1. SHEAR AND BENDING MOMENT DIAGRAMS IN HORIZONTAL BEAMS WITH
DOWNWARD VERTICAL LOADING

2. When a horizontal beam, supported at each end, is subjected to vertical loads along its length, internal stress develops within the beam, which is the measure of the beam’s resistance to loading in order that equilibrium may be maintained.
The loads may be intermittent point loads, or they may be distributed over either a partial or full length of the beam.
A beam made of elastic material, with vertical downward loading has a TENDENCY to bend downward. Such downward bend causes three types of stress develop within the beam:
Shear stress, which is variable along its length, but is maximum at the neutral axis.
Compressive stress, also variable along its length, but maximum at the top surface of the beam’s cross section.
Tensile stress, variable along its length, and maximum at the bottom surface of the beam’s cross section.

3. Elevation of concrete block wall with opening
A beam loaded as shown, might have simple support at the ends, but for analysis, the assembly would be shown as a FREE BODY DIAGRAM as shown, where the supports are replaced by FORCES that will hold the beam in
EQUILIBRIUM.
Elevation of concrete block wall with opening

4. Because of vertical, downward loading, the elastic beam bends in the direction of the load:
The shape of the beam then assumes a portion of the circumference of a circle:

5. The same is true for the tensile force

9. Remember that BENDING MOMENT is equal to a force times a distance, and units are foot-pounds, foot-kips, inch-pounds, etc., whatever the units of loading and distance on the beam.
The value of SHEAR may also vary along the length of the beam, depending upon the way loads are placed on the beam.
The values of both SHEAR and BENDING MOMENT can be demonstrated graphically, by determining the difference between the vertical upward loads, and the vertical downward loads at any point along the beam.
First, draw a diagram (in reasonable proportions) of the beam as a free body diagram, showing all the vertical forces on the beam, upward or downward as applicable, with the beam supports shown as reaction forces. Second, find the values of the reaction forces - - -

10. Calculate the values of the
end reactions at A and B:
To find B, sum the moments
about point A, and write:
-(200x5) – (300x10) –(400x18) + By x 24 = 0
1000 – 3000 – x By = By = 11,200
By = 11,200 ; By = lb 24
Then, by summation of vertical forces,
Ay = 900 – = lb

11. Show the calculated forces at A and B.
Below the free body diagram, draw a line that will be a reference of zero, for values. All numbers below the line will be negative,
and all above the line will
be positive.
Then extend a vertical line downward from each load
on the free body diagram, including the reaction forces.
433.33
466.67

12. the reaction at A – an upward force of
Beginning at the left, from the reference line, directly under point A, draw a line upward that represents
the reaction at A – an upward force of
lb. Select a suitable scale for the length of the values, or use judgment
in comparative
proportion.
Then move from that point above the reference, to the right, and realize nothing happens to change the upward value until you get to the 200 lb load – which is downward. Draw a line that represents the 200 lb force downward, and subtract from the lb load.
433.33
466.67
433.33
233.33

13. From that point, draw a line downward that represents the 300 lb
Then from that point, draw a horizontal line to the right and realize that nothing happens to change the – until you get to the 300 lb force, downward.
From that point, draw a line downward that
represents the 300 lb
downward – then calculate the sum of the forces,
– 300 = , and if the lines are drawn to a suitable scale, see that your line goes BELOW the reference line, which indicates the negative value of lb.
433.33
466.67
433.33
233.33

14. From that point, draw a line downward that represents the 400 lb
From that point, draw a line horizontal to the right and realize that nothing happens to change the – 66.67, until you get to the downward 400 lb force.
From that point, draw a line downward that
represents the 400 lb
downward – then calculate the sum of the forces
– 400 = , which then indicates the negative value below the line of lb.
433.33
466.67
433.33
233.33

15. From that point, draw a line vertically upward that represents the
From that point, draw a line horizontal to the right and realize that nothing happens to change the – until you get to the UPWARD reaction at B, which is a force.
From that point, draw a line vertically upward that
represents the lb
reaction and see that the sum of the forces
equals zero, at the right
end of the reference line.
433.33
466.67
433.33
233.33

16. This demonstrates graphically that all the downward forces must equal all the upward forces, and satisfies the equilibrium formula that the sum of vertical forces = 0.
Also, important to the graphic values of bending moments, is that the sum of the AREAS above the reference line of the shear diagram must equal the sum of the AREAS below the reference line.

17. It is called a SHEAR DIAGRAM of the beam loading. 433.33
The diagram drawn below the free body diagram is a graphic illustration of the values of vertical SHEAR across the length of the beam.
It is called a SHEAR DIAGRAM
of the beam loading.
433.33
466.67
433.33
233.33

18. Now take a clean sheet of paper and do this exercise.
With the reactions calculated, SKETCH A SHEAR DIAGRAM.
Leave space below your diagram to draw a moment diagram later.
Ay = 30 lb
By = 20 lb

19. Ay = 280 lb. By = 420 lb EXERCISE TWO: Sketch a shear diagram.
NOW ON THE BACK SIDE OF THAT SAME SHEET OF PAPER DO THIS
EXERCISE TWO: Sketch a shear diagram.
Ay = 280 lb By = 420 lb

20. The shear diagram is a graphic illustration of the values of vertical SHEAR across the length of the beam. The values vary because of the position of loads. It is easy to see that if one were to cut the beam just to the right of the 200 lb load, then isolate the left side of the cut, that it would take a force of lb to keep that part in equilibrium.
Realize that any
portion of a body
may be isolated,
provided the
supports for that
portion of the
body are replaced
with forces that
will maintain
equilibrium for
the object.

