ef 202, module 3, lecture 3 frames ef 202 - week 11
TRANSCRIPT
EF 202, Module 3, Lecture 3
FramesEF 202 - Week 11
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EF 202, Module 3, Lecture 3
Multiforce Member
• A multiforce member has either
• three or more forces acting on it,
• or two or more forces and one or more moments acting on it.
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EF 202, Module 3, Lecture 3
Frames
•A frame
•has at least one multiforce member,
•is stationary,
•and is designed to support/resist loads.
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EF 202, Module 3, Lecture 3
Machines
•A machine
•has at least one multiforce member,
•is not stationary,
•and is designed to transmit loads from one thing to another.
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EF 202, Module 3, Lecture 3
Why?
•By definition, a truss is not a frame, because ....
•a truss has only two-force members and a frame must have at least one multiforce member.
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EF 202, Module 3, Lecture 3
Frame Examples
Example 6.9, p. 288
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EF 202, Module 3, Lecture 3
Frame Examples
PS 2-2, Problem 4
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EF 202, Module 3, Lecture 3
Frame Examples
PS 2-2, Problem 5
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EF 202, Module 3, Lecture 3
Frame Examples
Online Homework 2-2-3
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EF 202, Module 3, Lecture 3
Frame Examples
Test Two, Problem 2
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EF 202, Module 3, Lecture 3
Analysis of Frames•Because some forces are not
directed along members,
•the method of joints is not sufficient;
•the method of sections is not sufficient.
•But Newton’s first and third laws are still sufficient
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EF 202, Module 3, Lecture 3
x
y
+
Ax
Ay
Dx
Dy1000 lbStatically indeterminate:3 equations4 unknowns
Let’s try to solve without recognizing that BD and ABC are a 2-force and a 3-force member, respectively.
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EF 202, Module 3, Lecture 3
Ax
Ay
Dx
Dy
1000 lb
Bx
By
By
Bx
x
y
+
Statically determinate:6 equations6 unknowns
lbDlbDlbBlbBlbAlbA
y
x
y
x
y
x
250033332500333315003333
+=−=+=−=−=+=
Now, let’s draw a separate FBD for each member and recount equations and unknowns.
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EF 202, Module 3, Lecture 3
Observation
•Sometimes the only way to find all the loads on a frame is to draw a separate FBD for each member.
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EF 202, Module 3, Lecture 3
Ax
Ay
1000 lb
6’ 4’
BD
BDBD
x
y
ij
k^
^^
+
ABC is now statically determinate, because it has only three unknowns.
Now, let’s use the fact that BD is a 2-force member.
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EF 202, Module 3, Lecture 3
Observation
•Using what we know about two-force and three-force members reduces the work.
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EF 202, Module 3, Lecture 3
Why is this a frame, and not a truss?ADC and BGC are 3-force members.
Any 2-force members?DH and GH
Statically determinate?No.
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EF 202, Module 3, Lecture 3
x
y
ij
k^
^^
+
0 cos cos 0HD HGxF F Fθ θ+→ =⇒− + =∑0 sin sin 800 0HD HGyF F F Nθ θ+↑ =⇒+ + − =∑
800 N
FHDFHG
€
θ
€
θ
€
∴FHD = FHG =800 N
2sin θ( )
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EF 202, Module 2, Lecture 3
Suppose that one pair of lines form an angle, and a second pair of lines form an angle. If each line in the second pair is perpendicular to a line in the first pair, then the two angles are equal.
Nixon(?) Theorem
€
∴FHD = FHG =800 N
2sin θ( )=
800 N
2 15( )
= 894 N
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EF 202, Module 3, Lecture 3
Ax
Ay
Bx
By
Cx
Cy
Cy
Cx
894 N 894 N
Statically determinate:6 equations6 unknowns
x
y
ij
k^
^^
+
3004003004005000
xyxyxy
A NA NB NB NC NC
=−=+=+=+=+=
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EF 202, Module 3, Lecture 3
Given: The 150 N squeezeapplied to the pliers shownat right.
Required: Find forceapplied to the round rodand the magnitude of theforce at pin A.
Why is this a frame, and not a truss?3-force membersAny 2-force members?No.Statically determinate?Yes. (So, what?)
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EF 202, Module 3, Lecture 3
00 =⇒=→+∑ xx AF01500 =−−⇒=↑+∑ NFAF yy
0)160(150)30(0 =−+⇒=+∑ mmNmmFMA
( ) NNmmmmNFNAA yx 800333.515030
160150,950,0 ==⎟⎟⎠⎞⎜⎜⎝
⎛===
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EF 202, Module 3, Lecture 3
Given: The 150 N squeezeapplied to the wire cuttersshown at right.
Required: Find forceapplied to the round wireand the magnitude of theforce at pin A.
Why is this a frame, and not a truss?3-force membersAny 2-force members?Yes.Statically determinate?Yes. (So, what?)
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EF 202, Module 3, Lecture 3
150 N
150 N
F
F
Ay
Ay
Ax
Ax
Cy
Cx
Cy
Cx
Dy
Dy
Statically determinate?
Yes.
10 equations 7 unknowns
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EF 202, Module 3, Lecture 3
0 0x xxF AC+→=⇒−−=∑0 150 0y yyF ACF N+↑=⇒−−−− =∑( )0 (30)150(160)50 0Ay
M Fmm NmmCmm+ =⇒− +− =∑0 0xxF A+→=⇒=∑0 0y yyF ADF+↑=⇒−+=∑ ( )0 (30) 80 0yAM FmmDmm+ =⇒ + =∑0 0y yyF DB+↑=⇒−=∑0 0xxF C+→=⇒=∑0 150 0y yyF CB N+↑=⇒++ =∑ ( )0 150(110) 30 0yCM NmmBmm+ =⇒− − =∑
150 N
150 N
F
F
Ay
Ay
Ax
Ax
Cy
Cx
Cy
Cx
Dy
Dy
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EF 202, Module 3, Lecture 3
F = 150 N (9.8) = 1470 N
F = 150N (5.33) = 800 N