Solution-HW_1

Solution-HW_1 - A simply supported beam AB supports a...

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Unformatted text preview: A simply supported beam AB supports a trapezoid- aiiy distributed ioad (see figure). The intensity of the load varies linearly from 50 kN/m at support A to 30 kN/m at support 8. Calculate the shear force V and bending moment M at the midpoint of the beam. so kN/m ‘ h30kN/m AA i , _, , .\ B ré 3 m )u‘ RA: ZMe=O C) ~EA(3m)r(30Wm)(3m)(LS‘m) +(20 kN/mXBmXJ/ZXZM) =30 RA: 65kN Zva=o 1+ RA+R8~%g(szm+aok~/m>tsm> =0 R = 8 55kN4OIW/m v ~ so kN/m ' B M ( " $55k», ZFVEET =0 f+ V— (30 Wm)(l.5m) ~- }§(lo kNAnXLSM) “ +55k~a=o Vm-~2.5kN‘—“ Z.Mc =0 rip —M ~ (30Wm)(l.5'm)(a75h1) e fiéao m/m)(1. 5m)(0~5‘m>‘+( ‘WW 57”} 0 f M: 4.50 mm During construction 0:921 highway bridge, the main :rders are cantilevered outward from one pier toward the ex: {see figure}. Each girder a cantiiever length of 170 t; and I—shaped cross section with dimensions as shown ? “core. The load on each girder {during construction) ‘umed to be 750 ib/i’t, which includes the weigh: of the it (If). rmigmdgrs, ‘b’= 750 um g I“: 170;: W ~ 1 =(',5a(b/muzo€n: M “3% J 2 rift/12m) = [.30. 0 x [06 “Vin. _ i I: J. (24in) (76303 "””..i'12,(23iyar) in.)3 4 I2 . = X [03”],4 c; 2: “Ln” :3 1 GM emu,“ c :: (.lleleb“9714-1148 in)“ I .277. Ox 10"”th4 : 1?.2,500F’51_ Clawlj u. Win *M-W-. A bracket ABCD of solid circular cross section the shape and dimensions shown in the figure. A V' pg _ vettzeai éoati P = 8 1b acts at the free end D. Determine the minimum permissibie diameter dmin of the bracket if the zeiiowab‘te bending stress in the material is 3500 psi and ‘ 7L0; 4/ a = :5 in. (Disregard the weight ofthe bracket itself.) I ~ ' i __M____ .1 Fiber-glgss brgcket. P: glb b = L5 in. 03...“, :4 3500 psi d 3 diameter" Mmax 2‘ P ( 3 O‘mwx 2: fl’ffi’ft l: I 1rd3 36qu 0.10477 1,1,3 0. i“ :t 77' O‘a'lold at..." 0.47; in, , Calculate the maximum shear stress Tm21X and the maximum bending stress (rum in a simply supported wood beam (see figure) carrying a uniform load of 25 kN/m (which includes the weight of the beam) if the length is 1.75 m and the cross section is rectangular with width 200 and height 250 mm. (k s 25 KN/m L = 1.75 m b 200 mm h =' 250 mm M ax [mum shear st ress 7:31;: « L—LS W3" "-7 m = 21.32 kA/ A :- bh = (200 mm)(250mm)= 50x103mm2 ax. Natzmamlfl S _- Tm“ 2n = 2(50KI03mm2) 656kpa" ll 5 = = 2.033 x106 mm“: :2 J1- I, -flflimcim.‘ 5M 5 2.083x10‘f mma 2: 4-. Sq MPa «m- w q, :35oN/m Diameter d = 300 mm A simple log bridge in a remote area consists of two 0'3"” c 7. 5 M Pa parallel logs with planks across them (see figure). The logs Tan" 5 0. 3 M Pa are Douglas fir with average diameter 300 mm. A truck F. I moves slowly across the bridge, which spans 2.5 m. Assume ""1 0" OW“ ble “”d w that the weight of the truck is equally distributed between I. L = 1- 5 m 4 the two logs. Because the wheelbase of the truck is greater . than 2.5 m, only one set of wheels is on the bridge at a time. r—(a) Based upon bend Inge—Strgsesa- «4. W ‘ ‘ vmamt Luivok Thus. the wheel load on one log is equivalent to a concen— Max i mum mome n‘t Occurs when wheel is ; trated load Wacting at any position along the span. In addi- 0.17 midspon (x = L/ 2). lion. the weight of one log and the planks it supports is wL _ (bLl g V L 2 equivalent to a uniform load of 850 N/m acting on the log. Mm“ 3’ 4 f "3"" ‘ "117(2-5 m) (WM/Wm 5"") Det ' th .' "‘bl hl‘de"d ermme e maxtmum permissi e w ee roa ase a: 0. 