Question

The piece of plastic is originally rectangular. De

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7-98

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Answer #1
Concepts and reason

Shear strain is the change in angle that occurs between two line segments that are initially perpendicular. Shear strain is denoted by and measured in radians.

Shear stress can be calculated using known values of shear modulus and shear strain.

Normal strain is a measure of longitudinal deformation that is elongation or contraction in the body. Normal strain is denoted by .

Normal stress can be calculated using known values of Young’s modulus and normal strain.

Both the normal stress and shear stress are used in various real time applications.

Fundamentals

The shear strain is calculated as follows:

ABCD represents initial position of the block and AECD represents final position of the block.

From the figure,

tan ø=

For small values of , tan
=
.

The angle is called shear strain.

The relation between shear stress and shear strain is as follows:

T=GY

Here, G is shear modulus.

The normal strain is calculated as follows:

From the figure, initial length is and the final length is .

The normal strain is,

5-

7-98P

Draw the deformed diagram of the rectangular plastic piece.

5mm
2mm
-
-
V
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
3 4mm
2mm
11
11
11
300mm
-
-
DI
3m

Draw the deformed diagram of the triangle CBB
.

mm
(400+5-2)

From the geometry of the triangle,

tan φ (400 +5-2)
φ =α
= 0.28430

Convert the units degree to radians as follows:

TT
=a
= 0.2843°x rad
= 4.963x10-3 rad

Draw the deformed diagram of the triangle CCD
.

300

From the geometry of the triangle ,

tan o–(300+2)
0=B
= tan
_2
300+2)
= 0.37940

Convert the units from degrees to radians.

0=B
= 0.3794°x
24 * 1800 ta
rad
= 6.6225x10-rad

Write the relation to calculate the shear strain at corner C.

(rc), =-(+B)

Substitute (6.6225x10-rad)
for and (4.963x10-  rad)
for .

pel c-0149TI-=
pes S8S1100-=
(4_01x7960+q_01457799) - = **(34)

Draw the deformed diagram of corner D and triangle CCD and DATA

2mm
300
2mm
Dk
400
3mm

Calculate the shear strain at corner D.

( ) , : # 0-a
Ꭲ
Ꭱ ,Ꮎ +a
2 2*

Substitute for and for .

(YD), = (6.6225x10 %)+(4.962x10-)
= 0.011585 rad
11.6x10-rad

7-99P

Figure shows original and deformed rectangle piece:

2mm
-
-
-
-
-
-
-
-
--------
-
-
-
4mm
2mm
ERASE
I
11
11
11
11
II
II
II
300mm
11
11
400

Calculate the diagonal length from the figure of AC and BD.

AC = BD
= VAD? +CD2

Here, and are sides of the rectangle.

Substitute 400 mm for , and 300 mm for .

AC = 4002 +3002
= 250000
= 500 mm

Consider the triangle Ꭰ AC
to find the deformed length of the diagonal.

AC= VAD+2+DC2

Here, is deformed length of the AD, is deformed length of DC.

Substitute (400+3-2) mm
for , and (300+2-2) mm
for .

AC= VAD2+DC2
= (400–2+3)2 + (300+2-2)
= 500.8mm

Find the normal strain along the diagonal.

AC- AC
EACAC

Substitute 500.8mm
for and 500 mm
for .

EAC
AC-AC
AC
500.8-500
500
= 1.6x10-mm/mm

Consider the triangle ADB
.

300 +4
400 +5

Consider the triangle, use Pythagoras theorem to calculate the length of .

DB=VAD+ BAm2

Substitute (400+5) mm
for and (300+4) mm
for .

DB= VAD? +BAM2
=(400 + 5)2 +(300+4)
= 506.4 mm

Calculate the normal strain along using the equation.

DB-BD
EBD
BD

Substitute 506.4 mm
for and 500 mm for .

506.400-500
EBD =
500
6.400
500
=0.0128 mm/mm
=12.8x10 mm/mm

Ans:

The shear strain at corner C and D are -11.6x10-3 rad
and 11.6x10-3 rad
respectively.

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