MEE 320 Problems Book Chapter 1, Spring 2018

Chapter | 1.1 Two solid cylindrical rods AB and BC are welded together at B and loaded as Problem 1.1 shown. Determine the magnitude of the force P for which the tensile stress in rod AB is twice the magnitude of the compressive stress in rod BC. 75 mm Aaa? $y =1963-5 tne 50 mm 10kN pg -P 2. PP P ona = Arg 1963-5 = $09-3x10" oP | 750 mm i 1000 mm > Kae = HF CISY = 4417-7 mame _ (220) -P See ne = 240 - FP = 0-0543 ~ 226-4010 P 4417-9 Equating Sa, to 2 Ce - -6 $09-3 yi" Px Q(0-0543 ~ 226-400 P) P=Hu2aen Problem 1.2 1.2 In Prob. 1.1, knowing that P = 160 KN, determine the average normal stress at the midsection of (a) rod AB, (4) rod BC. 1.1 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Determine the magnitude of the force P for which the tensile stress in rod AB is twice the magnitude of the compressive stress in rod BC. ( a) Rod AB. 50 mtn 75 mm 120kN B P= /60 WW Cheusion) A 2. No = Tae - TEM « 1962-5mm2 P=160kNE 3 Gre 7 - 462%". ens mp, <t TROKN | : Ang 1963-5 750 mm | 1000 mm |; (be) Red BC. Fr 160~(2)020) =-f0 kn Le. 80 kN compression. 2 2 Mec * Tolec : aes) = 4417-9 mint 3 Oa. * _E — _~8Oxso < On. 2 ~/9+1 MPa <th Aw F4IT-F a Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. Problem 1.3 1.3 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that the average normal stress must not exceed 175 MPa in rod 4B and 150 MPa in rod BC, determine the smallest allowable values of d; and dp. Rod AB. P = 4O+ 30 = 7OkN = Jo xlo’ N 6, 300 mm im - P - 4qP. dy se” Ans : Ede wd,* B up (eo xto*) “3 : > fA = 6% an oe a= AE | UNIe HOY 22.610 m 40 kN a, = 22.6 mm =< 250 mm — — dy Rod ciel . c P = 30kN = 30x(O N CG, Pp... B.- HP 30 kN & Ase 44 a Tra” 4P 4)(20 x (03) 8 > v ~AGxXxlO mM ALG. Y masoxioey = 1S 96x! d. =|]5.96 mn <e@ Problem 1.4 1.4 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that d; = 50 mm and d; = 30 mm, find average normal stress at the midsection of (a) rod AB, (b) rod BC. > i (0) Road AB e Ps 40 +30 = 70 kN = 70x10°N “ee A = Bat = E (60) = 1.9635 xlome = 14635*10 -P 2 Toxo | Pp iB Gre = A 1.9635x1O 35.710 FS Cag = 3S.7 MPR 250 mm (bY Roe BC, P = Zo kN = 30x10°N i A = Ea, = | ()* = 706.86 mm = 706. 86%15° m* 30 kN Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. PB. _30 x10? = = 4 4 lo* RP. Cee = A 706, 861 ae “ 6a: 42.4MPa = 1.5 A strain gage located at C on the surface of bone AB indicates that the average Problem 1.5 normal stress in the bone is 3.80 MPa when the bone is subjected to two 1200-N forces as shown. Assuming the cross section of the bone at C to be annular and knowing that its outer diameter is 25 mm, determine the inner diameter of the bone’s 1200 N cross section at C. A _-f. - ~ & ion - A we A SC Geometry : A= r Ca* - a,”) 2 2 ‘A z 4P a =a - B= dt- te 2 poe ps \t_ (4)(12 90) dy = (28eIO*)— ra easy B = 222.9 *10° 1200 N a, * 14.43% 107m d,= 14.93 yam “<el Problem 1.6 1.6 Two steel plates are to be held together by means of 16-mm-diameter high- strength steel bolts fitting snugly inside cylindrical brass spacers. Knowing that the average normal stress must not exceed 200 MPa in the bolts and 130 MPa in the spacers, determine the outer diameter of the spacers that yields the most economical and safe design. =ai— ae At each bolt Pocatyran the Upper plate is pudled down by the tens fe force Py of the bolt. At the same time the spacey pushes that P P plate Upward with a compressive force Ts. In ovder +o maintain equidilbecum Bos the on Peo in = - _4Ps - a For the Spacen, SSF AL T(de-de) o P= E 6s (d,’- aA, ) Equvating Py and Ps, oa = if Se (d'- Ae) des fi + Sb) A= fl+ BE 0) dg = 25.2 mmm <a Problem 1.7 1.7 Each of the four vertical links has an 8 x 36-mm uniform rectangular cross section and each of the four pins has a 16-mm diameter. Determine the maximum value of the average normal stress in the links connecting (a) points B and D, (d) 0.4m points C and E. Use bac ABC asa free body, 20 #8 0.025 ae ONO ’ cacee| A B c Fan Fee oO! (0,080) Fy, — (0.025+0.040 )(20x10*) = © M < o “ Feo = 32.5 jo" N Link BD js m tension. 0, ~(0,04%)F, - (o.0zs)(20x0') =O M 3s is w Fay = -125 ")0 N Link Ce as ia Com pression, Net area of one Dink For pension = (0,008)(0,036.