Chapter 6 - Analysis of Structures

Sections 6.9 - 6.11

Sections 6.9 - 6.11 - Frames and Machines

Frames differ from trusses because they contain at least one multiforce member.

A multiforce has at least 3 forces acting on it. Remember that a truss is made entirely of 2 force members!

In Fig 6.20, the frame shown contains two multiforce members: AD and CF.

Sections 6.9 - 6.11 - Frames and Machines cont.

With any multiforce member it can no longer be assumed that all forces follow the axis line of the member. This makes the analysis more difficult.

The procedure for analyzing the forces in a frame is usually to first solve for the reactions by considering the frame as a single rigid body.

Sections 6.9 - 6.11 - Frames and Machines cont.

Once you have the external reactions you must separate the members and examine each with a free body diagram.

In the free body diagram, show all forces acting on the pins in each member, all external support reactions and all external applied loads.

Sections 6.9 - 6.11 - Frames and Machines cont.

When you solve for an unknown force acting on a pin, you can apply an equal and opposite force acting on the same pin in a connected member. (See Fig 6.20)

For a multiforce member you have three equilibrium equations to work with.– Usually MP, Fx, Fy but you may use more

than one moment equation.

Sections 6.9 - 6.11 - Frames and Machines cont.

To determine whether a frame is

statically determinate, break it into

members and count the unknowns.

If there are more unknowns than there

are independent equations, the frame is

indeterminate.

Sample Problem 6.4Determine the force in DE and the force at pin C.

160 mm

100 mm

480 kN

Ay

Ax

B

B

MA = 0 = B x 160 - 480 x 100

B = (480 x 100)/160 = 300 N

Fx = 0 = 300 N - Ax

Ax = 300 N

Fy = 0 = - Ay - 480

Ay = - 480 N

Sample Problem 6.4 cont.

170

150

80

100 mm

80 mm

220 mm

Ax = 300 N

Ay = 480 N

Cx

Cy

DE

MC = 0 = 300 N x 220 mm + (150/170)DE x 80 mm + (80/170)DE x 100 mm

0 = 300 x 220 + 70.6 DE + 47 DE

DE = -66,000/117.6 = -561.2 N (Compression)

Sample Problem 6.4 cont.At pin C:

Fx = 0 = - Cx - 300 N - (150/170)(561)

Cx = 795 N

Fy = 0 = - Cy + 480 N - (80/170)(DE)

Cy = 216 N

•Now these are the forces exerted at C on member ACE.

•However for member BCD these forces would have to be turned around.

480

B

Cy = 216 N

Cx = 795 N

Sample Problem 6.5

A

B

CD

E F

3.6 m

4.8 m

2.7 m

2.7 m

Determine the components of the forces acting on each member of the frame shown.

Sample Problem 6.5 cont.

A

B

CD

E F

3.6 m

4.8 m

Ex

Ey F

2400 N

Free Body: Entire Frame

Since the external reactions involve only three unknowns, we compute the reactions by considering the free-body diagram of the entire frame.

Sample Problem 6.5 cont.

+ ME = 0: -(2400 N)(3.6 m) + F(4.8 m) = 0

F = + 1800 N F = 1800 N

+ Fy = 0: -2400 N + 1800 N + Ey = 0

Ey = + 600 N Ey = 600 N

+ Fx = 0: Ex = 0

Members. The frame is now dismembered; since only two members are connected at each joint, equal and opposite components are shown on each member at each joint.

Sample Problem 6.5 cont.

+ MB = 0: -(2400 N)(3.6 m) + Cy(2.4 m) = 0

Cy = + 3600 N

+ MC = 0: -(2400 N)(1.2 m) + By(2.4 m) = 0

By = + 1200 N

+ Fx = 0: -Bx + Cx = 0

We note that neither Bx nor Cx can be obtained by considering only member BCD. The positive values obtained for By and Cy indicate that the force components By and Cy are directed as assumed.

Free Body: Member BCD B C D

1.2 m2.4 mBy

Bx Cx

Cy

2400 N

Sample Problem 6.5 cont.Free Body: Member ABE

+ MB = 0: Bx(2.7 m) = 0 Bx = 0

+ Fx = 0: +Bx - Ax = 0 Ax = 0

+ Fy = 0: -Ay + By + 600 N = 0

-Ay + 1200 N + 600 N = 0 Ay = +1800 NA

B

E

2.7 m

2.7 m

600 N

Ax

Ay

By

Bx

Free Body: Member BCD

Returning now to member BCD, we write

+ Fx = 0: -Bx + Cx = 0; 0 + Cx = 0

Cx = 0

Sample Problem 6.5 cont.Free Body: Member ACF (check)

All unknown components have now been found; to check the results, we verify that member ACF is in equilibrium.

+ MC = (1800 N)(2.4 m) - Ay(2.4 m) - Ax(2.7 m)

= (1800 N)(2.4 m) - (1800 N)(2,4 m) - 0 = 0

0 = 0 (checks)A

F

C2.4 m

1800 N

Ax

Ay

Cy

Cx

Sample Problem 6.6Determine the forces acting on the vertical members of the frame.

600 lb

F

Fy

Fx

Ex

Ey

10'

6'

E

(1) Since this frame has 2 pin supports, it is technically indeterminate.

However you can still solve for Ey and Fy, which will get you started.

ME = 0 = - 600 x 10' + Fy x 6'

Fy = 6000/6 = 1000 lb (1)

Fy = 0 = 1000 + Ey

Ey = -1000 lb (1)

Sample Problem 6.6 cont.

Now write a FBD for AE.

Fy = 0 = -1000 + (2.5/6.5)CD + (2.5/6.5)AB

ME = 0 = (6/6.5)AB x 10' -(6/6.5)CD x 2.5 - 600 x 10

CD = (1000 - .385)AB/.385

CD = 2597 - AB

6.5

6

2.5

2.5

6

6.5

600 lb

Ey = 1000 lb

Ex

CD

AB

2.5

7.5

(2)

Sample Problem 6.6 cont.0 = 9.23 AB - 2.31 CD - 6000

0 = 9.23 AB - 2.31(2597 - AB) - 6000

0 = 11.54 AB - 12,000

AB = 12000/11.54 = 1040 lb (force as drawn in FBD (2))

CD = 2597 - 1040 = 1557 lb (force as drawn in FBD (2))

and Fx = 0 = 600 lb - (6/6.5)AB + (6/6.5)CD - Ex

0 = 600 - 960 + 1437 - Ex

Ex = 1077 lb (2)

Sample Problem 6.6 cont.Now for BF

2.5 6.5

6

6

2.5 6.5

2.5

5.0

D

B

F Fx

Fy

AB

CD

Fy = 0 = -((6/6.5)AB x 7.5) + ((6/6.5)CD x 5.0)

6.92 AB = 4.62 CD

AB = (4.62/6.92)CD = .67 CD

Fy = 0 = -(2.5/6.5)AB - (2.5/6.5)CD +1000

1000 = .385 AB + .385 CD

1000 = .258 CD + .385 CD

CD = 1556 lb

AB = 1040 lbFx = 0 = (6/6.5)AB - (6/6.5)CD + Fx

Fx = 480 lb