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# The stress components are shown in their posirve

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Unformatted text preview: wice using rules for the disconRnuity funcRons to get M. No constants of integraRon needed. •  Integrate M twice to ﬁnd dυ/dx and υ. Two constants of integraRon will be needed •  Apply boundary condiRons to determine the constants of integraRon. Be very careful to evaluate the disconRnuity funcRons properly. < x-a >n = 0 for x < a! •  Write a ﬁnal expression for υ in terms of the disconRnuity funcRons including the values of the constants. Macaulay FuncRons •  Macaulay FuncRon DeﬁniRon: where n ≥ 0 •  Its integral: •  Examples: a x w0 •  C = 0 in both integraRons here. m a ax x Macaulay FuncRons a x •  Examples: 1 1 m m a ax x w0 a L x •  Construct equivalent loads that can be represented by Macaulay FuncRons! •  How? w0 w0 w0 a a x x L L a x w0 w0 P w0 w0 Singularity FuncRons w a x 0 •  For handling point forces/couples we use singularity func(ons and P M a x a x M0 a x •  Its integral: where n = -1 or n = -2 Boundary CondiRons Support Type: Pin or roller at end of beam Fixed end Boundary CondiRon: Displacement: and Moment: Displacement: and Slope: Only condiRons on displacement or slope are needed to ﬁnd constants of integraRon. 16- 28) Hibbeler 4th ed The beam is subjected to the load shown. Determine the equaRon of the elasRc curve. EI is constant. 3 kip/ ft 5 kip ft 5 kip ft A B x 4 ft 8 ft 4 ft Given: applied loads, EI Find: υ(x) Plan: Use disconRnuity funcRons to ﬁnd M(x) then integrate to ﬁnd υ(x). 24 kip 5 kip·ft 5 kip·ft 3 kip/ ft 16- 28) Hibbeler 4th ed EnRre FBD 5 kip ft 4 ft Ay 24 kip 5 kip·ft By 5 kip·ft 4 ft Ay Load: 4 ft Ay 3 kip/ft 8 ft 5 kip·ft B y 4 ft 3 kip/ft B 4 ft 8 ft 4 ft By symmetry 5 kip·ft By x 4 ft 8 ft 5 kip ft 4 ft A 5 kip·ft 4 ft Ay 8 ft 3 kip/ft 8 ft 5 kip·ft B y 4 ft 3 kip/ft 3 kip/ ft 16- 28) Slope and DeﬂecRon Hibbeler 4th ed 5 kip ft 5 kip ft A B x 4 ft Boundary CondiRons: 8 ft 4 ft 3 kip/ ft 16- 28) Slope and DeﬂecRon Final Result Hibbeler 4th ed 5 kip ft 5 kip ft A B x 4 ft 8 ft 4 ft General State of Plane Stress •  The element below shows a general state of Plane Stress: •  Two normal stresses, sx and sy, and one shear stress txy. •  The stress components are shown in their posiRve senses. •  The components sx, sy and txy deﬁne the stress state. •  Their values are dependent on the orientaRon of the element. Rotated State of Plane Stre...
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