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When a beam is loaded by forces or couples, stresses

and strains are created throughout the interior of the

beam.

To determine these stresses and strains, the internal

forces and internal couples that act on the cross sections

of the beam must be found.

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Types of Loads

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Members that are slender and support loads applied

perpendicular to their longitudinal axis.

Span, L

Distributed Load, w(x) Concentrated Load, P

Longitudinal

Axis

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Depends on the support configuration

M

Fv

FH Fixed

FVFV

FH

Pin

Roller

PinRoller

FVFV

FH

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Continuous Beam

Propped Cantilever

Beam

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At any cut in a beam, there are 3 possible internalreactions required for equilibrium:

normal force, shear force,

bending moment.

L

P

a b

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At any cut in a beam, there are 3 possible internalreactions required for equilibrium:

normal force, shear force,

bending moment.

Pb/L

x

Left Side of Cut

V

M

N

Positive Directions

Shown!!!

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At any cut in a beam, there are 3 possible internalreactions required for equilibrium:

normal force, shear force,

bending moment.

Pa/L

L - x

Right Side of CutVM

N

Positive Directions

Shown!!!

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A Happy Beam is +VE A Sad Beam is -VE

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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. Its 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.

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20 ft

P = 20 kips

12 kips8 kips

12 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.

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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 kips

96

4864

4872

24

80

1632

x

x

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20 ft

P = 20 kips

12 kips8 kips12 ft

V(kips)

M(ft-kips)

8 kips

-12 kips

96 ft-kips

x

x

V & M Diagrams

What is the slopeof this line?

a

b

c

96 ft-kips/12 = 8 kips

What is the slope

of this line?-12 kips

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20 ft

P = 20 kips

12 kips8 kips12 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 ofthe blue rectangle?

96 ft-kipsWhat is the area of

the green rectangle?

-96 ft-kips

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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.

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Load0 Constant Linear

Shear

Constant Linear Parabolic

Moment

Linear Parabolic Cubic

Common Relationships

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Load0 0 Constant

Shear

Constant Constant Linear

Moment

Linear Linear Parabolic

Common Relationships

M

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Examples of Devices under ending Loading