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2.001 - MECHANICS AND MATERIALS I
Lecture #3
9/13/2006
Prof. Carol Livermore

Recall from last time:

FBD:

Solve equations of motion.

Fx = 0

Fy = 0

MA = 0
See 9/11/06 Notes.

Solution:

1

Draw each component separately.
FBD of Bar 1

Fx = 0
FAx + FCx = 0

Fy = 0
FAy + FCy = 0

MA = 0
l cos θFCy − l sin θFCx = 0
Solve.

FAx = FCx
FAy = −FCy
FCy sin θ
= = tan θ Forces track with angle.
FCx cos θ
Two Force Member:
If forces are only applied at 2 points, then:
1. Forces are equal and opposite.
2. Forces are aligned (colinear) with the vector tha connects the two
points of application of forces.

So the FBD could be rewritten as:

2

Now look at the FBDs of 1, 2, and C:

Solve equations of equilibrium for Pin C.

Fx = 0
P + F2 sin φ − F1 sin φ = 0

Fy = 0
−F2 cos φ − F1 cos φ = 0 ⇒ F1 = −F2

M = 0 ⇒ No additional info.(Γ = 0)

P + 2F2 sin φ = 0
Note φ = 30◦ , so:

F2 = −P
F1 = P
Now look at FBD of 1, 2, and A:

3

Look at equilibrium of Pin A:

FX = 0
P cos θ + F3 = 0
F3 = −P cos θ

Note θ = 60◦ , so:

P
F3 = −
2

So:

EXAMPLE:

4

Q: What are the reactions at the supports?

FBD:

Fx = 0
RAx + RDx = 0

Fy = 0
R A y + R Dy − P = 0

MA = 0
−2dP + dRDx = 0
Solve.

RDx = 2P
RAx = −2P
RAy + RDy = P
Note: BD is a 2-Force member.

Draw FBD of BD and Pin D.

5

Solve for Pin D.

Fx = 0
2P + FBD cos θ = 0

Fy = 0
RDy + FBD sin θ = 0
Solve.
2P
FBD = −
cos θ
2P
R Dy − sin θ = 0
cos θ
RDy = 2P tan θ
Note: θ = 45◦ , so:

RDy = 2P
Substituting:

RAy + 2P − P = 0
RAy = −P
Check:

Statically Determinate: A situation in which the equations of equilibrium de-
termine the forces and moments that support the structure.

EXAMPLE:

6

FBD

Fx = 0
R Ax = 0

Fy = 0
R Ay + R By − P = 0

MB = 0
−aRAy + MA − (l − a)P = 0
Note: 4 unknowns and 3 equations.

This is statically indeterminate.

7

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