Transcript Beams Shear & Moment Diagrams

Beams
Shear & Moment Diagrams
E. Evans
2/9/06
Beams
• Members that are slender and support
loads applied perpendicular to their
longitudinal axis.
Distributed Load, w(x)
Concentrated Load, P
Longitudinal
Axis
Span, L
Types of Beams
• Depends on the support configuration
FH
Pin
FH
FV
FV
Fixed
Roller
M
Fv
Roller
FV
Pin
FV
FH
Statically Indeterminate Beams
Continuous Beam
Propped Cantilever
Beam
• Can you guess how we find the “extra”
reactions?
Internal Reactions in Beams
• At any cut in a beam, there are 3 possible
internal reactions required for equilibrium:
– normal force,
– shear force,
– bending moment.
P
a
b
L
Internal Reactions in Beams
• At any cut in a beam, there are 3 possible
internal reactions required for equilibrium:
– normal force,
– shear force,
– bending moment.
M
Left Side of Cut
N
V
Pb/L
x
Positive Directions
Shown!!!
Internal Reactions in Beams
• At any cut in a beam, there are 3 possible
internal reactions required for equilibrium:
– normal force,
– shear force,
– bending moment.
M
V
Positive Directions
Shown!!!
Right Side of Cut
N
Pa/L
L-x
Finding Internal Reactions
• Pick left side of the cut:
– Find the sum of all the vertical forces to the
left of the cut, including V. Solve for shear, V.
– Find the sum of all the horizontal forces to the
left of the cut, including N. Solve for axial
force, N. It’s usually, but not always, 0.
– Sum the moments of all the forces to the left
of the cut about the point of the cut. Include
M. Solve for bending moment, M
• Pick the right side of the cut:
– Same as above, except to the right of the cut.
Example: Find the internal reactions at
points indicated. All axial force reactions
are zero. Points are 2-ft apart.
P = 20 kips
1
2
3
4
5
8
9
10
6 7
8 kips
12 kips
12 ft
20 ft
Point 6 is just left of P and Point 7 is just right of P.
P = 20 kips
1
2
3
4
5
8
9
10
6 7
8 kips
12 kips
12 ft
20 ft
8 kips
V
x
(kips)
-12 kips
M
(ft-kips)
16
32
48
64
80
96
72
48
24
x
V & M Diagrams
8 kips
P = 20 kips
12 kips
12 ft
20 ft
8 kips
V
x
(kips)
What is the slope
of this line?
96 ft-kips/12’ = 8 kips
M
(ft-kips)
-12 kips
96 ft-kips
b
What is the slope
of this line?
-12 kips
a
c
x
V & M Diagrams
8 kips
P = 20 kips
12 kips
12 ft
20 ft
8 kips
V
x
(kips)
What is the area of
the blue rectangle?
96 ft-kips
-12 kips
96 ft-kips
b
What is the area of
the green rectangle?
-96 ft-kips
M
(ft-kips)
a
c
x
Draw Some Conclusions
• The magnitude of the shear at a point
equals the slope of the moment diagram at
that point.
• The area under the shear diagram
between two points equals the change in
moments between those two points.
• At points where the shear is zero, the
moment is a local maximum or minimum.
The Relationship Between Load, Shear and
Bending Moment
w( x )  the load f unction

M( x )   V( x )dx
V( x )  w( x )dx
Common Relationships
0
Constant
Linear
Constant
Linear
Parabolic
Linear
Parabolic
Cubic
Load
Shear
Moment
Common Relationships
Load
0
0
Constant
Constant
Constant
Linear
Linear
Linear
Parabolic
M
Shear
Moment
Example: Draw Shear & Moment
diagrams for the following beam
12 kN
8 kN
A
C
D
B
1m
RA = 7 kN 
3m
1m
RC = 13 kN 
12 kN
8 kN
A
C
D
B
1m
3m
8
7
V
1m
8
7
-15
(kN)
-5
7
M
(kN-m)
2.4 m
-8