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Unformatted text preview: __Rkﬁ ﬂgngLEM 53.1;
The components of plane stress acting on an element of a bar ,
SLbjiected to torsion are shown in Fig. 57. Assuming the stress 1,, to be known. use Mohr’s circle to determine the principal ‘ st: asses and the maximum inptane shear scess and show them
' acting on properly oriented elements. i ﬂLQTION: The center of the circle is at :
i i Oat0', 0',+0 a; i a": = =— i 2 2 2
Tc=0 MPa Th 3 radius of the circle is: r=V(ax_dc)2+(fxy)2 =V(_0'x“0)2+02=0'1 T We see from Mohr’s circle that: ANS 01:55: 0'2 =0 Irwl=crx i E; V i ‘3‘
The components of plane stress acting on an element of a bar
subjected to torsion are shown in Fig. 58. Assuming the stress :1, to be known. use Mohr’s circle to determine the principal
stresses and the maximum inplane shear stress and show them
‘ i acti '19 on properly oriented elements. i i
' seems: . 1 The center of the circle is at :
Q
at =JX+Uy =_0_+_Q=0 ‘ 2 2 " rc = 1,, O I; Fhe radius of the circle is: ll  H r=‘i(crJr ~ac)2 +(r‘1tv)2 =1i(0—0)2 +33? =er I lie see from Mohr’s circle that: P Ii {NS J] = T”, 0'2 2—“), [TMAXI = 1") —— ’i‘ “‘ 271 III ___________—__—____.___—] EBQW  I
F or tr e state of plane stress c. = 8 ksi, 0,, = 6 ksi, and 11, =  6 ksi, 3e Mohr’s circle to determine the absolute maximum shear stress. 3 ;TRATEGY: Use Mohr‘s circle to determine the principal stresses
31d then determine the absolute maximum shear stress from the a <pressions (524)} O (ksi)
§ ﬂlTlON:
. o + o ' 
T we center of the circle is located at: ' 06 = ’_2.L = w = 71m O (6.6)
1,.y = 0 he radius of the circle is: r = (8161' — 7131'): + (* 61602 = 6.081(51' 1706;)
T 13 principal stresses are: 01 = etc + r = 7 ksi + 6.08 ksi = 13.08 ksi 52 = cc  r = T ksi  6.08 ksi = 0.92 ksi
L sing equations (5.24) to determine the absolute maximum shear stress: 13.08ks‘i 0.92k51‘ 2 6.54ksi — (0.46131) 2 0'1 0'2
2 a, —o'. 2 = 6.54161? = 3.04ksi = 0.46131' ANS WSW=654 k§i ER_0I3LEM 53.;1
For the state of plane stress c,= 240 MPa, cr, = 120 MPa. and 1x, = 240 MPa, use Mohr’s circle to determine the
a )solJte maximum shear stress. ggigy'nou
T 1 c enter of the circle is located at: JC ._. 3% = W = 60MB:
xx, = 240 MPa he radius of the circle is: r = 1[moms  son/119.2)2 + (240MHz)2 = 3OOMPa 1' 1e principal stresses are:
c1=cc+r =60MPa+300MPa=360MPa 0’2 = O";  r = 60 MPa — 300 MPa = 240 MPa 1’ ie ansolute maximum shear stress is: a. vs W = 39g MPa (240,240) TWPO) 275 ...
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This note was uploaded on 03/15/2009 for the course ENGR 213 taught by Professor Keil during the Spring '06 term at Oregon State.
 Spring '06
 Keil

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