Class 10corrected

Class 10corrected - ENGR 111 Week 5 Wednesday Class 10 T....

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1 ENGR 111 Week 5 Wednesday Class 10 T. Pollock Fall, 2009
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2 Homework 4 Due next MONDAY!
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3 Unit Loads Set the largest load to 1 unit and scale other loads accordingly. In this example, the magnitude of the external force applied at C is known to be 5/7ths of the magnitude of the external force applied at B. • First: Sketch the structure FBD. • Second: Calculate the reactions at the supports.
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4
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5 Now use the Method of Joints to solve for the internal forces in each element. Start with a joint which has no more than two unknown magnitudes. • Draw the joint FBD. • Calculate angles. Choose the best order of solution. • Solve for the unknown magnitudes.
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6
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7 1 . 0 0 0 l b f B 1 . 3 4 7 l b f ( C ) ) 1 . 0 0 0 l b f ( C ) 1 . 0 7 7 l b f ( T ) C 0 . 7 1 4 l b f ) A 1 . 4 3 4 l b f ( T ) 0 . 8 0 9 l b f 0 . 3 5 7 l b f )
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8 The complete solution:
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9 Factor of Safety A factor of safety is a number that: reduces maximum allowable loads to values that guarantee a low failure rate is generally agreed to by  industry government professional societies In other words, we usually don’t want the structure to be loaded up to the very  limit of failure. Factors of safety vary depending on many factors. For example: buckle on a parachute harness:  15 wing of a cargo aircraft:  1.5 wing of a fighter aircraft: 1.2 air-to-air missile:  1.1
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10 The complete solution: Now suppose that the maximum tensile force that each member can withstand without failing is 450 lbf. What applied load set would produce this maximum with a factor of safety of 1.5? (Remember that the external load at C is 5/7ths of the external load at B.) Factor of safety
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This note was uploaded on 04/04/2011 for the course ENGR 111 taught by Professor Walker during the Fall '07 term at Texas A&M.

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Class 10corrected - ENGR 111 Week 5 Wednesday Class 10 T....

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