PLANE STRUCTURES FRAMES â€“ is a rigid structure which is made up of members at least one of which is not a two-force member. Â‰ To simply put it, in order to solve for the unknown forces in the frame, you need to draw the FBD of the frame and individual members in the right sequence.

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EXAMPLE 6-15 The compound beam shown below is pinned connected at B. Determine the reactions at its supports. Neglect its weight and thickness. 10 kN 4 kN/m 4 3

C A

B

2m

2m

2m

Dismember the beam into 2 segments Entire Beam Bar BC Bar AB 2

i. Dismember the beam into 2 segments: (For Bar AB) 10 kN

4

MA

Ay

3

Bx

Ax By 2m 4m

Entire Beam For bar BC Back given problem 3

i. Dismember the beam into 2 segments: (For Bar BC) 8 kN

Bx

By

Cy 1m 2m

Entire Beam (Summary) For bar AB Back given problem 4

i. Dismember the beam into 2 segments: 10 kN

8 kN

4

MA

Ay

3

Bx

Ax By

Bx

By

Cy

2m 4m

1m 2m

For bar AB

For bar BC Compute for the reactions Segment AB Segment BC

5

ii. Equations of Equilibrium (For segment AB) + → ∑ Fx = 0;

10 kN

A x − (10 kN )(

4

MA

Ay

3

∑M

Bx

Ax By 2m

M

A

A

3 ) + Bx = 0 5

= 0;

− (10 kN )(

4 )( 2 ) − B y ( 4 m ) = 0 5

+ ↑ ∑ Fy = 0; 4m

4 A y − (10 kN )( ) − B y = 0 5 Back Segment BC 6

ii. Equations of Equilibrium (For segment BC) + → ∑ Fx = 0;

8 kN

Bx = 0

∑M

Bx

B

= 0;

− 8 kN (1 m ) + C y ( 2 m ) = 0 By

Cy 1m

+ ↑ ∑ Fy = 0;

B y − 8 kN + C

y

= 0

2m

Back Final Answer 7

iii. Solving these equations, we obtain: 3 ) + Bx = 0 5 4 − (10 kN )( )( 2 ) − B y ( 4 m ) = 0 5

A x − (10 kN )( M

A

A y − (10 kN )(

4 )− By = 0 5

Ax = 6kN

Ay = 6kN

Bx = 0

B y = 4kN

Bx = 0

− 8 kN (1 m ) + C y ( 2 m ) = 0

B y − 8 kN + C

y

= 0

M A = 32kNm

C y = 4kN Back Review Problem Next Problem

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EXAMPLE 6-16 Determine the horizontal and vertical components of force which the pin at C on member ABCD of the frame. 1.6 m

0.4 m

D

Draw the FBD of the frame

0.4 m

See the FBD F C

E

1.6 m

B 0.8 m

100 kg A 9

i. FBD of the frame ABCD: 2m D

Dx

Equation of Equilibrium for Entire Frame Dismember the beam into 3 segments

2.8 m

Bar CF F=981 N

Bar AD Bar BE

A Ay

Ax

10

ii. Equations of Equilibrium (Entire Frame)

2m D

∑M

Dx

A

= 0;

− 981 N ( 2 m ) + D x ( 2 . 8 m ) = 0 + → ∑ Fx = 0;

2.8 m F=981 N

A x − 700 . 7 N = 0 + ↑ ∑ Fy = 0;

A Ay

Ax

A

y

− 981 N = 0

Dismember the beam into 3 segments Bar BE Bar AD Bar CF 11

ii. Equations of Equilibrium (For Bar CEF)

1.6 m Cx

0.4 m E

C

F

45º Cy

+ ccw∑ M C = 0;

FB

981 N

− 981 N ( 2 m ) − ( F B sin 45 ° )( 1 . 6 m ) = 0 + → ∑ Fx = 0;

− C

x

C x = 1230 N − ( − 1734 . 2 cos 45 ° N ) = 0

+ ↑ ∑ Fy = 0;

C

y

C y = −245 N

− ( − 1734

. 2 sin 45 ° N ) = 0

Entire Frame Bar ACD Bar BE 12

FBD of bar ACD

D Dx

0.4 m Cx Cy 1.6 m

FB

0.8 m A Ay

Ax

Back Bar BE Entire Frame 13

FBD of bar BE

FB

45ยบ

FB Bar ACD Bar CEF Entire Frame 14

1.6 m E

C

Cx

0.4 m F

45ยบ

D Dx

0.4 m

Cy

FB

981 N FB

Cx Cy 1.6 m

FB 45ยบ

0.8 m A

Ax

FB

Review Final Answer

Ay

15

ii. Final Answers for Problem 6-16

C x = 1230 N C y = âˆ’245 N

Review Next Problem 16

EXAMPLE 6-17 The smooth disk shown below is pinned connected at D and has a weight of 20 lbs. Neglect the weights of the other members, determine the horizontal and vertical components of reaction at pins B and D.

r=0.5 ft

D

C

3.5 ft A

3 ft B

FBD of Entire Frame Member AB Disk

17

ii. Free Body Diagram (Entire Frame) Dx

Cx

20 lb

Dy Dx Bx

Dy ND

By ND

A Bx

EQ of Equilibrium 3 ft

3 ft

By

Member AB Disk

18

ii. Equations of Equilibrium (Entire Frame) 20 lb D

C

Cx

Back 3.5 ft

Member AB Disk

A Ax

B

+ CCW ∑ M A = 0;

Ay

3 ft

− 20 lb ( 3 ft ) + C x ( 3 . 5 ft ) = 0 + → ∑ Fx = 0;

C x = 17.1lb

Ax = 17.1lb

A x − 17 . 1 lb = 0 + ↑ ∑ Fy = 0; A

Ay = 20lb y

− 20 lb = 0

19

ii. Equations of Equilibrium (Member AB) + → ∑ Fx = 0;

Bx = 17.1lb

17 . 1 lb − B x = 0

ND

+ CCW ∑ M B = 0; A Ax

− 20 lb ( 6 ft ) + N D ( 3 ft ) = 0

Bx

Ay

3 ft

3 ft

N D = 40lb

By

+ ↑ ∑ Fy = 0; 20 lb − 40 lb + B

B y = 20lb y

= 0

Back Disk 20

ii. Equations of Equilibrium (Disk) + → ∑ Fx = 0;

20 lb

Dx = 0 Dx

Dy

Dx = 0

+ ↑ ∑ Fy = 0;

D y = 20lb

40 lb − 20 lb − D

y

= 0

ND

Back Review Summary of Answers 21

ii. Final Answers for Problem 6-17

C x = 17.1lb

Bx = 17.1lb

Dx = 0

Ax = 17.1lb

N D = 40lb

D y = 20lb

Ay = 20lb

B y = 20lb

Review 22

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Solution:

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Solution:

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Solution:

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Solution:

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Solution:

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Solution:

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Solution:

32