Hw11 - Combined Stress: 3. The cantilever wooden post is...

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Unformatted text preview: Combined Stress: 3. The cantilever wooden post is supporting two off-axis loads as shown. (a) Find a", and Tm at point A. (b) Repeat the calculation of 0'", and Tm if only one of the off-axis \ \ loads acts on the post. The radius of the post is 10 in. and Em", = 1-106 psi 20° 2000 lb 2000 lb 7 240 in -000 lb 240 in y C ross-section A Solution: Each off-axis force can be decomposed into a horizontal FX and a vertical F r components. F‘ = 2000' Sin(20") = 684.0411) FY = 2000 - C0s(20”) = 1879.4 lb (a) For case (a). the two horizontal forces cancel each other. thus only a compressive force of 2R. acts. which produces _ 2F? _2-1879.4 A 7:402 =11.96 psi 0.)) Here. in addition to the compressive force a bending moment acts at the fixed end. M Moment at the fixed end: M = F; -(10 in)+ Fx -(240 in) = 182,960 lb—in Normal stress 0", resulting from M: _n.D‘_3.14-204_78540m4 oA_M-r_182,960-10 64 64 ' " 1 7854.0 1 = 232.95 psi FI'Olll 110111131 fOI‘CCI Total normal stress: o"°"" = 5.98 + 232.95 = 238.93 psi ." Combined Stress 4 A round table 4 ft in diameter is supported by a 3 in. diameter pipe with 0.25 in. wall thickness. A 200 lb person sits on the edge of the table. Find the maximum nonnal stress 0': in the pipe. .\'ote: neglect the weight of the table and the pipe. and assume the pipe is fixed on the floor. 2001b I 3 in. Solution: The pipe is subjected to a compressive force of 200 lbs. and moment of 200*24=4.800 in-lb. For the pi e: A=3.14(3--2.753) 4=1.12844 in2 I=3.14(34-2.754) 64:1.16811 in4 Max tensile stress: arm" 2 —£+fl = —2L0+M = 6, I 46psi ‘ A 1 1.12844 1.16811 Max compressive stress: o"""" = —£ = —£— = 6,182psi ' A 1 1.12844 1.16811 Combined Stress: 9 A beam with a distributed load and a cross section as shown is given. The allowable normal and shear stress is also given. Each nail can resist 1.5 kN shear. Check the safety of the section and provide the required nail spacing. 2 kN/m F—zoo mmgpl i A B 3.6 m '.6 kN 3.6 kN N 00,, = [0 MPa _ l 200 mm 1",, = 0.8 MPa Y=157 5 mm Solution.- —>{ id—30 mm Mm =i = 3.24 kN - m From the beam loading: 8 Vmu = = 3.6 kN i=157.5 mm=0.158m From the section r‘o erties: p p 1:60.125XIO'6 m4 114C 3240 X 0.158 0 : — — m I -60125 10_b=8.SMPaSIOMPa :>ok , X 3600x(0.|58x0.03)x 0'58 rm“ = Q = —62 = 0.75 MPaS 0.8 MPa = 0k ' lb 60,125x10' x0.03 V . . . . Tb = _Q: 3600x(0 2x0 03X60 015): 5.4 MW 1 60.125x10' Spacing of nails: 1.5 . S = a = 0.278 m say 25 cm spacmg Combined Stress: 11 A traffic light pole carries the weight of the traffic light and signs. D. and a horizontal wind load. \V. acting along the z-axis. Characterize the state of stress at point A at the base of the pole by drawing appropriate stress arrows on each stress block indicated. For each case. indicate the equation to use to determine the corresponding stress value expressed in terms of D. W. h. L. E. A. I. a. b. Ip. and or G. as appropriate. For the bending shear case. show on a sketch the area you would use to compute Q. Note that you do not need to evaluate A. Ip. or I. but you do need to use a and b for those stress expressions that include location terms or thickness terms. Axial Bending Normal Stress y |y No shear (Q=0) .3. rnm. x Torsion Bending Shear Combined Stress: 16 The tank shown is filled up with petrol of density p = 720g/lt (grams per liter). Its height is 17m, its internal radius is r=12m, and its thickness is 2.5cm. At the bottom of the tank, the petrol creates a hydrostatic pressure of 120.07 kPa, while the hydrostatic pressure at the top of the tank is zero. The top cover weight is W=l 1.0 kN, which is distributed as compressive stress on the tank wall. Find the maximum shear stress anywhere on the tank wall, examining material elements at the