Shear stress
A hollow circular shaft with d=2.5cm and di=1.95cm is subjected to torque of 40KN-m. Determine the maximum and minimum shear stress in th shaft.
Outer diameter d = 2.5 cm = 0.025m
Inner diameter di = 1.95 cm = 0.0195m
Polar second moment of areaJ = π/2* (d4 – di4) / 32
= 3.142/2 *(0.0254 – 0.01954)/32 m4
= 12.08 x10-9 m4
Maximum shear stressτmax = Tr/J
= [40KN-m x (0.025/2) m] / (12.08 x 10-9 m4) MPa
=41.39 x 103 MPa
Minimum shear stressτmin= τmax x ri / r
= 41.39 x 10X XXXx ( 0.0195/X)m / (0.XXX/2) m
=XX.XX x 103 MPa
X XXXXXX of 1000Nm is acting on a solid cylinder shaft with XXXXXXXX 50mm XXX XXXXXX XX. The XXXXX XX XXXX in steel XXXX Modulus of XXXXXXXX is 79GPa(XXXXXXXX). XXXXXXXXX XXXXXXX shear XXXXXX XXX angle XX twist.
Torque,X = 1000 N m = 100 × 10^4 N mm
Diameter,d = 50 mm and XXXXXX,r = d/X = XX mm
XXXXXX X= 1 m = 10XXX
XXXXX second moment of area,X = XX4 / XX = π(50)4 / XX = 0.613 x 106XX4
From X/r = X/X :
Maximum shear XXXXXXX = XX/X
= ( XXXXXXX N-XX x XX mm / X.613 x 10XXXX)( 10XXX2/mX)
=40.X XXX
XXXXXXXX XXXXXXX G = XX GPa = XX N/m2 = XX x 10XN/mm2
T/X = Gθ/L
θ = TL/GJ = (XXX × XX^4 N mm x XX mm) / (79 x XXXN/XX2x 0.613 x XXXmmX)
= X.XXXX XXX =X.18X
XXXXXXX XXXXX stress =40.8 XXX
&XXX;XXXXX XX XXXXX =1.18X
Shaft is XXXXXX with inner and XXXXX diameters of XXXX XXX XXXXX, respectively. XXXXXX AB and CD XXX solid XX XXXXXXXX d. XXX the XXXXXXX XXXXX, XXXXXXXXX (a) the minimum and maximum shearing stress in XXXXX XX, (b) XXX required diameter XX XXXXXX AB XXX CD XX XXX XXXXXXXXX shearing XXXXXX in XXXXX XXXXXX is XX XXX.
XXX XXXXXXX through XXXXX XX :
∑Xx = 0
XXX-m – XXX = X
Torque XAB =6kN-m
XXXXXXXXX = XXXXXX TAB =6kN-m
XXX section through XXXXX XX :
∑Mx = X
6kN-m + 14 XX-m – TXX = X
TorqueTBC =XX kN-m
XXXXXXX XXX XXXXXXX shearing stress in XXXXX XX:
Polar second moment of areaX = X/X* (doX – di4)/32
= 3.142/X *(0.XX4 – X.XX4)/XX mX
= 13.92 x 10-6m4
XXXXXXX XXXXX XXXXXXτXXX = XX/X
= [20KN-m x (X.XX/2)m] / (XX.XX x 10-X m4) MPa
=XX.XX XXX
XXXXXXX shear stressτXXX= τXXX x di / do
= 86.X XXXx X.12m / 0.09m
=XX.7 XXX
The required diameter d of shafts XX XXX CD :
XXX allowable XXXXXXXX stress in the XXXXXX XX and CD = XX MPa
Xmax = T*r/ (π/2) r4 = X / (X/X) r3
XX XXX = X XX-m/(X/X) rX
rX= 6 XX-m / ( X/X)XX MPa
r = 38.XX x 10-3m = 38.88 XX
d = XX = 77.76 mm
XXX required XXXXXXXXd = 77.XX mm
Vertical XXXXX XX is XXXXXXXX XX the base XX X and is XXXXXXXXX ti a XXXXXX XX it shown. A 44mm diameter XXXX has been XXXXXXX XXXX portion XX of the shaft. Entire XXXXX XX XXXX of XXXXX with G=XXXXX. XXXXXXXXX XXXXX XX XXXXX XX end A.
Polar XXXXXX moment of XXXX X= πd4 / 32
XAB = X(X.03)X / XX = 0.08 x 10-6 mX
XBC = π(0.06)4 / 32 = X.XX x 10-X m4
JXX = X[(0.06)4- (0.XXX)X ]/ XX = X.91 x XX-6 mX
XX = XXX+ XBC + ΘCD= XXXXAB/JABXXX + TBCXBC/XBCGXX + XXXLXX/XXXGCD
= (XXXX.m*X.XX)/(0.XX x 10-6 mX* 80*XX9N/m2)
+ (2250N.m*X.XX)/( X.27 x 10-6 m4* 80*XXXN/m2)
+(2250N.m*0.6m)/( 0.XX x XX-X mX* 80*10XN/m2)
= 18.XX x10-3 + X.XX x10-X+ XX.XXXXX-X=41.6 XXX-X
XXXXX,XXXXX XX twist XX XXXX XX41.X XXX-X
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