AMS 03 Samp 10 - Shear Diagrams

nutritiouspenΜηχανική

18 Ιουλ 2012 (πριν από 5 χρόνια και 28 μέρες)

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Applied Mechanics Dr. Pogo
Handout: Relationship between distributed load, shear force, and bending moment


Find the shear force and bending moments in the beam having the loading shown.

First, find w
AB
(x): w
AB
(x) = w
max
(1

x/L
AB
)

Finding the shear force for section AB:
V
AB
(x)

V
A
=
xd
L
x
x
x
w








 





AB
max
1
0


V
AB
(x) =

























AB
2
maxAB
AB
2
max
2
)(
0
2 L
x
xwxV
xx
x
L
x
xw


Evaluate this at point B:
22
ABmax
AB
2
AB
ABmaxB
Lw
L
L
LwV













V
B
=

45N

Finding the shear force for section AB. In this region, w = 0:
V
BC
(x)

V
B
=
00
AB






xd
x
Lx
, so V
BC
(x) =
2
ABmax
Lw



Finding the bending moment for section AB:

M
AB
(x)

M
A
=
x
x
L
xx
wxd
L
x
x
x
x
wxd
x
x
V
0
6
3
2
2
2
00
AB
max
AB
2
maxAB













































AB
maxAB
6
3
2
2
)(
L
xx
wxM


Evaluate this at point B:
362
2
ABmax
AB
3
AB
2
AB
maxB
Lw
L
LL
wM













M
B
=

90Nm
Finding the bending moment for section BC:
M
BC
(x)

M
B
=
 
22222
ABABmax
2
ABmaxABmax
AB
ABmax
AB
AB
max
AB
BC
LxLwLwxLw
x
Lx
xLw
xd
x
Lx
L
wxd
x
Lx
V


















M
BC
(x) =


 
xL
Lw
xM
LxLwLw
BC
3
6
)(
23
AB
ABmaxABABmax
2
ABmax






The shear and bending moment diagrams are shown:


x
y
w
max

L
AB

L
BC

Given:
w
max
= 30N/m
L
AB
= 3m
L
BC
= 2m
A

C

B

B
C
x
A
V

45N

B
C
x
A
M

90
N
m


18
0
N
m