1.

• IV .) Shear and Bending Moment in Beams

A.) Reaction Forces (Statics Review)

1.) Replace Supports with unknown

reaction forces (free body diagram)

2.

a.) Roller - produces a reaction forceperpendicular to the support plane.

R Y 3.

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

R y R x 4.

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

R y R x M 5.

2.) Apply laws of equilibrium to findR AX,R AY,R BY

F x= 0 F y= 0 M z= 0

R AX R AY R BY 6.

B.) Internal Shear

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

R y R x F.B.D. V 7.

B.) Internal Bending Moment

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

R y R x F.B.D. V M 8.

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.

9. 4 4 4 8 8 2k/ft 3k 4k A B V(k) 10. 4 4 4 8 8 2k/ft 3k 4k A B M (k-ft) 11.

C.) Shear Diagram - Simpler Way to Draw

1.)Sketch the beam with loads andsupports shown (this is the LOADDIAGRAM).

2.)Compute the reactions at the supportsand show them on the sketch.

12.

3.)Draw Shear Diagram baseline

(shear = zero) below the load diagram

a horizontal line.

4.)Draw vertical lines down from the load

diagram to the shear diagram at:

a.)supports

b.)point loads

c.)each end of distributed loads

13.

5.)Working fromlefttoright , calculate

the shear on each side of each

support and point load and at each

end of distributed loads:

a.)For portions of a beam that have

no loading, the shear diagram is

a horizontal line.

14.

b.) Point loads (and reactions) cause

a vertical jump in the shear

diagram.

- The magnitude of the

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

- Downward loads cause a negativechange in shear.

15.

c.)For portions of a beam under

distributedloading:

i.) the slope of the shear diagram isequal to the intensity (magnitude) ofthe uniformly distributed load (w).

ii.) the change in shear between two

points is equal to the area underthe load diagram between thosetwo points.

16.

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)

17. 4 4 4 8 8 2k/ft 3k 4k A B 18.

D.) Moment Diagram - Simpler Method

1.)Moment = 0 at ends of simplysupported beams.

2.)Peak Moments occur where the shear

diagram crosses through zero.There

can be more than one peak moment

on the diagram.

19.

3.)Extend the vertical lines below the

shear diagram and draw the Moment

Diagram baseline (moment = 0), a

horizontal line.Also, extend vertical

lines down from points of zero shear.

4.)Working left to right, calculate the

moment at each point the shear was

calculated and at points of zero shear:

20.

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:

21.

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 22. Positive, constant slope Negative, constant slope (+) 0 (-) M 23.

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 24. Positive, decreasing slope Negative,decreasing slope (+) 0 (-) M 25. 4 4 4 8 8 2k/ft 3k 4k A B 26.