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# HW 16 - Hw it.a a-2 Suppose that the length of the beam in...

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Unformatted text preview: Hw it: .a a .. ' .-2- Suppose that the length of the beam in Example 8- 2' Is L: 8 ﬂ and it is made of ASTM- A36 structural steel The maximum anticipated magnitude of the distributed load is WI): 2400 lblft. Determine the dimension n so that the beam has a factor of safety 8= 3. FTQQ 39d}: Diagram: ’ ,.. "l 2,400llo/Ft /,#«’ I A B A y B OLUTION: Y Summing moments about point B to ﬁnd the reaction at point A: ' 1 lb 3 2M, = o = [5[2400E](8 ML 19]— A118 fr) Ay=3200|b T Summing vertical forces to ﬁnd 3,: 2Fsf = 0 = - 1/: (2400 Ibi‘ft) (8ft) 4- 3200 lb + By =6400Ibt Draw the F80 for the left-hand portion of the beam and ﬁnd the location where the shear force is zero (M is max): ft F. , M 4619ft X_.1Tj Summing moments about a cut through the beam at x = 4. 619 ft: M = 0 =(32001b)(4.619ﬁ}[—[300%)(4 691ﬁX4 ﬁlgﬁ)][4.6;9ﬂ] + M M = 9853 ft-lb y 1 w (300xxb/Ft as; = o = 32001.5 — E[300—400 (x) Using the factor of safety (given) to ﬁnd the maximum allowable normal stress: 36, oooib crm=%=——-—— 3/ =12,ooo!/ The moment of inertia for the square cross-section would be: I _ Mk 3 # h‘ 12 12 Using equation (8-12) to determine the dimensions of the square (and remembering to convert ft-lb to in-lb): (9853 f: mung— 11—12:] 1b = _..__ —> h3 = 59.12 in3 12,000 All *7 12 ANS h = as in 41? ..13.?.— 1-.-__1.‘- -.'"'.;:_'_‘_'.:'. ‘F’F; ._._ .._. -: \ l .-. a'j- Ii. 1: |- I The maximum anticipated load on the beam is shown. Choose a material from Appendix B and design a cross section for the beam so that it has a factor of safety 5‘ = 2. Free Body Diagram: / «1 EU’UUOWN ﬂ !‘}~20kNim A ’ " / I t .I B I . ' l-I—3nI—‘-'-f-'——Jrn . 3m——4-—3m B sgcynon: y i From Appendix B for 70T5-T6 aluminum. o, = 480 MPa. A A cross-section is chosen such that its height is twice its width. The moment of inertia for the cross-section is: 3 I = ﬂ = ‘0'"): = 0.667x‘m‘ 12 12 Ex Summing moments about point B to determine the reaction at point A: l 1 2MB 1 o 1 [3(2000001 f mksm)](4m) — Mam) l-—x—-l A, = 20.000 N t The bending moment is maximum at the location where the shear force is zero, or the sum of vertical forces is zero. Summing vertical forces on the FBD: 2F), = 0 = 20,000N —[~1—(<5,i567mm2 xm)](xm) 1 x = 2.45 m 2 Summing moments on the FBI): W = 0 = (20,000NX2-45m) - [%(5,66?N im)(2.45mX2.45m)12':5m] + M M=32.659 N-m The maximum allowable normal stress due to bending is found using the factor of safety. a 10W; 2 a...” = :3: = “if—m = 240::10‘th2 Now using Equation (8:12) to determine the dimension. x: 240x10iiwm2 = W 1 x3 = 0.000204 m3 0.667x‘m“ ANS x = 9.0589 m = 5&3 mm 421 ...
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