HW 9 - 404 CHAPTER 6 Stresses in Beams (Advanced Topics)...

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404 CHAPTER 6 Stresses in Beams (Advanced Topics) Problem 6.8-4 Solve the preceding problem for the following data: b = 145 mm, h = 250 mm, t w = 8.0 mm, t f = 14.0 mm, and V = 30 leN. Solution 6.8-4 Wide-flange beam (b) CALCULATIONS BASED ON MORE EXACT ANALYSIS (SECTION 5.10) See Figure 5-38. Replace h by h z and t two h = h + t = 264 mm hi = h - t = 236 mm z f f Moment of inertia (Eq. 5-47): b = 145 mm _ I 3 3 2 I - - (bh 2 - bh 1 + twh I) h = 250 mm 12 t = 8.0 mm w 4 4 t f = 14.0 mm 1=- 1 (867.20 X 10 6 mm ) = 72.267 X 10 6 12 V = 30 leN Maximum shear stress in the web (Eq. 5-48a): (a) CALCULATIONS BASED ON CENTERLINE DIMENSIONS _ V 2 2 2 T max --(bh 2 -bh l +twh l ) (SECTION 6.8) 8It w 3 2 I = twh + bt J h = (6.4864 X 10- 6 N/mm 5 )(2.4756 X 10 6 3 ) Moment of inertia (Eq. 6-57): z 12 2 = 16.06 MPa +- I = 10.417 X 10 6 4 + 63.438 X 10 6 4 z = 73.855 X 10 6 4 NOTE: Within the accuracy of the calculations, the maximum shear stresses are the same. Maximum shear stress in the web (Eq. 6-54): J h) Vh T max = ( t+"4 2i w z = (316.25 mm) (0.050775 N/mm 3 ) = 16.06 MPa y t f1 E.: j c hI il il z t w t ft z Shear Centers of Thin-Walled Open Sections When locating the shear centers in the problemsfor Section 6.9, assume that the cross sections are thin-walled and us e centerUne dimensions for all calculations and derivations. Problem 6.9-1 Calculate the distance e from the centerline of the web of a C 12 X 20.7 channel section to the shear center S (see figure). (Note: For purposes of analysis, consider the flanges to be rectangles with thickness t r equal to the average flange thickness given in Table E-3, Appendix E.) Z __ S---++,C e-1 Solution 6.9-1 Channel section y C X 20.7 d = 12.00 in. t = 0.282 in. w b = 2.942 in. t = average flange thickness f f t = 0.501 in. b = b - twl2 = 2.801 in. f f h = d - t = 11.499 in. S f Z--~+iI--" rl 3b 2 t h d Eq. (6-65): e = J = l.0l1 in. '27 ht w + 6bt r
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SECTION 6.9 Shear Centers of Thin-Walled Open Sections 407 Solution 6.9-5 Channel beam with double flanges Shear force V acts through the shear center S. ... 2:Ms = ~F3e + F,h: + F1h l = 0 Fzh j + Fjh z bZt z z e = = - (h j + h z ) F 3 41, th 3 I, = ~ + 2 [bt(hz/2)z + bt(h l l2)z] - 12 = ~ [h 3 + 6b(h z + h Z )] z I z 3b 2 (hi + h~) e= - t = thickness h~ + 6b(hi + h~) VQA V(bt)( ~2) bh V TA=--=----=-- z Izt Izt 2I z 1 bZhztV F j = -TAbt = -- 2 bhjV T =-- B 2I z F = V 3 A F j z B Fl S F 3 r h z hI e-l i'-- F 1 1'1 - Problem 6.9-6 The cross section of a slit circular tube of constant y thickness is shown in the figure. Show that the distance e from the center of the circle to the shear center S is equal to 2r. s Slit circular tube z----+l-----!", ~e'-+'r---I V(l - cos 0) 7Trt r Z----+l---~.---.w. . ~e-+'r---l s Q A = J ydA At point A: dA = rtdO T c = moment of shear stresses about center C. = (rsin cf; )rtdcf; o T c = TArdA cos 0) dO J = f1T: (l - = 2Vr o = r 2 t(1 - cos 0) Shear force V acts through the shear center S. r = radius Moment of the shear force V about any point t = thickness must be equal to the moment of the shear stresses about that same point. T c 2: Me = Ve = Tee = - = 2r - V
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SECTION 6.9 Shear Centers of Thin-Walled Open Sections 409 Problem 6.9-8 The cross section of a slit rectangular tube of constant thickness is shown in the figure. Derive the following formula for the IY distance e from the centerline of the wall of the tube to the shear center S: b(2h + 3b) e= 2(h + z I h 2 S ~e >- IC h 2 b b 2 2 Solution 6.9-8 Slit rectangular tube t = thickness ts 2 VQ S2V FROM A ToB: Q =- T =-=- 2 l,t 21, = t..
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This note was uploaded on 11/14/2011 for the course ECIV 310 taught by Professor Huckelbridge during the Spring '11 term at Case Western.

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HW 9 - 404 CHAPTER 6 Stresses in Beams (Advanced Topics)...

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