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..t. A ars A and B have solid circular cross sections and consist of
aterial with shear modulus G = 17 GPa. Bar A is 150 mm ng and its radius varies linearly from 10 mm at its left to mm at its right end. The prismatic bet E! is 100 mm long 1d its radius is 5 mm. There is a small gap between the bars.
1e end of bar A is given an axial rotation of one degree and e bars are welded together. What is the torque in the bars terward? V WM rod
gamma: m:
n expression for the radius of bar A is: '3. = [0.010 ”[00:550Hm = ((0.010 —0.0333x)m)
n expression for the polar moment of inertia for bar A is: J = g,” = )2:(0.01m _ (0.033;)m)‘ titer ioining, the resulting bar will be subjected to a moment at end A and a moment at end B, or:
MA  M3 = 0
MA = M3 [1} Vhen the two bars are joined. the angle of twist for bar A will be reduced and an angle of twist will be introduced
1to bar 3. The sum of these two angles of twist will be one degree (0.0175 radians). or: ta + ¢B = 0,0175 radians [2] the angle of twist in bar A can be described by: 0.IS MAdx 0.I5 p ___ J‘________________= 2M4 I_____JE___.
A o £(001m(0033xmr(17x10°N/mz) #117110“mesz (0.01m—(0.033x)m)‘
2 it. = {8.445x10" A ism this expression for M and ””3”“ [1] in equation [2]: MA (0.1m)
’25(o.oosm)‘(17xto’ N rm!) (2.553;:103 )M;I + =0.0175rad “"5 MA: N 213 was
The components of plain stress at a point p of a material are ox= 20 MPa. cry: 0 and 1“,: o. If e= 45° what are the
stresses 0',“ o; and 1,, at point p? m:
Using Equation (57) to ﬁnd 6;:
ax + a", a —a a"; = 2 + ‘ 2 ’ (cos2€)+r,y (513129)
a; = ZOMZH + 0 + ZOMPC —0 (C08900)+ 0
ANS g; = 10 MP: Using Equation (5.9) to ﬁnd a}:
a} + a, a" — a 0;, = 2  2 y (cosh?) raisin 219) a} = 20110:: + 0 _ ZOMPa  0 c0590°)—0
ANS gx' = 1 P
Using Equation (58) to ﬁnd txy': . 0' a3, 1;? == — 2 J'(sin219)+ : ”((33526) r13, = ________20MPa __ 0 (sin 90°)+ 0
ANS a; :MEa Thecomponents of plane stress at a point p of a material are ox= 0 =0 and 1x, * 25 ksi. if 9“ 45“, what are the stresses 0,, o,‘ and
I; at point p? W:
Using Equation (57) to ﬁnd 0,:
. + — a" = a} 2 a", + a} 2 0" (cos 26)+ rxy(si112€) 0+0 0—0
+ _ 2
ANS = i 0": (cos 90°)+ (zshdsin 90°) Using Equation (59) to find 0;: a; = ”X :0? _ a” _ 6” (cosza)— ry(sin 29)
a = 9:43 — o g 0 (cos9o°)— (ZShiIsin 90°)
ANS g; =  ‘k i Using Egggtg‘ n (SQ) to ﬁnd gag: 2i),  — 6120’ (sin 249)+ r ”(cos 26)
7;? = —°—‘—° (sin 90°)+ (251:3:ch 90") NW 9. : PR E 2.7
g The components of plane stress at a point p of a material referred to the x‘y'z’ coordinate system are a“, = 8 MPa,
o", = 6 MPa and r'xy = 16 MPa. If e = 20°, what are the stresses a", cry. and Txy at point p?  SQLQTION: ‘ Adding equations (57) and (59): a,
o',+cr'y=o'x+oy .V ’ I
~8MPa+6MPa=ax+oy .V \ .
o,=2MPac, [1]  Using equation [1] in equation (58):
r' = — a" £0", (sin 260+ 19(cos26’) 11' — l6MPa = {Mysm 40°)+ 7:,(005 40°)  16 MPa = (1 MPa + a.) (sin 40°) + 1., (cos 40°)
1,, =  21.73 MPa — 0.839 a, [2] 4‘
I‘Using equations [1] and [21 in Equation (58): 0’ :‘7’ ‘7‘ i 0’ (cos 29)— :4, (sin 2(9) (—ZMPa  cry) + cry (—ZMPa — a'y) — a3, 2 2
 7.736 MPa = 1.305 c, g, =  as; MP3 (Ty: 6MPa = (cos 40°)~ [— 21.73MPa — 0.839a'yIsin 40°) 92.3w t 395.. .41 .  5 r
point p of the car's frame is subjected to the components of plane stress a’, = 32 MPa. 0’, = 16 MPa and
' 'xy '= 24 MPa. If e = 20°, what are the stresses 0,. cry, and 1,, at point p? " QLQTIQN:
 ding Equations (5?) and (5—9): ' o’x+o'y=ox+oy 32 MPa + (—16 MPa) = 6x + cry  I oy=16MPaox ing equation [1] in Equation (58):
a" — cr ti? = — 2 y (sin 260+ rxy (cos 29)
— 244m: .—. {Wﬁsm 40°)+ ry[cos4o°)  24 MPa = (csx + 8 Mpa) (sin 40°) + 1., (cos 40°)
,' 1,, =  3004 MPa._— 0.839 a, sing equations [1] and [2] in Equation (58): ax+cry ax—ay a; = 2  (cos 2(9)— rxy (sin 26)
" 32MPa = W—ﬂiﬁfz—mﬁlﬁm 4o°)+ [— 38.04MPa + 083903150: 40°)
. 4‘ 32 MPa = 8 MPa + 0.766 a, — 6.128 MPa — 24.45 MPa +0539 0,.
‘ s 9:41. MPa g,=—25.§ MPa 3x,= . 7MPa 247 ...
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 Spring '06
 Keil

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