- 0.016 ) = 160 xlo” we For two paratle) Din Ks And = 320 xjo"* ws Tensile stress m Drak BD. re 32.5 oF o (Geo * ran . hee = 101.56 "10 Spyz l0l.6 MPa aa net z Area fon one Dir le in Com pression = (0.008 )(0.036) & 2 = 288x 10° m* For two paraDlel Paks, A= S76IO m aT) t Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited’distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. Problem 1 8 1.8 Knowing that the cehtral portion of the link BD has a uniform cross-sectional area of 800 mm”, determine the ‘magnitude of the load P for which the normal stress in that portion of BD is 50 MPa. P A — O.lm Ve 6 Feo = SA + 2 = (50 * 10%) (300% ic) _ 3 Lam Fep ° 70 *10 N AZ BD =-/(0.56)"+ (1.92)" \ | > 2,00 m —p Use Free Boaly AC for statics, 056m = . 56 “ ODM. =o: ce (Yo lo2YC.4) + LE (Hom 1?) C1. #) - Plo.741.4) = 0 P= 33.110" N P= 38)kN <a Problem 1.9 1.9 Knowing that link DE is 25 mm wide and 3 mm thick, determine the normal stress in the central portion of that link when (a) 6 = 0, (b) 6 = 90°. eo 300 mn — fot Use member CEF as a free body it HIM, = oO | 7 il. —P | -6°3 Fyz-(0+2) (24.0 sin®)- (0-4) (240 e036) = O 7 Foe = —/b6o sin® —3200e03 0 N A : 240N | | j ~G 2 2 9 A ye = (01025) (01003) = 75X10 mM Ope = fe @) 8-0: Foe = -320 N —320 > ad oJ XN eg, Soe jena 4.27 MPa (b) 0 = 40°: Foe = - (60N -60 Goe = E75 we” = ~-2-12 MPa <a Problem 1.10 1.10 Link AC has a uniform rectangular cross section 3 mm thick and 12 mm wide. Determine the normal stress in the central portion of the link. Free Body Diagram of Piate 7 960 N t ae 150 mm 175 mm | 960 N 30° Note that the two F60-N -fovres Form a coupte of wmenent (960 N)\ (015m) = 144 .N.m DIM,=0: 144 Nin ~ (Fae cog 30° (925m) = O Fae 7 6651 W Aven ot Dink t Aac = (3 mm \(12 mm) = 36 mmZ < 665: Stress Sxe = rs = —s = 18-475 MPa Sac 18:5 MPa <a Ac Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. P 1.7% The rigid bar EFG is supported by the truss system shown. Know- roblem 1 11 ing that the member CG is a solid circular rod of 18 mm diameter, determine the normal stress in CG. Using portion EFGCB as a free body Ht pay -o: 24 Fas -/5 =O Fae = 25 KA k~1.2 m—>}—1.2 m—>|<- 1.2 m—>| Using beam EFG as a Fvee booty Fre C OM.s0: -Uez)et Fe + cn) (St Fd = © Fee = Fae = 25kW Cross sectioned aurea of member CG Reg = Ed? = Eloy = 2544x109 Pm Novmal stress im CG. F ZS mn 6 = 2! - 42 - G92 MP “ Re. 254-Ay 9 © 4 Zz Problem 1.12 1.12 The rigid bar EFG is supported by the truss system shown. Deter- mine the cross-sectional area of member AE for which the normal stress in the member is 105 MPa. Using portion EFGCB as a Fre body HZ Fy =o: 24 Ire ~ (=O Fae = 2ZS5kN Stress in meaber AE Gace = /05MPy Fae F, 25x / é 10 5 -b 2 Ane * a = ~L7 236/Xi10 m =m re 05% 0 Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. Problem 1.13 1.13 A couple M of magnitude 1500 N - m is applied to the crank of an engine. For the position shown, determine (a) the force P required to hold the engine system in equilibrium, (b) the average normal stress in the connecting rod BC, which has a 450- mm” uniform cross section. P Use piston, rod, and évank together as free body eld wall peaction H and beawing peoections Ay awed Ay. M opm 3) 2M, = O2 | (0.280 m)H - ISOO Nem = © 60mm He 5.3571 10° N Use piston alone as free body. Note that yod is a two-fovree mem ber; hence the divection oF Force Fac is known. Draw the fovee triangle ane sodve for Pane Fre by proportions. Q =.) 2Q00* + 607 = 208.81 mm Po 202 + pe 17, 26x10? N P= 17. 3¢ kW <= H Fee 2 20838) 2 pf ay Ae = ee ae = 18.643 x/0 Rod BC is & Compression member. Tts area is 4SO mm = 4S0 x10" mn - ~Fe _ —/8.643x/o? __ & Stress, Ope A 7 Taso xioe Al.4xlo” Pa (b) 6a. = - 41.4 MPa = Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission. 1.14 An aircraft tow bar is positioned by means of a single hydraulic cylinder Problem 1.14 connected by a 25-mm-diameter steel rod to two identical arm-and-wheel units DEF. The mass of the entire tow bar is 200 kg, and its center of gravity is located at G. For the position shown, determine the normal stress in the rod. Dimensions in mm —— | <->} 850 500 675 825 FREE Goby ~— ENTIRE Tow BARE W = (200 kg)(4.8t m/st) = 1162.00 N HVEM,= Ot & SS50R- 150(1462.00 N) = 0 B50mm) R R= 2654.5 N EREE Boor - Both ARM FE wueew units: 166 tan ~& = GTS X= 8.4270° (Fig cos XSS0) — R(Sod) = 0 S00 Fov™ 550 cos BAzTOR (2654-5 N) 2439.5 N Coome.) 500mm GTS mm w=z- Feo _ _ _2439.5,N R=2654.5 kN co Kew (0.0125 yn)" = - 4.9697 * 10° Po Tz, = 4.97 MPa 4 Proprietary Material. © 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.