Mechanics of Materials

1
Mechanics of Materials

• Internal Forces Distributed Loads

I don't care to belong to a club that accepts
people like me as members. Groucho Marx
2
Review

• The sign convention for shear is

Downward on the right hand face is considered as
positive shear. This does not hold for the FBD.
3
Review

• The sign convention for the bending moment is

Counterclockwise rotation on the right end of the
section is considered as a positive bending
moment.
4
Review

• A general heuristic for finding the internal
  forces and moments at a point in a simple beam
• Choose one end of the beam as the reference
• Convert all the distributed loads that act on the
  beam into forces
• Solve for the reaction at the reference end of
  the beam

5
Review

• A general heuristic for finding the internal
  forces and moments at a point in a simple beam
• Draw a new FBD of the beam section under
  consideration including all distributed loads as
  distributed loads
• Draw the shear and axial forces, and the bending
  moment at the cut surface following our sign
  conventions
• Solve for these three components using the
  equations of equilibrium

6
Distributed Loads

• There is one point that we need to make in
  determining the internal forces
• When we have a distributed load that spans the
  point where we want to calculate the internal
  forces, we need to be careful that we only
  include that part of the distributed load which
  is on the section under consideration.

7
Distributed Loads

• The is best illustrated by working a simple
  example problem.

8
Example

• If we have the system shown below that is loaded
  as shown

9
Example

• We want to know the value of the internal forces
  at a point 2 ft from the left end of the beam

10
Example

• We will have to decide if we want to look at the
  left section (from the red line to A) or the
  right section (from the red line to B)

11
Example

• If we choose the left section, we will need the
  reaction at A
• If we choose the right section, we will need the
  reaction at B

12
Example

• For this example, choose the left section
• The reason may become easier to see later in the
  analysis

13
Example

• Similar to what was done in analysis in Statics,
  we can convert the distributed load to an
  equivalent point load before we solve for the
  reaction

14
Example

• The magnitude of the point load is the area of
  the loading diagram

15
Example

• The line of action of the equivalent point force
  is through the centroid of the loading

16
Example

• For a triangular loading, this is 2/3 of the
  distance from the minimum intensity

17
Example

• So we can replace the distributed load with an
  equivalent point load

18
Example

• Now we can use the equations of equilibrium to
  solve for the reaction at A

19
Example

• We replace the support at A by the reaction
  provided by the support

20
Example

• Here is where the critical mistakes are made
• It would seem that you could just cut the
  section, draw a new FBD and solve that FBD

21
Example

• However, since the point of concern lies under
  the span of a distributed load, we have to go
  back and replace the 30 kip point load with the
  original distributed load

22
Example

• If we had not done this, we would have assumed
  that the effect of the distributed load would not
  have been felt until we reached 4 feet into the
  beam

23
Example

• When we draw the FBD at the section, we can see
  that this isnt so
• We will now cut the beam at the red line and draw
  a new FBD

24
Example

• The FBD of the two sections are as shown

25
Example

• Notice that we have what amounts to new
  distributed loads on each of the sections

26
Example

• It should be apparent why we choose the left hand
  section now
• We are dealing with a triangular loading only
  rather than a triangular loading on top of a
  rectangular loading

27
Example

• We do need to solve for the intensity of the
  loading at 2 feet to proceed with the analysis

28
Example

• We can use similar triangles to solve for the
  intensity
• At 6 feet, the loading intensity is 10 kips/foot
• We need the loading intensity at 2 feet

29
Example

• So using the similar triangles

30
Example

• We can work with our left hand section and put
  the right hand section aside

31
Example

• I have assumed that the shear, the axial force,
  and the bending moment are all positive

32
Example

• We can now take this new distributed loading and
  convert it to an equivalent point load

33
Example

• Using the equilibrium expressions

34
Example

• Using the equilibrium expressions

35
Example

• Using the equilibrium expressions

Notice that we assumed the shear was positive and
drew it pointing downward on the right hand face.
36
Example

• Using the equilibrium expressions

When we used it in the equilibrium expression, it
has a negative sign because it is directed
downward.
37
Example

• Using the equilibrium expressions

When we solve for the shear the sign in the
solution is positive so our original assumption
for the shear was correct.
38
Example

• To solve for the bending moment, we choose a
  moment center and solve

39
Example

• If you go back now and look at what would have
  happened if we had not reconsidered the
  distributed load acting on the section you can
  see that we would have overstated the shear and
  understated the moment

40
A Final Note

• When dealing with a system which has distributed
  loads acting on it take care when you want to
  calculate the internal forces on any point within
  the area covered by the distributed load
• When dealing with points that are not under the
  distributed loading, you can treat all the
  distributed loadings as equivalent point loads

41
Homework

• 7-1.13, 7-1.16,7- 1.20