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Unformatted text preview: Shear: 2. The axle of a pulley is subjected to double shear as shown in the ﬁgure. The width of the pulley is 1 inch and the bracket is 0.4 inches thick Find: (a) The shear stress in the axle; (b) The bearing
stress in the bracket; (c) Bearing stress in the pulley. 0.5 inl 0.4 ' F/ 2 1.0 in F 0.4 in F/ 2
D=1/2 in Pulley Axle under Double Shear Solution: The load F is the one resulting from the two tension forces, T. Thus, since the two cables are at 90° to
each other, F=JSOOZ +5002 =707.lllb 0W = Force= 707.11
W Area 0.40.5 = 353.6 psi Shear: 3. A truss joint connecting bars 0A, OB, and 0C is shown. The connecting plate is 20mm thick,
and the diameter of the connection bolts is D=10nnn. Find: (a) the shear stress for each of the six bolts
connecting bar CC to the joint and (b) the beating stress on the plate. Note that the bars are only on one
side of the plate, thus the bolts are subjected to single shear. Shear: 9 The two 0.75 by 4.0 inch tension bars are riveted together through two 0.8 by 6.0 inch splice
plates, and rivets of 1.0 inch diameter. The allowable stress for the bars and splice plates is 03"” = 20.0k8i , the allowable shear stress on the rivets is TERM = 9.0k8i , and the allowable beating stress on the rivets and splice plates is m = 25.0ksi . Find the maximum load P such that none of the allowable stresses will be exceeded. Which part of this connection would you ﬁrst replace
to make it more eﬂicient ? Splice Plate: .=o.75m 33* Solution For tension in the bar: 025'” = ﬁ —> P = or?“ Aw = 20.0 (0.75  4.0)= 60.0kips
l7'or tension in the plate: ow“ = :_/2 —> P = mtgP“ Am = 2 20.0 (0.8 6.o)=192.0kips For shear in thtrivets: ﬁ”=:—:—>P=2ﬂmAﬂm =29.0[n¥]=14.12kips For bearing in the rivets: cﬁﬁﬂﬁ = t —> P = ﬁg Am = 25.0  (1.0  0.75): 18.75kips The rivets are the weakest. Shear 10 The steel bar (E=200GPa) is of rectangular crosssection (4cm x 10cm) and is to be bolted on a
structure by 10 bolts on each side; the diameter of each bolt is 3.5cm. Due to an error, the length L shown was 1.795n1 instead of 1.80m, thus there is a gap of 0.5cm, which makes it difﬁcult to bolt the bar on the
structure. The engineer suggests a tensile force is imposed in order to close the gap. (a) Find the force F required to close the gap. Aﬁer the gap is closed and the bar is bolted, F is released. (b) Find the shear
stress on each bolt. (c) Find the bearing stress on the bar. _— L —_ cross—section: 4cm x 10cm
Solution 'I'heforcerequiredtocompressthebarby 5 is
EA 2001090.040.10 F=—5= 0.005=2.216106N=2228kN L 1.795
The shear force on each bolt is F 2. 228 10‘
t=—=—=23l.6MP 10A...» 10 u/4 0.0352 a
The bearing force on the bar is
s a. _L=&=159 2MPa 10.4,“, 100.04 0.035 Shafts: l. A hollow steel shaft, depicted in the following diagram, must transmit a torque of 300,000
inlb. The maximum shear stress allowable (rum) is 21,000 psi. Ifthe safety factor is 3.0 and the inner
diameter is half of the outer diameter, ﬁnd the following: @0 k—L—a What is the outer diameter (D,)? What is the polar moment of inertia (1p)? Ifthe shaft is spinning at 1200 rpm, how many horsepower (hp) is it transmitting? If the shear modulus of elasticity (G) is 11 x 106 psi and the length (L) is 44 inches, what is the torsional angle of rotation, 0? 90 9'.» Solution:
a. Finding the outer diameter (D,) 1,,=—(D‘—Dt‘) D, =Do/2
Df =D;/16 7r . . 7:150"
I=—D—D16=— °
” 32(° °/)32[16] Tr T(Do/2) _ T*Do *16*32 1' :—=— m 1,, 12y —n*15*2*D:
32 16 ° 256T [256*300,0001b—in]y3 D : 3 = .
157”: 157!“ * 7000 [bl/in2 b. Finding the polar moment of inertia (Ip). 4
1,:1 150° = ”*15 *(4.27m)‘
32 16 32*16
 4 I, = 30.6 m c. Finding how many horsepower the shaﬁ is transmitting. P = Tn = 300,000 in—Ibnzooﬂa: “m“ * 2” rad = 3,770x107 ”—15
mm 60 sec 1 rev sec
Pm, =3.77x107 ”"1” *LJ)
SCC 6600 m
860
P”, = 5710 hp (1. Finding the torsional angle of rotation.
9 _ TL _ 300,000 in—lb*44in ‘E‘ 30.6in‘*11x10° psi 9 = 0.039 rad = 2.25° Shaft 4 The ﬁgure shows a street sign supported by a
16’ hollow cylindrical pole of internal diameter 3.2” and
external diameter «4.0". The sign has an area of2 112
and its centroid is 4’ from the center of the pole. While
the sign is subjected to a wind load of 40 lb/ﬁz,
determine: a) The absolute rotation of the pole at the top, in degrees.
b) The maximum shear strain the pole experiences. G=10x106 psi Dw=4 in Din=3'2 in
Solution Total wind force on the sign: Force = Pressure " Area, or F=40x2=80 lb. Torque the pole is subjected at the tope: T=80x4=320 lbﬁ = 320x12=3840 lbin. TL
Rotation ofthe pole at the top: 4) = 6—1, where, T=3840 lbin, L=16x12=l92 in, P
6:107 psi, IP = 312(Dﬁm — D.‘In )= 14.84 in‘. This yields
(I) = 0.004968 rad = 0.245°
_ Tr _ 38404.0/2 Tm _— _—=517.58psi
I, 14.84 7...: =T£=L973=aoooosn =0.00517%
G 10 Shaft 8: The solid shaft of radius r=0.5 inches and shear modulus G=l0,000ksi is subjected to the
torques shown At the surface of the shaﬁ at point A, the angle between the “line before torques are
applied" and the same line “aﬁer torques are applied” was measured to be 3° (0.052 rad). Find the torque
M. Line before torques Linc after torques
are applied are applied Solution
At p=r,the shear strain is 7.... =0.052 . Then, cm =67... =10‘ .o.052 =52,358.3 psi. Since 1:".I =—, it follows that
P
4 , 4
t I 52,358.31 52,358.3ﬂ
T=ﬂ=—2=—2=10,275.31b—in
r r 0.5 Shaft 9: A cylindrical shaﬁ of length L=2.5m and diameter d=10cm is fixed at one end and the other end
is subjected to a torque T=10kNm. The rotation of the crosssection at the loaded end was measured to be
0.2 degrees (0.00349 rads). Find the shear modulus G of the material the shaft is made of. Solution We have,
TL TL 10 ooo.oo2.5
=_—)G=—=—’ =73GP
¢ GI ¢I 0.00349no.1‘/32 a P ...
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