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

This preview shows pages 1–4. Sign up to view the full content.

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 by 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 mm 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 mm 3 ) Moment of inertia (Eq. 6-57): z 12 2 = 16.06 MPa +- I = 10.417 X 10 6 mm 4 + 63.438 X 10 6 mm 4 z = 73.855 X 10 6 mm 4 NOTE: Within the accuracy of the calculations, the maximum shear stresses are the same. Maximum shear stress in the web (Eq. 6-54): bt 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 12 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 )] 12 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 41, 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 Solution 6.9-6 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
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 + 3b) 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern