Assignment Engr. Materials & Practices; Me-201 Submitted

8/14/2019 Assignment Engr. Materials & Practices; Me-201 Submitted

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ASSIGNMENT

ENGR. MATERIALS & PRACTICES; ME-201

Submitted by;

Name: Muhammad Riyaz- ul- Islam

Semester: Fall-2007

Program: B.S.T.E

Batch: 4th

ID: 06313241

Submitted to;

MS. NAZIRA USMANI

Course Advisor.

Lecturer of B.S.T.E Department.

CITY UNIVERSITY.

Date of Submission:1 3 th N O V E M B E R , 20 0 7

Shearing strain:

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It is defined as the angular change between perpendicular faces of

differential element.

It is denoted by gamma.()

s

L

Shearing strain, tan = s/L

Tan is negligible for very small angle,

= s/L

Assuming Hookes law applied to shearing stress & shearing strain.

Shearing stress Shearing strain

where,

= Shearing stress

=Shearing strain

G= Modulus of rigidity.

V/As = G.s/L

Poisson ratio:

It is the ratio of lateral strain to the axial strain. It is denoted by nue ().

Poisson ratio = Ilateral strain/axial strain I

Poisson ratio for steel 0.25 to 0.30

Poisson ratio for Concrete 0.20

Poisson ratio for most other metals is 0.33

Torque Turning moment:

2

= G

s =VL/G.As


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The product of the turning force & distance the point of application of the

force & the axis of the shaft known as torque turning moment or twisting

moment.

In during the torsion formula some assumptions are:

a) The material of the shaft is uniform.

b) The twist along the shaft is uniform.

c) Normal sections of the shaft are plane of circular after twist.

d) Shaft is loaded by twisting couples in plane that is perpendicular to

the axis of the shaft.

e) Stress does not exit the proportional limit.

Formula for math:

Angle of twist, = TL/GJ (in radian form)

Or, = 180/ * TL/GJ (in degree form)

Where,

T = Torque.

L = Length.

G = Modulus of rigidity.

J = Polar moment of inertia.

For solid shaft = J = .d4/32

Where,

d = diameter of the shaft.

Problem:

An aluminum shaft with a constant diameter of 50mm is loaded by torque

applied to gears attached to it as shown in figure. Using G = 28GPa diameterthe relative twist of gear D relative to gear A.

800Nm 1100Nm 900Nm 600Nm

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D C B A

2m 3m 2m

Solution:

D C B A

800Nm 1100Nm 900Nm 600Nm

TDC = 800Nm (clock wise)

TCB = 1100(ACW) 800(CW)

= 300Nm (Anti clock wise)

TBA= 800(CW) + 900(CW) 1100(ACW)

= 600Nm (clock wise)

Angle of twist of gear D relative to gear A.

D/A = TL/GJ (radian) For solid shaft = J = .d4/32

1*32 TDC * LDC + TBC * LBC + TBA *LBA

28109** (0.05)4 = 800 * 2 300 * 3 + 600 * 2= 1600 - 900 + 1200

= 0.11085 (cw) (radian) = 1900 Nm2 (cw)

= 0.11058* 180/

= 6.3363 (cw)

Ans:

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Problem:

A shaft with a constant diameter of 2 inch is loaded by torques applied to

gears fastened to it. Using G = 12106 PSI, compute in degrees the relative

angle of twist between gears A relative to gear D.

600lb-ft 1000lb-ft 900lb-ft 500lb-ft

D C B A

4ft 3ft 5ft

Solution:

D C B A

600lb-ft 1000lb-ft 900lb-ft 500lb-ft

TAB = 500lb-ft (clock wise)

TBC = 900(ACW) 500(CW)

= 400lb-ft (Anti clock wise)

TCD= 1000(CW) + 500(CW) 900(ACW)

= 600lb-ft (clock wise)

Angle of twist of gear A relative to gear D.

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A/D = TL/GJ (radian) For solid shaft = J = .d4/32

32*T TAB * LAB + TBC * LBC + TCD *LCD12106** (2)4 = (500*5) (400*3) + (600*4)

= 2500 - 1200 + 2400

= 5.3110-8*532800 = 3700 lb-ft2 (cw)

= 0.0283 (cw) (radian) = 3700*12*12 lb-inch2 (cw)

= 0.0283* 180/ = 532800 lb-inch2 (cw)

= 1.62 (cw)

Ans:

REFERENCE BOOKS:

Andrew Pytel & Ferdinand L. Singer; Strength of Materials; Fourth

Edition; Harper & Row, Publishers, Inc; India; 2001; Page: (01-73)

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Assignment Engr. Materials & Practices; Me-201 Submitted