bottom, midpoint, and top of the tank only. Solution Stresses: From internal pressure: 0'l = fl, 0'2 = fl . From the cover: 0’2 = Z = fl = 5838.64Pa t 2! A 27: ~12-0.025 At the top of the tank, only the compressive stress is applied, thus rm =% = 5838'“ = 2919.32Pa At the mid-depth: a, =fl=w =2.88168-10"Pa, t 0.025 a, =fl_Z=W—5,838.64 = 1.44026-107Pa 21‘ A 2 ~ 0.025 3 At the bottom: a, = fl =W = 5.76336-107Pa, t 0.025 _fl_Z_120.07~103~12_ 2: A 20.025 The stress state at the bottom provides the maximum semi—difference of principal stresses 0'2 5,838.64=2.8811-107Pa _+_= 2 2t 2 120.07-10’ .12 + 120.07403 2-0.025 2 T p—'+ ="I‘_(‘P)=t_p=P’ P 2 = 2.88768 - 107 Pa = 28.87MPa Combined Stress: 14 A cylindrical pressure vessel of inner diameter d=4ft (48 inches) and wall thickness t=1/2 inch is fabricated from a welded pipe with helix angle tp=38°. Someone claims that by imposing an appropriate torque T as shown, the total shear stress parallel to the weld, 1w , i.e. the one from the combined effects of the internal pressure and the torque, can be reduced to zero. Is this possible? Explain your answer, and, and, if the answer is that it is possible, evaluate the magnitude of T that would do that. if V L _ V- * T Solution 0, =fl=—2°°'24 =9,600psi t 0.5 02 =fl=—2°°'24 = 4,800psi 2: 2-0.5 Considering the x-axis to be in the vertical (along the cylinder axis) direction, at the weld, 9 = -(p= -38°, 0', =4800, O”, =9600,‘txy = 0 . From T, the stress state changes to 0=-(o=—38",0'x =4800,ay =9600,er =? P ,P = %_T = 5(245‘ _ 24‘ )= 44805.5in‘ 0—0' _ __ x y- _ 1w] —¢'w— 2 sm20+rxycos29— _wsin(_2 .380) +7 24 44805 5cos(—2.38")=O—) T=1.79~107Ib—in Beam Deflections 2 The wood beam (E=11 GPa) is 4 cm thick. 30 cm wide and in equilibu'um. The moment at A is 300 N-m. Find the radius of cm'vatlu‘e p in span AB Solution Part AB of the beam is subjected to constant moment equal to -300 N-m. Then. 3 E1 E1 “.109 0.3 0.04 M: Elx=——>p=—=—‘2=58.67m p M 300 Beam Deflections 3 The 200 lb person is standing at the end of the cantilever beam made of wood (E=l .5x106 psi). Determine the shape of the deformed beam. fix A 12inx2in B — L=12 ft Cross-section U [j 3 1:12 2 = 8in‘, E1 =12-106 Ib-in2 EIu'"=0—) E11)": 9', F0, V=200 lbs—) §=2001bs Elu”=200x+ q, x= 0, M: -200~l44=-288001b- ('71—) q =-28800 EIu'=1001’ -28800x+ q, x=0, if=0—) (5 =0 EIKx)=33.3f—14400)?+ q, x=0, u=0—> q =0 33.36 i_ 14402 2 12-10 12-10 EIKx)=33.3£ —l4400x3 —> t(x)= Beam Deflections 4. A 100 111.111 square pipe with a wall thickness of 5 111111 is fixed at one end and a load ofSSOO N is applied to the free end. The beam is 3 111111 long. E = 210 GPa. Find the defouned shape of the beam. 3500 N 5 mm 1:3:m 100 mm H Solution: Start with Load-Shear-lV'Ioment diagram 3500 N M(x) = -10,500 + 3500X -10,500 N*m Now find the deformed shape using the integration method d 2 y M = E] (x) «Ix2 E1? = j M(x)1x = I: —10,500 + 3500x dx = —10,500x +1750):2 + c x — EIy(x)=I—10,500x +1750x2 + CI dx = —5250x2 + 583.3x3 + C,x + C Now. fmd C1 and C 2 by applying bomldary conditions y(0)= 0 because displacement at left end is 0 = 0 because slope is 0 at left end Y(0)= é [—5250(0)2 + 585.3(0)‘ + C1(0)+ C2 ]= 0 1C, =0 El “ E L 2=0XEI so C2=0 E] Q dx = [—10,500(o)+ 1750(0)’ + C, ]= 0 so C. = o x=0 Y(x)= i [— 5250;:2 + 583.3x’] I = i0.10x0.103 —i0.09x0.09~‘ El 12 12 = 1 .662x10" [—szsox2 + 583.3x’] = 2.866x10‘6 m“ 51 =2.866x10"’ x210x10" =601,860 -: _ ‘ ‘ ':’ a 1:3ij 2 __1 __ E : - .4 _.-.: I 3.5 I 1.5 I 2. x Posiziczim) ...
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This note was uploaded on 04/28/2010 for the course CE 215 taught by Professor Frantziskonis during the Spring '08 term at University of Arizona- Tucson.

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Hw11 - Combined Stress: 3. The cantilever wooden post is...

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