sm13 30

________________________________________________________________________
PROBLEM (*13.30) A propped stepped-cantilever beam is subjected to a concentrated load P at
node 2 and a bending moment M at point 3 as shown in Fig. P13.30. Determine:
(a) The displacements ,
2
v2
θ
, and 3
θ
.
(b) The support reactions R1 , M1 , and R3.
Given: L = 2 m, P = 30 kN, M = 10 kN
⋅
m, EI = 2 x 106 N
⋅
m2
*SOLUTION .
We have
11 2
33 33
48
() () 2
EI EI EI EI L
LL L
LL LL
== =
2
=
Equations (13.22):
22
13
22
12 6 12 6
64 62
4
[] 12 6 12 6
62 64
LL
LL LL
EI
kLL
L
LL LL
−
⎡⎤
⎢⎥
−
⎢⎥
=⎢⎥
−− −
⎢⎥
−
⎣⎦
22
23
22
12 3 12 3
33
8
[] 12 3 12 3
323
LL
LL LL
EI
kLL
L
LL L L
−
⎡⎤
⎢⎥
−
⎢⎥
=⎢⎥
−− −
⎢⎥
−
⎣⎦
2
2
Substituting the given numerical values and rearranging:
112
vv
θ
θ
6
1
3333
3432
[ ] 4(10 ) 3333
3234
k
−
⎡⎤
⎢⎥
−
⎢⎥
=⎢⎥
−− −
⎢⎥
−
⎣⎦
223
vv
3
θ
θ
6
2
6363
3231
[ ] 4(10 ) 6363
3132
k
−
⎡⎤
⎢⎥
−
⎢⎥
=⎢⎥
−− −
⎢⎥
−
⎣⎦
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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6
33 3 3 00
34 3 2 00
33(36)(33)63
[ ] 4(10 ) 32(33)(42) 31
00 6 3 6 3
00 3 1 32
K
−
⎡⎤
⎢⎥
−
⎢⎥
⎢⎥
−− + −+ −
=⎢⎥
−+ + −
⎢⎥
⎢⎥
−
−−
⎢⎥
−
⎣⎦
Equation (13.21a):
vv
112233
v θθ
⎥
θ
1
1
6
3
3
33 3300
34 3200
3390 63
4(10 )
03206 31
00 636 3
0031 32
R
M
P
R
M
−
⎧⎫ ⎡⎤
⎪⎪ ⎢⎥
−
⎪⎪ ⎢⎥
⎪⎪ ⎢⎥
−−− −
⎪⎪
=
⎨⎬ ⎢
−
⎪⎪ ⎢⎥
⎪⎪ ⎢⎥
−
−−
⎪⎪ ⎢⎥
−
⎪⎪ ⎣⎦
⎩⎭
(1)
(a) Boundary conditions are : vv
113
000==
θ
=
Then from Eqs.(1), we have
2
62
3
903
04(10)061
312
Pv
M
θ
θ
−
⎧⎫ ⎡ ⎤⎧
⎪⎪ ⎪
⎢⎥
=
⎨⎬ ⎨
⎢⎥
⎪⎪ ⎪
⎢⎥
⎩⎭ ⎣ ⎦⎩
⎫
⎪
⎬
⎪
⎭
N Inverting and substituting 30Pk
=
and 10 :
M
kN m
=
⋅
3
26
23
3
22 6 36 30(10 ) 2.83
10 6 18 18 0 0.001
360 36 18 108 10(10 ) 0.006
vm
rad
rad
θ
θ
−⎧⎫
−− −
⎧⎫ ⎡ ⎤ ⎧ ⎫
⎪⎪
⎪⎪ ⎪ ⎪
⎢⎥
=−=−
⎨⎬ ⎨ ⎬⎨ ⎬
⎢⎥
⎪⎪ ⎪ ⎪⎪ ⎪
⎢⎥
−−
⎩⎭ ⎣ ⎦ ⎩ ⎭
⎩⎭
m
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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(b) Equation (1) gives
6
12
12(10 )( )R2
ν
θ =−+
6
12
4(10 )( 3 2 )M2
ν
θ =−+
6
32
12(10 )(2 )R23
ν
θθ =− + +
or 6
112(10 )(0.00283 0.001) 22
R
kN=−=
6
14(10 )(0.008 0.002) 24
M
kN m=−=⋅
6
312(10 )( 0.006 0.001 0.006) 12
R
kN=− − − + =
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States
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