sm10 77 78

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
Copyright Act without the permission of the copyright owner is unlawful.
________________________________________________________________________
PROBLEMS (10.77 and 10.78) A beam AB made of a W150 x 30 wide-flange section
supported and loaded as shown in Figs. P10.77 and P10.78. Determine:
(a) All of the reactions for this beam.
(b) The deflection at C for the given data:
P = 20 kN, M = 10 kN⋅m, L = 5 m, E = 200 GPa, I = 17.2 x 106 mm4 (see Table B.8)
SOLUTION (10.77)
(a)
'' 2
AL
EIv R x P x
=−<−>
22
1
1
'222
APL
EIv R x x C
=−<−>+
33
12
1
662
APL
EIv R x x C x C
=−<−>++ (1)
Using boundary conditions:
2
(0) 0: 0vC==
22
111
'( ) 0: 82
A
vL C PL RL==−
3332
1111 5
() 0: 0,
64882 16
AAA
v L R L PL PL R L R P=−+−==↑
Then, from statics:
11
0: 16
yB
FRP==↑
∑
3
0: 16
AB
M
MPL==
∑
(b) Equation (1) becomes
322
[5 16 12 15 ]
96 2
PL
vxxLxLx
EI
=−<−>+−
Making x=L/2:
3
7
768
CPL
vEI
=↓
33
96
7(20 10 )(5)
768(200 10 )(17.2 10 )
−
×
=××
6.62 mm
=
Continued on next slide
L/
2
L/
2
R
A
A
C
B
y
x
RB
MB
P

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
Copyright Act without the permission of the copyright owner is unlawful.
SOLUTION (10.78)
0
0
'' 2
AA L
EIv M R x M x
=− + − < − >
201
1
'22
AA L
EIv M x R x M x C
=− + + < − >+ (1)
23 2
012
262 2
AA
M
Mx Rx L
EIv x C x C
=− + + < − > + + (2)
Using boundary conditions:
201
11
'( ) 0: 0
22
AA
vL ML RL ML C=− + + +=
12
'(0) 0: 0 (0) 0: 0:vCvC== ==
23 2
0
111
() 0: 0
268
AA
vL ML RL ML=− + + =
Solving
0
3
2
AM
RL
=↓ 0
1
4
A
M
M=
Statics:
0
3
2
BM
RL
=↑ 0
4
B
M
M=
(b) Equation (2) for x=L/2:
23
00
30
32 12 8
CML M L
vL
=− =
L/
2
L/
2
R
A
A
C
B
y
x
RB
MB
M0
M
A