21. Cut the beam just to the right of the 200 lb load and isolate the left side. Then sum the vertical forces and write:
lb lb - shear value = 0
Shear value = – 200 = lb and must be downward

22. WITH THE SAME SECTION OF THE LEFT END OF THE BEAM, CONSIDER THAT BENDING MOMENT IS PRESENT IN THE BEAM AS WELL AS SHEAR. AT A POINT JUST BELOW THE 200 LB LOAD, SUM THE MOMENTS THERE AND WRITE:
x 5 = ft lbs of moment must exist to hold the assembly in equilibrium. Realize that the 200 lb force will not cause rotation about the moment point, because perpendicular distance is 0.

23. NOW CUT A SECTION THROUGH THE BEAM JUST TO THE RIGHT OF THE 300 LB LOAD - - -

24. sdfg

25. ONCE MORE, JUST TO THE RIGHT OF THE 400 LB LOAD

27. ONE CAN READILY SEE THAT A GREAT DEAL OF WORK IS REQUIRED TO FIND THE SHEAR AND MOMENT VALUES AT EACH POINT, PARTICULARLY FOR A BEAM WITH A LARGE NUMBER OF LOADS. THE PROCESS IS EVEN MORE INVOLVED WHEN THE LOADS ARE DISTRIBUTED, RATHER THAN POINT LOADS.
WITH A SHEAR AND MOMENT DIAGRAM, ONE CAN SEE THE ENTIRE LENGTH OF THE BEAM, AND THE VALUES OF SHEAR AND MOMENT AT EACH POINT
SINCE A RELATIONSHIP EXISTS BETWEEN SHEAR AND MOMENT, A GRAPHIC ILLUSTRATION OF THE MOMENT VALUES CAN BE MADE FROM THE SHEAR DIAGRAM.

28. Bending moment within the beam is created by the loads, and the value also varies with the position and intensity of the loads.
Since moment is
a force times a
distance, the
units of moment
are foot-pounds
in this example.
The area of the
shear diagram is
also in units of
foot-pounds.
Draw a line below
The shear diagram that represents a reference for moment.

29. Consider this statement and follow the procedure to construct and illustrate the moment diagram;
The CHANGE in value of the bending moment between any two points along the length of a beam is equal to the AREA OF THE SHEAR DIAGRAM between the same two points.

30. To construct A MOMENT DIAGRAM, Begin at the left End of the
reference line,
And realize . . .
That the value
Of Bending
Moment
Must begin at 0
and
End at 0
Then calculate the individual areas of the shear diagram, both above and below the reference line.
Moment reference line

31. 533.36
2800

32. Beginning at the left end of the Moment reference line, realize the area of the shear diagram for the first 5’ is ft lb, which is the CHANGE in the value of MOMENT. Draw a line from the beginning to a point above the reference that will represent the value of the moment. The line will be in an upward direction because the shear value is POSITIVE
533.36
2800

33. Then, realize from the point of 2166.65, the CHANGE in
moment for the next 5’ of beam is the area of the shear
diagram for that 5’ of beam, ADD this sum to the
to get Continue the moment line UPWARD to
that point, because the shear is POSITIVE.
533.36
2800
3333.3

34. Then, realize from that point of 3333.3, the CHANGE in
moment for the next 8’ of beam is the area of the shear
diagram. But note that the next segment of shear is below the
line, NEGATIVE, so the moment line will continue DOWN
WARD ; SUBTRACT the from to get 2800.
533.36
2800
3333.3
2800

35. The last value of shear area is 2800, which must be that
amount since the last value of moment was also 2800, in
order that the last segment of line on the moment diagram
must return to zero on the reference line. Moment values must BEGIN with zero, and END with zero.
533.36
2800
3333.3
2800

36. FROM THE SHEAR AND MOMENT DIAGRAMS, ONE CAN SEE, the
Maximum Shear =
pounds
Maximum Bending Moment =
foot pounds
533.36
2800
3333.3
2800
The purpose of a shear and moment diagram is to determine the maximum values of shear and moment in a beam. These are two reactions to loading that must be known so that a beam of suitable material, strength and size can be chosen for the task.

37. NOW FOR THE TWO BEAMS FOR WHICH YOU DID SHEAR DIAGRAMS, SKETCH A MOMENT DIAGRAM FOR EACH ONE.
ON YOUR DRAWINGS, JUST BELOW THE SHEAR DIAGRAM, DRAW A LINE TO REPRESENT A REFERENCE LINE FOR MOMENT.
THEN CALCULATE THE AREAS OF THE SHEAR DIAGRAM.
THEN, BEGINNING AT THE LEFT END OF THE MOMENT REFERENCE LINE, DRAW A LINE FOR EACH SEGMENT THAT REPRESENTS SHEAR AREA
REMEMBER, THE MOMENT LINES GO UPWARD FOR POSITIVE SHEAR, AND DOWNWARD FOR NEGATIVE SHEAR

38. Ay = 30 lb By = 20 lb 30 SHEAR DIAGRAM - 20
YOUR FIRST SHEAR DIAGRAM SHOULD LOOK LIKE THIS:
Ay = 30 lb
By = 20 lb 30
SHEAR DIAGRAM
AREA = 120
AREA = 120
- 20

39. SHEAR DIAGRAM YOUR SECOND SHEAR DIAGRAM SHOULD LOOK LIKE THIS:
Ay = 280
By = 420
280
AREA = 560 80
AREA = 1960
SHEAR DIAGRAM
AREA = 2520
-420
OBSERVE THAT A MAXIMUM MOMENT OCCURS WHERE THE SHEAR LINE CROSSES THE ZERO REFERENCE LINE