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Frames, Statics

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Presentation for frames in statics
5
1 1 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. 2 The compound beam shown below is pinned connected at B. Determine the reactions at its supports. Neglect its weight and thickness. EXAMPLE 6-15 2 m 2 m 2 m 10 kN 4 kN/m Dismember the beam into 2 segments A B C 3 4 Entire Beam Bar AB Bar BC 3 i. Dismember the beam into 2 segments: (For Bar AB) 10 kN 3 4 A x A y M A B x B y 2 m 4 m For bar BC Entire Beam Back given problem 4 B x B y 8 kN 1 m 2 m i. Dismember the beam into 2 segments: (For Bar BC) C y For bar AB Entire Beam (Summary) Back given problem 5 10 kN 3 4 A x A y M A B x B y 2 m 4 m B x B y 8 kN 1 m 2 m C y For bar BC For bar AB i. Dismember the beam into 2 segments: Compute for the reactions Segment BC Segment AB 6 10 kN 3 4 A x A y M A B x B y 2 m 4 m 0 ) 5 3 )( 10 ( = + x x B kN A ; 0 = + y F ; 0 = + x F 0 ) 5 4 )( 10 ( = y y B kN A 0 ) 4 ( ) 2 )( 5 4 )( 10 ( = m B kN M y A ; 0 = A M ii. Equations of Equilibrium (For segment AB) Segment BC Back
Transcript
Page 1: Frames, Statics

1

1

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

q  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.

2

The compound beam shown below is pinned connected at B. Determine the reactions at its supports. Neglect its weight and thickness.

EXAMPLE 6-15

2 m 2 m 2 m

10 kN 4 kN/m

Dismember the beam into 2 segments

A B

C 3

4

Entire Beam Bar AB

Bar BC

3

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

10 kN

3 4

Ax

Ay MA Bx

By 2 m

4 m

For bar BC

Entire Beam

Back given problem 4

Bx

By

8 kN

1 m

2 m

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

Cy

For bar AB

Entire Beam (Summary)

Back given problem

5

10 kN

3 4

Ax

Ay MA Bx

By 2 m

4 m

Bx

By

8 kN

1 m

2 m

Cy

For bar BC For bar AB

i. Dismember the beam into 2 segments:

Compute for the reactions

Segment BC Segment AB

6

10 kN

3 4

Ax

Ay MA Bx

By 2 m

4 m

0)53)(10( =+− xx BkNA

;0=↑+ ∑ yF

→ ;0=∑+xF

0)54)(10( =−− yy BkNA

0)4()2)(54)(10( =−− mBkNM yA

;0=∑ AM

ii. Equations of Equilibrium (For segment AB)

Segment BC Back

Page 2: Frames, Statics

2

7

Bx

By

8 kN

1 m

2 m

Cy

0=xB

;0=↑+ ∑ yF

→ ;0=∑+xF

08 =+− yy CkNB

0)2()1(8 =+− mCmkN y

;0=∑ BM

ii. Equations of Equilibrium (For segment BC)

Final Answer Back

8

iii. Solving these equations, we obtain:

0)53)(10( =+− xx BkNA

0)54)(10( =−− yy BkNA

0)4()2)(54)(10( =−− mBkNM yA

0=xB

08 =+− yy CkNB

0)2()1(8 =+− mCmkN y

kNAx 6=

0=xB

kNCy 4=

kNAy 6=

kNBy 4=

kNmMA 32=

Review Problem Back

Next Problem

9

EXAMPLE 6-16

Determine the horizontal and vertical components of force which the pin at C on member ABCD of the frame.

0.8 m 100 kg

1.6 m

0.4 m

1.6 m 0.4 m

F E C

B

A

D Draw the FBD of the frame

See the FBD

10

i. FBD of the frame ABCD:

2 m

2.8 m

Dx

Ax

Ay

A

D

Dismember the beam into 3 segments Bar CF

Bar AD

Bar BE

Equation of Equilibrium for Entire Frame

F=981 N

11

07.700 =− NAx

;0=↑+ ∑ yF

→ ;0=∑+xF

0)8.2()2(981 =+− mDmN x

;0=∑ AM

ii. Equations of Equilibrium (Entire Frame)

0981 =− NAy

2 m

2.8 m

Dx

Ax

Ay

A

D

Dismember the beam into 3 segments

Bar CF Bar AD Bar BE

F=981 N

12

F

981 N

ii. Equations of Equilibrium (For Bar CEF)

0)45cos2.1734( =°−−− NCx

;0=↑+ ∑ yF

→ ;0=∑+xF

0)6.1)(45sin()2(981 =°−− mFmN B

;0=+ ∑ CMccw

0)45sin2.1734( =°−− NCy

C Cx

Cy

E

FB

45º

1.6 m 0.4 m

Bar ACD

Bar BE

Entire Frame

NCy 245−=

NCx 1230=

Page 3: Frames, Statics

3

13

Dx

Ax

Ay

A

D

FB

Cx

Cy

0.8 m

1.6 m

0.4 m

FBD of bar ACD

Back Bar BE Entire Frame

14

FBD of bar BE

45º

FB

FB

Bar ACD

Entire Frame

Bar CEF

15

Dx

Ax

Ay

A

D

C Cx

Cy

E F

981 N FB

45º

45º

FB

FB

FB Cx

Cy

0.8 m

1.6 m

0.4 m

1.6 m 0.4 m

Review

Final Answer

16

NCy 245−=

ii. Final Answers for Problem 6-16

NCx 1230=

Review Next Problem

17

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.

3.5 ft

3 ft

C D r=0.5 ft

A B

Member AB FBD of Entire Frame

Disk 18

3 ft

Cx

A

Dx

Dy

Bx

By

Dx

Dy

20 lb

ND

Bx

By 3 ft

ND

ii. Free Body Diagram (Entire Frame)

Member AB EQ of Equilibrium

Disk

Page 4: Frames, Statics

4

19

3.5 ft

3 ft

C D

A

B

ii. Equations of Equilibrium (Entire Frame)

01.17 =− lbAx;0=↑+ ∑ yF

→ ;0=∑+xF

0)5.3()3(20 =+− ftCftlb x

;0=+ ∑ AMCCW

020 =− lbAy

lbCx 1.17=

lbAx 1.17=

lbAy 20=

Member AB Disk

Back

20 lb

Cx

Ax

Ay

20

ii. Equations of Equilibrium (Member AB)

3 ft

A Bx

By 3 ft

ND 01.17 =− xBlb

;0=↑+ ∑ yF

→ ;0=∑+xF

0)3()6(20 =+− ftNftlb D

;0=+ ∑ BMCCW

04020 =+− yBlblb

lbBx 1.17=

lbND 40=

lbBy 20=

Disk Back

Ax

Ay

21

Dx

Dy

20 lb

ND

ii. Equations of Equilibrium (Disk)

0=xD

;0=↑+ ∑ yF

→ ;0=∑+xF

02040 =−− yDlblb

0=xD

lbDy 20=

Summary of Answers Review

Back

22

ii. Final Answers for Problem 6-17

lbCx 1.17=

Review

lbAx 1.17=

lbAy 20=

lbBx 1.17=

lbND 40=

lbBy 20=

0=xD

lbDy 20=

23 24

Solution:

Page 5: Frames, Statics

5

25

Solution:

26

27

Solution:

28

Solution:

29

Solution:

30


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