625 w + 664' I (N'm) (Wz newto upon: (a) an allowable bending stress of 7.5 MPa, and (b) an 3 allowable shear stress of 0.8 MPa. 5 r. z 2. 65’ K ,0 w3 m 3 "max =2 502nm» = = [9,880 /V'rn 0.625W +661] == 19,880 W “‘ l Kb), fiofifs’d, upon, Shear'st its; Maximum shear force occurs, when wheel is adjacent to support (x a: 0)_ vm, m w + = w +"é‘1850N/mV2bmi == w + maze/vi (w r: newtona) H == 1’14»? x 04070636r111‘ I} cum, '“ 3f} ‘ - A M. 3H..;gltgv_, WWB.‘ be; . gm” vm -—r-~~‘f m 4(0,070636m1)(0.3/‘1pa) 1: m 42, 4:0 N \A , w + tool. 5 N 42,4:0 N ( i" 6} fl y" I ' 3A} 2}. W L: 4 I 3 K N Hm,“ I I V: fly-ti»; e z . . \J L/ I ( Bending stress governs.) v ~"—i"’\v’liw’\ :x’ “\k‘)! ‘3’“ A Innhcr beam/ll? 01‘spun l0ft2mdnommal \\'l(l[ll'—lll].(21([ll£1lWldth i 3.5 in) lg 10 suppm/t the thrce concentratul luttds shown, Knowing. that for the, grade ml tlmhct used (rm : 1800psi 11nd rd” 120 psi. dctcnmnc thc minimum 1'67 qtm‘cd depth (i of the beam. SOLUTION 1 Maximum Shear and Bending Moment. »\l'tct' drztxxtng the <hcztr and lu’fndmfymomvnt dtagrttmx \\'<‘ HUlL‘ that M 75 km - l1 ’ 90 kip ‘ 1n, mttx \' '1 3 ktps ma \ Design Based on Allnwzthle Normal Stress. Wc‘lnixt cxpt‘csx‘ thc «1th In \‘nix‘ll‘flll mmdttltm S m tct’ms ol thc depth {i \\'c haw l l l ‘, l , t , l / ' Inf 5' 'i ilnl’ " 'l35lti 415mm [ l: t 0 0 L lim' MMK I ()0 l<1p ‘ 1n, and (rm : l>l00 ps1, \‘uc \vntc _ 7 _ ,, an x 10311) - 1n .8 7 038sz :' H i "W ' l "m l800 p51 t (/3 >437 d r: 9 261m We have. Snttshed [llc‘ t‘cqtm’fiment that (rm :1 lll00 [NV t (1'11ka Shearing 511135.? Fm \“mx W 3 klps‘ and d 9.30111‘ WC find 3 2 mm lb W l I 7,, ’ ’ W’ V’f’pif", f Tm : ln‘u’h’ pm 1 ’ 2 ,r\ 3 l1.) 1n.,ll¥).2(w m.) l Smt‘c‘ Tab 7 110 pm. the dtipth d 1 0,3011]. 1.\ rm] ztcccpttthlc zmd wc must rt» l design the hvztm 0n the husns ml lllt‘ t’cqtm'cmcnt that T“, a 120 pm l l l Design Based on Allowable Sh‘uring Stress. Snnic we mm knuw that tho allowablt‘, shaming. mess CDHU‘OlE) thc (lCSlg’lL’ \vc erh‘ '3, \7Hm I _ '1, 3000 llw ‘ T,” : TM ' A" l20 pst “7 f: _ if? 1 sl 2 (33111,)(1 d I’ l07l m. 4 The nmmal stress IR 01‘ cumsct low than all“ "r 1800 1m. and the, dt-pth <th 10.7l m. 1\ fully :tt‘t:cpt;thlc, ,' mu IlL mm mtlwr \‘ nd tall}, l ‘;i hlc iydcpt int‘n‘ tents 0t :1 4 l29n, no Hltéll s [,L‘ tin mt Slit tld c uscd 'l‘h :lClLlL L‘t'msIASCL’ Ill. ttm W0 ltl th 25 n ’ 7ThccantilevcrbeamABhasarcaangularamsscctionofISOX200mm. ‘ ,\_ V ‘ Knowing that the tension in cable ED is 10.4 kN and neglcding the weight ofthc beam1 determine the normal and shearing stresses at the three points indicated SOLUTION 6—8 = .751+1.8" t LQS m Ver+fc~p Co-«fohewf a; 1—09 21$ 7 (fig )(Iofl) ‘4 W) “I HDVIZOMja/P COMPOMQWT 0‘? Tog ((93% ’. Af seo‘hon cantaim'nn F0043 OJ 5) and C p:—‘7_él(N Ig—4:t2w M ? (1.5)(43-(o.e)(ze) = — 3.6 kN-m a 850+;0V) ,Pwpewlriesv c 4.4101 A : (alrovo’zoo) : 0-030 ml 3A W M ‘ 3 1 L 12 kw I : E(O.ISO)(O.200) = 100 x/O‘L m C r 0.100 M ‘" _ _ 3 fig _ _ 7,0403 3.6.xlo3MoJoo) _ ‘ A+ Falfif 0. 