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PLANE STRUCTURES FRAMES – is a rigid structure which is made up of members at least one of which is not a two-force member. ‰ 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

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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

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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

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ii. Final Answers for Problem 6-16

C x = 1230 N C y = −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

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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

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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

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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:

27


Solution:

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Solution:

29


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Solution:

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Solution:

32


Es11 lecture 21