Beams Shear & Moment Diagrams

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15 Νοε 2013 (πριν από 3 χρόνια και 6 μήνες)

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Beams

Shear & Moment Diagrams

E. Evans

2/9/06

Beams


Members that are slender and support
loads applied perpendicular to their
longitudinal axis.

Span, L

Distributed Load, w(x)

Concentrated Load, P

Longitudinal
Axis

Types of Beams


Depends on the support configuration

M

F
v

F
H

Fixed

F
V

F
V

F
H

Pin

Roller

Pin

Roller

F
V

F
V

F
H

Statically Indeterminate Beams


Can you guess how we find the “extra”
reactions?

Continuous Beam

Propped Cantilever
Beam

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.

L

P

a

b

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.

Pb/L

x

Left Side of Cut

V

M

N

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.

Pa/L

L
-

x

Right Side of Cut

V

M

N

Positive Directions
Shown!!!

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.

20 ft

P = 20 kips

12 kips

8 kips

12 ft

1

7

10

6

2

3

9

4

5

8

Point 6 is just left of P and Point 7 is just right of P.

20 ft

P = 20 kips

12 kips

8 kips

12 ft

1

7

10

6

2

3

9

4

5

8

V

(kips)

M

(ft
-
kips)

8 kips

-
12 kips

96

48

64

48

72

24

80

16

32

x

x

20 ft

P = 20 kips

12 kips

8 kips

12 ft

V

(kips)

M

(ft
-
kips)

8 kips

-
12 kips

96 ft
-
kips

x

x

V & M Diagrams

What is the slope
of this line?

a

b

c

96 ft
-
kips/12’ = 8 kips

What is the slope
of this line?

-
12 kips

20 ft

P = 20 kips

12 kips

8 kips

12 ft

V

(kips)

M

(ft
-
kips)

8 kips

-
12 kips

96 ft
-
kips

x

x

V & M Diagrams

a

b

c

What is the area of
the blue rectangle?

96 ft
-
kips

What is the area of
the green rectangle?

-
96 ft
-
kips

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.







dx
)
x
(
V
)
x
(
M
dx
)
x
(
w
)
x
(
V
function
load
the
)
x
(
w
The Relationship Between Load, Shear and
Bending Moment

Load

0

Constant

Linear

Shear

Constant

Linear

Parabolic

Moment

Linear

Parabolic

Cubic

Common Relationships

Load

0

0

Constant

Shear

Constant

Constant

Linear

Moment

Linear

Linear

Parabolic

Common Relationships

M

Example: Draw Shear & Moment
diagrams for the following beam

3 m

1 m

1 m

12 kN

8 kN

A

C

B

D

R
A

= 7 kN


R
C

= 13 kN



3 m

1 m

1 m

12 kN

A

C

B

D

V

(kN)

M

(kN
-
m)

7

-
5

8

8 kN

7

-
15

8

7

-
8

2.4 m