Shear and Bending Moment in Beams

IV.) Shear and Bending Moment in Beams

A.) Reaction Forces (Statics Review)

1.) Replace Supports with unknown reaction forces (free body diagram)

a.) Roller - produces a reaction force perpendicular to the support plane.

RY

b.) Pin (or Hinge) - produces a vertical and horizontal reaction.

Ry

Rx

c.) Fixed - produces a reaction force in any direction and Moment.

Ry

Rx

M

2.) Apply laws of equilibrium to find RAX,

RAY, RBY

Fx = 0 Fy = 0 Mz = 0

RAX

RAYRBY

B.) Internal Shear

1.) Shear - find by cutting a section at the point of interest and Fy = 0 on the FBD.

Ry

Rx

F.B.D.

V

B.) Internal Bending Moment

2.) Moment - find by cutting a section at the point of interest and M = 0 on the FBD.

Ry

Rx

F.B.D.

V

M

If you were to find the internal shear and moment at several locations along the length of a beam, you could plot a graph shear vs. length and a graph of moment vs. length and find where the maximum shear and moment occur.

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V(k)

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M (k-ft)

C.) Shear Diagram - Simpler Way to Draw

1.) Sketch the beam with loads and supports shown (this is the LOAD DIAGRAM).

2.) Compute the reactions at the supports and show them on the sketch.

3.) Draw Shear Diagram baseline(shear = zero) below the load

diagrama horizontal line.

4.) Draw vertical lines down from the load diagram to the shear diagram at:a.) supportsb.) point loadsc.) each end of distributed loads

5.) Working from left to right, calculate the shear on each side of each support and point load and at eachend of distributed loads:

a.) For portions of a beam that have

no loading, the shear diagram isa horizontal line.

b.) Point loads (and reactions) cause a vertical jump in the shear

diagram.

- The magnitude of the jump is equal to the magnitude ofthe load (or reaction).

- Downward loads cause a negative change in shear.

c.) For portions of a beam under distributed loading:

i.) the slope of the shear diagram is equal to the intensity (magnitude)

of the uniformly distributed load (w).

ii.) the change in shear between two points is equal to the area under

the load diagram between those two points.

V = wL (uniformly distributed)

V = (wL)/2 (triangular distribution)

Note: If the distributed load is acting downward “w” is negative.

6.) Locate points of zero shear using a known shear value at a known location

and the slope of the shear diagram(w)

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D.) Moment Diagram - Simpler Method

1.) Moment = 0 at ends of simply supported beams.

2.) Peak Moments occur where the sheardiagram crosses through zero.

Therecan be more than one peak momenton the diagram.

3.) Extend the vertical lines below theshear diagram and draw the MomentDiagram baseline (moment = 0), ahorizontal line. Also, extend verticallines down from points of zero shear.

4.) Working left to right, calculate the moment at each point the shear wascalculated and at points of zero

shear:

a.) the change in moments between two points is equal to the area

under the shear diagram between those points.

b.) determine the slope of the moment diagrams as follows:

i.) if the shear is positive and constant, the slope of the moment diagram is positive and constant.

Negative, constant shear(-)

(+)

Positive, constant shear

(+) 0(-)

V

Positive, constant slope

Negative, constant slope

(+) 0 (-)

M

ii.) if the shear is positive and increasing, the slope of the moment diagram is positive and increasing.

Positive, decreasing shear

Negative, decreasing shear

(+) 0 (-)

V

Positive, decreasing slope

Negative, decreasing slope

(+) 0 (-)

M

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