6x “A “” I ’ W't‘L—HDOXlO-L ‘ A t —‘ 213 o . _ .3 L’LC : _ 7.6km“ ‘ {3.0103 Katee] _ _ At’ Fun‘} 5 6; ‘ A + I 0.030 ‘OOHLY‘ ~ 3-512 Mpg cm 2:] : O K 4 3 M Pan‘w‘) c S, = - § = — 21‘3")” : — 0-320 MPa. 4 J ‘ loo Q: (I50 )(IooMso) r 750w)3 m3 = 750 x/o‘C m3 7 7/ :_MVQ:-W- —0600 MP5; 4 ' It ‘(Ioovto“)(o.150\, ' ' L—ls’o —>l A circular post and a rectangular post are each compressed by loads that produce a resultant force P acting at the edge of the cross section (see figure). The diameter of the circular post and the depth of the rectangular post are the Same. (a) For what width b of the rectangular post will the maximum tensile stresses be the same in both posts? (b) Under the conditions described in part (a), which post has the larger compressive stress? - Circular 9051' _L4_ _7Td3 H“? 5‘32 t Tension: II I l“; + 3 73 c-£="’——\ ll AL A + 5 Camprtssion; _P A 4 [6 fi‘ 5 “ ‘ = ‘ 02. ‘ Rectangular past ; _ _ bdz _ 39L. 1 fl _‘ S _ 6 M — 2 Tens' : _ -B A _ —_P_ .13- .19. 1°" Gt',n+s“ bd+bd_b Compression; _£_A=_£_ie__j£ 02 n 5 bd bd bd For equal maximum ‘tensile stresses (a) IZP _ 29 -111 4—— ‘ 1rd _ bd b “ 6 (b) Circular ost: p = - .293. 0“ 1rd7' Rectangular post: ._ __4_E_ = ___4_P___ -fifl CT" " bd (nd/b)d _ wd" Rectangular post has “the larger compressive stress. ‘— A Vertical pole ofaluminum is fixed at the base and pulled at the top by a cable having a tensile force 7' (see figure). The cable is attached at the outer surface of the pole and makes an angle a = 28° at the point of attachments The P016 has length L = 2.0 m and 21 hollow Circular cross Section with outer diameter d2 2 250 mm and inner diam~ eter d1 = 200 mmi Determine the allowable tensile force Tallow in the cable if the allowable compressive stress in the 31urninum pole is 80 MPa. 7‘ r ' Aluminum pole A ,Tsma L=2‘0m a = 28° T605“ d.=200mm d2. d2=250mm (delallow a I A’c the base of the pole ! Iv: Tcoscz = 0.88295 T (T: newtonS) M = (Tcosoz)(%) + (Tsin «N‘Ll = 0.ll037T + 0.93894 = LO‘HST I (M = Newton meters)l l A =J<+L(d:-d3) =ne7mo’mm1 4 l =n.67lx10‘3 m2 I = fitd: —d,“)= 113.21 x106 mm‘ = 113.21 x 10‘6 m4 Comgressive stress __M M; _085235; (1.0193 l 0< “ n + I ‘nemzdfi film 10“ 4 . 7966T + H58.6T = 1208.6T (0; = pascals) CE = (CI-clam,” = x106 1:auow = H A single horizontal force P of magnitude 150 lb is applied to end D Of lever A81). Knowing that portion AB oi the lever has a diameter of 1.2 in, detep mine \CE the normal and shearing stresses on an element located at point H and having sides parallel to the x and y axes / 7;: ”\ l" 134) ll) We replace the force P by an equivalent force» Force—Couple System. ' the transverse section containing point H: eouple system at the center C Oi ’ : 150 lb T : (150 lb)(l8 in.) : 2‘7 kip" in. 13 lip m M}, : (150 lb)(l() in) : 1.5 kip - in. Stresses a“ (7” TH at Point H. Mr: (15 kip - mate in.) (n10 0“ : l : + r '1 j- 4 ' (r‘ : +884 ksi 4 I in (0.6 mi) I 1 Tc 1(27 kip ' in.)(()i6 in.)‘ r :: +77 : + reefer ~77 TH : +7.96 ksi 4 "l J 77 (0.6 in.)1 We note that the shearing force 1’ does not cause any shearing stress at point H. ...
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This note was uploaded on 11/04/2008 for the course CE 207 taught by Professor Elghandour during the Spring '06 term at Cal Poly.

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Solution-HW_1 - A simply supported beam AB supports a...

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