sm10 77 78

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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)
P
y
A
L/2
RA
C
MB
L/2
B
x
RB
L
>
2
1
P
L
EIv ' = RA x 2 − < x − > 2 +C1
2
2
2
1
P
L
EIv = RA x 3 − < x − >3 +C1 x + C2
6
6
2
EIv '' = RA x − P < x −
Using boundary conditions:
v(0) = 0 : C2 = 0
1
1
v '( L) = 0 : C1 = PL2 − RA L2
8
2
1
1
1
1
v( L) = 0 :
RA L3 − PL3 + PL3 − RA L2 = 0,
6
48
8
2
Then, from statics:
11
∑ Fy = 0 : RB = 16 P ↑
3
∑ M A = 0 : M B = 16 PL
(1)
RA =
5
P↑
16
(b) Equation (1) becomes
P
L
v=
[5 x 3 − 16 < x − > +12 L2 x − 15L2 x]
96 EI
2
Making x=L/2:
7 PL3
vC =
↓
768 EI
7(20 ×103 )(5)3
= 6.62 mm
=
768(200 ×109 )(17.2 ×10−6 )
Continued on next slide
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SOLUTION (10.78)
y
M0
MA
A
L/2
C
L/2
B
x
MB
RA
RB
L 0
EIv '' = − M A + RA x − M 0 < x − >
2
1
L
EIv ' = − M A x + RA x 2 + M 0 < x − > +C1
2
2
2
3
M x
R x M
L
EIv = − A + A + 0 < x − > 2 +C1 x + C2
2
6
2
2
(1)
(2)
Using boundary conditions:
1
1
v '( L) = 0 : − M A L + RA L2 + M 0 L + C1 = 0
2
2
v '(0) = 0 : C1 = 0
v(0) = 0 : C2 = 0 :
1
1
1
v( L) = 0 : − M A L2 + RA L3 + M 0 L2 = 0
2
6
8
Solving
3 M0
1
RA =
↓
M A = M0
2 L
4
Statics:
M
3 M0
RB =
↑
MB = 0
2 L
4
(b) Equation (2) for x=L/2:
M 0 L2 3 M 0 L3
vC =
−
=0
32
12 L 8
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
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