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Unformatted text preview: Combined Stress: 3. The cantilever wooden post is supporting two offaxis loads as shown. (a)
Find a", and Tm at point A. (b) Repeat the calculation of 0'", and Tm if only one of the offaxis \ \ loads acts on the post. The radius of the post is 10 in. and Em", = 1106 psi 20°
2000 lb 2000 lb
7
240 in 000 lb
240 in
y C rosssection
A
Solution: Each offaxis 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? _21879.4 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‘_3.14204_78540m4 oA_Mr_182,96010
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 inlb.
For the pi e: A=3.14(32.753) 4=1.12844 in2 I=3.14(342.754) 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 zaxis. 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: ~120.025
At the top of the tank, only the compressive stress is applied, thus
rm =% = 5838'“ = 2919.32Pa
At the middepth: a, =ﬂ=w =2.8816810"Pa,
t 0.025
a, =ﬂ_Z=W—5,838.64 = 1.44026107Pa
21‘ A 2 ~ 0.025
3
At the bottom: a, = ﬂ =W = 5.76336107Pa,
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=2.8811107Pa _+_=
2 2t 2 120.0710’ .12 + 120.07403
20.025 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: 20.5 Considering the xaxis 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 Nm. 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 Nm. 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 Crosssection U
[j 3
1:12 2 = 8in‘, E1 =12106 Ibin2 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
1210 1210 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 LoadShearlV'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.
 Spring '08
 FRANTZISKONIS

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