HW 15, 16 Revised

# HW 15, 16 Revised - .se Eq(74 and(7-5 to solve Problem...

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

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

View Full Document

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.

Unformatted text preview: .se__ Eq (74) and (7-5) to solve Problem 7-2.24. TION: -_ raw the F30 for the beam and determine the reactions at {rants A and B. ' EURN’N 4J-<N/m EMA = 0 = (6,000N)(6m) + 20,000N .- m — (4,000N)(6m)(3m) + By(6m) — B(4,000Nxm)(6m)](8m) B,=13.667N T 213 = 0 = —6,000N — (4,000Nf m)(6m) — [%(4,000me)(6m)] + A), +18,66’FN Ay = 23,333 N T fort} < x < 6 m: w= 0 X X V=sjwdx= 060:0“:1 @x=0,v=-6000N,so: C1=-6000N 0 0 V{x] =-6000 N I 1 M = J'm = J1 600%: = 4000; + c2 @ x = 0, M = — 20,000 N—m, so: 02 = — 20.000 N-rn .' o 0 _'ANS M): = - [Ix—20 -5For6 m < x <12 m: ' w=4000 Nim V=—J'wdx= —I4000dx= 4000): + C3 @ X: 6 v— — 17 333 N so: ca = 41333 N ;ANs Vx=11D—{ 4Qggx+41,3331N MI =I(—4000x + 41,3-33)dx 40003.:1 + 41,3333: + c4 @ x = 6m. M = 56.000 N-m. so: 0 C4 = -232,000 N-m g__ANs M =-2 00x2+4 33x—232 N-m For12m<x< 18m: w = (666.7 Nlm) (1B — x) m = (12.000 - 666.7 x) N I X = — [mix = « {(12,000 — 666.7x)dx = 42,0003: + 3333;:2 +18,66? + cs @ x = 12 m. v = - 666.7 N. 50: 1- c5 = 39,331 N. .- ANS V(x) = 1- 12.000 x + 333.3 x2 + 89.381) N M = Ide: J'(333.3x2 “12,000“ 89,331)dx=111.1x3 — 0,0003:2 +108,000x + Co @ 3: =18 m, M = 0. so: (35 = 648,000 _ ANS M(x1=(111.1 xa— 6,000 x2 +103,000 x + §4B,UOO} N-m 401 PROB M -1.1 The beam consists of material with modulus of elasticity E = 70 GPa and is subjected to couples M = 250 kN-m at its ends. (a) What is the resulting radius of curvature |._..__.' l.6m————-—-1I ‘ --l in- of the neutral axis? (in) Determine the maximum tensile c mam stress due to bending. you man §QLQTIQN: The moment of inertia for the cross-section is: 5&3 _ (0.16m)(o.32m)-“ 1:..._._ 12 12 (a) Using Equation (3.10) to determine the magnitude of the radius of curvature: = 4.369.210“ m“ i=£=—92§9W———-mﬁ28.174x10_3m4 p 51 (70x10 Nr'm )(4.369x10' m ) ANS g; 1.223 m (b)_The maximum normal stress due to the bending moment is: a H = Mym = (250,000N — m}(0.16m_) M 1 4.369x10“m“ ANS omx = 91.6 MPQ WLE . 4.2 The material of the beam in Problem 8-1.1 will safely support a tensile stress of 180 MPa and a compressive stress of 200 MPa. Based on these criteria. what is the largest couple M to which the beam can be subjected? §QLQTIQN: The symmetry of the cross-section tells us that: (UWMNSILE = (0m)coMF-ResswE- The moment of inertia for the cross-section is: 5.53 (0.16m)(o.32m)3 I=—= 12 12 = 4.369x10‘4m‘ The maximum normal stress due to the bending moment is: a, _ Mymx _) M “ 0M! _ (130x105Nrm=)(4.3s9xlo*m4) W — I - ———---—--~ yMAX 0.16m ANS M = 492 kN-m W - The beam is subjected to a uniformly distributed load W0 = 300 Iblin. Determine the maximum tensile stress due to bending at x = 20 in. if the beam has the cross section (a); if it has the cross (b). (The two cross-sections have approximately the same area.) 85in Y By we: The moments of inertia for the cross-sections in the two cases are: (4.47m)(4.47in)3 . 4 (6in)(6m)3 (490mm)3 , 4 I =——-—----——=33.27 I =———-—=86.6? _.____ ' a 12 m b 12 12 m r— 300lb/ll’i l Summing moments about the cut through the beam at x = 20 in: 1‘ j” M = (12,750 lb) (20 in) + (6,000 lb) (10 in) = 195,000 in-Ib EOin——lv We see that the maximum tensile occurs at the bottom of each cross-section. p, y . (1 95,0001): - rah-47%) In case (a), the maximum tensile stress is: (aﬁmx = -——;‘—— 33.27:}: ANS [QIJMMH .1 k iatth b ttom ofth r 5—5 in case (b). the maximum tensile stress is: (0-?)JW = W 86.67:}: ANS [£2th = .75 k i the bott fthe r - ction 410 W19 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 8 = 2. Free 39d! Diagram: , /| ED’DDON/n Q] It?” kam I ’ i _ / I 3m—-L—3m A! Byr SQLQTIQN: From Appendix B for 7'075-T6 aluminum, or, = 480 MPa. A cross-section is chosen such that its height is twice its width. The moment of inertia for the cross-section is: 3 I =3}:- = x(2.r)3 = 0.6I57x4m4 12 12 Ex Summing moments about point B to determine the reaction at point A: i 1 EM = 0 = [420mm mlJm)](4m) — A (6m) 3 2 ’ |-—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), = o = 20,000N — [-1-(6,66?Nim2 xm)](xm) ~+ x = 2.45 m 2 Summing moments on the F30: 2'” = 0 = (20.000N)(2.45m) - [%(6,667NFMX2.45mX2.45m) 1435”]+ M M=32,659 N-m The maximum allowable normal stress due to bending is found using the factor of safety. 0' 6 J 2 arm =§=ﬂxliﬂl=z40xw6mmz Now using Equation (8:12) to determine the dimension, x: 240x106Mm2 = w —» x3 = 0.000204 m3 0.667x‘m‘ ANS X = 9.0589 rn = 539 mm 42‘! ...
View Full 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