
3

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SOLUTION (10.121)
23 23
12
11 1
() ()
22 224
wa wa wa a wa
Aa A
EI EI EI EI
== = =
23 2 3
34
11
() ()
32 6 34 12
wa wa wa wa
Aa Aa
EI EI EI EI
=− =− =− =−
Thus,
/1 2 3 4
4253
() () () ()
3344
BA aaaa
tA A A A
=+++
3
/
19
232
ABA
wa
t
aEI
θ =− =−
44
9
(32 8 10 3)
48 16
wa wa
EI EI
=+−−=
(a) 33
/13
11
263
CA wa wa
AA EI EI
θ =+= −
33
/9
332
CCAA
wa wa
EI EI
θθ θ =+=− 3
5
96 wa
EI
=
Continued on next slide
Spandrel
p
arabola
x
tangent at A
x
C
C
v
B
C’
A
C’
’
tC/A
A
wa
B
y
M/EI
-wa2/4EI
wa2/2EI
wa2/2
Spandrel
parabola
t
B
/A
wa
x
2EI
w
a
a
C
A
M
wa2/EI
-wa2
/
2E
I
A
θ
A
1
A
2
A
4
A
3
4
a/
3
5a/4
3a/4
2a/3
B
EI
C

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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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(b) /1 3
() ()
34
CA aa
tA A
=+
44 4
1
6248
wa wa wa
EI EI EI
=− =
44
//
19
2832
CCA BA
wa wa
vt t EI EI
=− = − 4
5
32 wa
EI
=
↓