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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.
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.
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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
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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
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
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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
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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.
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
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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
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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
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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=
3
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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
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FBD of bar BE
45º
FB
FB
Bar ACD
Entire Frame
Bar CEF
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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
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NCy 245−=
ii. Final Answers for Problem 6-16
NCx 1230=
Review Next Problem
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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
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
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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
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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=
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Solution:
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Solution:
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27
Solution:
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Solution:
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Solution:
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