BeamsShear & 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

Fv

FHFixed

FV FV

FH

Pin

Roller

PinRoller

FVFV

FH

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/Lx

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 CutVM

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 kips8 kips12 ft

1

7

10

6

2 3 94 5 8

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

20 ft

P = 20 kips

12 kips8 kips12 ft

1

7

10

6

2 3 94 5 8

V(kips)

M(ft-kips)

8 kips

-12 kips96

4864

4872

24

80

1632

x

x

20 ft

P = 20 kips

12 kips8 kips12 ft

V(kips)

M(ft-kips)

8 kips

-12 kips96 ft-kips

x

x

V & M Diagrams

What is the slope of this line?

a

b

c

96 ft-kips/12’ = 8 kipsWhat is the slope of this line?

-12 kips

20 ft

P = 20 kips

12 kips8 kips12 ft

V(kips)

M(ft-kips)

8 kips

-12 kips96 ft-kips

x

x

V & M Diagrams

a

b

c

What is the area of the blue rectangle?

96 ft-kipsWhat 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

functionloadthe)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 m1 m

12 kN 8 kNA C

B

D

RA = 7 kN RC = 13 kN

3 m 1 m1 m

12 kNA C

B

D

V(kN)

M(kN-m)

7

-5

8

8 kN

7

-15

8

7

-82.4 m