Vector Mechanics for Engineers: Statics Analysis of...
Transcript of Vector Mechanics for Engineers: Statics Analysis of...
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Vector Mechanics for Engineers: Statics
Professor Nikolai V. Priezjev, Ph.D. Tel: (937) 775-3214 Rm. 430 Russ Engineering Center Email: [email protected]
Textbook: Vector Mechanics for Engineers: Dynamics, Beer, Johnston, Mazurek and Cornwell, McGraw-Hill, 10th edition, 2012.
Analysis of Trusses
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Chapter 6 –
Terms and
Analysis of Trusses,
Frames, Machines and
other Structures
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Analysis of Trusses by the Method of Joints
• Dismember the truss and create a free-body
diagram for each member and pin.
• The two forces exerted on each member are
equal, have the same line of action, and
opposite sense.
• Forces exerted by a member on the pins or
joints at its ends are directed along the member
and equal and opposite.
• Conditions of equilibrium on the pins provide
2n equations for 2n unknowns. For a simple
truss, 2n = m + 3. May solve for m member
forces and 3 reaction forces at the supports.
• Conditions for equilibrium for the entire truss
provide 3 additional equations which are not
independent of the pin equations.
FBD:
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Analysis of Trusses by the Method of Sections
• When the force in only one member or the
forces in a very few members are desired, the
method of sections works well.
• To determine the force in member BD, pass a
section through the truss as shown and create
a free body diagram for the left side.
• With only three members cut by the section,
the equations for static equilibrium may be
applied to determine the unknown member
forces, including FBD.
• Similarly, for FCE write moment about B.
• What about FBE ?
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Analysis of Frames
• Frames and machines are structures with at least one
multiforce member. Frames are designed to support loads
and are usually stationary. Machines contain moving parts
and are designed to transmit and modify forces.
• A free body diagram of the complete frame is used to
determine the external forces acting on the frame.
• Internal forces are determined by dismembering the frame
and creating free-body diagrams for each component.
• Forces between connected components are equal, have the
same line of action, and opposite sense.
• Forces on two force members have known lines of action
but unknown magnitude and sense.
• Forces on multiforce members have unknown magnitude
and line of action. They must be represented with two
unknown components.
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Equilibrium of a Two-Force Body
• Consider a plate subjected to two forces F1 and F2
• For static equilibrium, the sum of moments about A
must be zero. The moment of F2 must be zero. It
follows that the line of action of F2 must pass
through A.
• Similarly, the line of action of F1 must pass
through B for the sum of moments about B to be
zero.
• Requiring that the sum of forces in any direction be
zero leads to the conclusion that F1 and F2 must
have equal magnitude but opposite sense.
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ATTENTION QUIZ
2. For the above problem, imagine that you have drawn a FBD
of member BC. What will be the easiest way to write an
equation involving unknowns at B?
A) MC = 0 B) MB = 0
C) MA = 0 D) FY = 0
1. When determining the reactions
at joints A, B and C, what is the
minimum number of unknowns
in solving this problem?
A) 6 B) 5
C) 4 D) 3
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Example Problem
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FBD for each member separately
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CONCEPT QUIZ
1. The figures show a frame and its FBDs. If an additional couple
moment is applied at C, how will you change the FBD of member
BC at B?
A) No change, still just one force (FAB) at B.
B) Will have two forces, BX and BY, at B.
C) Will have two forces and a moment at B.
D) Will add one moment at B.
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FBD for each member separately
Moment about A: 4 ft * 500 lb – 10 ft * E = 0 gives E = 200 lb
Ax= 0
Ay= 300 lb
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2. The figures show a frame and its FBDs. If an additional force is
applied at D, then how will you change the FBD of member BC
at B?
A) No change, still just one force (FAB) at B.
B) Will have two forces, BX and BY, at B.
C) Will have two forces and a moment at B.
D) Will add one moment at B.
CONCEPT QUIZ (continued)
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ANALYSIS OF MACHNES
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65BG
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