RESOLUÇÃO RESMAT CAP 2

Prévia do material em texto

Problem 2-1
An air-filled rubber ball has a diameter of 150 mm. If the air pressure within it is increased until the
ball's diameter becomes 175 mm, determine the average normal strain in the rubber.
Given: d0 150mm:= d 175mm:=
Solution:
ε
πd πd0−
πd0
:=
ε 0.1667 mm
mm
= Ans
Problem 2-2
A thin strip of rubber has an unstretched length of 375 mm. If it is stretched around a pipe having an
outer diameter of 125 mm, determine the average normal strain in the strip.
Given: L0 375mm:=
Solution:
L π 125( )⋅ mm:=
ε
πL πL0−
πL0
:=
ε 0.0472 mm
mm
= Ans
Problem 2-3
The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the
end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD.
Given: a 3m:= LCE 4m:=
b 4m:= LBD 4m:=
∆LCE 10mm:=
Solution:
∆LBD
a
a b+
⎛⎜⎝
⎞
⎠ ∆LCE⋅:= ∆LBD 4.2857 mm=
εCE
∆LCE
LCE
:= εCE 0.00250
mm
mm
= Ans
εBD
∆LBD
LBD
:= εBD 0.00107
mm
mm
= Ans
Problem 2-4
The center portion of the rubber balloon has a diameter of d = 100 mm. If the air pressure within it
causes the balloon's diameter to become d = 125 mm, determine the average normal strain in the
rubber.
Given: d0 100mm:= d 125mm:=
Solution:
ε
πd πd0−
πd0
:=
ε 0.2500 mm
mm
= Ans
Problem 2-5
The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam is displace
10 mm downward, determine the normal strain developed in wires CE and BD.
Given: a 3m:= b 2m:= c 2m:=
LCE 4m:= LBD 3m:=
∆ tip 10mm:=
Solution:
∆LBD
a
∆LCE
a b+=
∆LCE
a b+
∆ tip
a b+ c+=
∆LCE
a b+
a b+ c+
⎛⎜⎝
⎞
⎠ ∆ tip⋅:= ∆LCE 7.1429 mm=
∆LBD
a
a b+ c+
⎛⎜⎝
⎞
⎠ ∆ tip⋅:= ∆LBD 4.2857 mm=
Average Normal Strain:
εCE
∆LCE
LCE
:= εCE 0.00179
mm
mm
= Ans
εBD
∆LBD
LBD
:= εBD 0.00143
mm
mm
= Ans
Problem 2-6
The rigid beam is supported by a pin at A and wires BD and CE. If the maximum allowable normal
strain in each wire is εmax = 0.002 mm/mm, determine the maximum vertical displacement of the load
P.
Given: a 3m:= b 2m:= c 2m:=
LCE 4m:= LBD 3m:=
εallow 0.002
mm
mm
:=
Solution:
∆LBD
a
∆LCE
a b+=
∆LCE
a b+
∆ tip
a b+ c+=
Average Elongation/Vertical Displacement:
∆LBD LBD εallow⋅:= ∆LBD 6.00 mm=
∆ tip
a b+ c+
a
⎛⎜⎝
⎞
⎠ ∆LBD⋅:=
∆ tip 14.00 mm=
∆LCE LCE εallow⋅:= ∆LCE 8.00 mm=
∆ tip
a b+ c+
a b+
⎛⎜⎝
⎞
⎠ ∆LCE⋅:=
∆ tip 11.20 mm= (Controls !) Ans
Problem 2-7
The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2
mm, determine the normal strain developed in each wire.
Given:
a 300mm:= θ 30deg:=
∆A 2mm:=
Solution:
Consider the triangle CAA':
φA 180deg θ−:= φA 150 deg=
LCA' a
2 ∆A2+ 2 a⋅ ∆A( )⋅ cos φA( )⋅−:=
LCA' 301.734 mm=
∆CA
LCA' a−
a
:=
∆CA 0.00578
mm
mm
= Ans
Problem 2-8
Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a
force is applied to the end D of the member and causes it to rotate by θ = 0.3°, determine the normal
strain in the cable. Originally the cable is unstretched.
Given:
a 400mm:= b 300mm:= c 300mm:=
θ 0.3deg:=
Solution:
LAB a
2 b2+:= LAB 500 mm=
Consider the triangle ACB':
φC 90deg θ+:= φC 90.3 deg=
LAB' a
2 b2+ 2 a⋅ b⋅ cos φC( )⋅−:=
LAB' 501.255 mm=
εAB
LAB' LAB−
LAB
:=
εAB 0.00251
mm
mm
= Ans
Problem 2-9
Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a
force is applied to the end D of the member and causes a normal strain in the cable of 0.0035 mm/mm
determine the displacement of point D. Originally the cable is unstretched.
Given: a 400mm:= b 300mm:= c 300mm:=
εAB 0.0035
mm
mm
:=
Solution:
LAB a
2 b2+:= LAB 500 mm=
LAB' LAB 1 εAB+( )⋅:= LAB' 501.750 mm=
Consider the triangle ACB':
φC 90deg θ+=
LAB' a
2 b2+ 2 a⋅ b⋅ cos φC( )⋅−=
φC acos
a2 b2+( ) LAB'2−
2 a⋅ b⋅
⎡⎢⎣
⎤⎥⎦:=
φC 90.419 deg=
θ φC 90deg−:= θ 0.41852 deg=
θ 0.00730 rad=
∆D b c+( ) θ⋅:= ∆D 4.383 mm= Ans
Problem 2-10
The wire AB is unstretched when θ = 45°. If a vertical load is applied to bar AC, which causes θ =
47°, determine the normal strain in the wire.
Given:
θ 45deg:= ∆θ 2deg:=
Solution:
LAB L
2 L2+= LAB 2 L=
LCB 2L( )
2 L2+= LCB 5 L=
From the triangle CAB:
φA 180deg θ−:= φA 135.00 deg=
sin φB( )
L
sin φA( )
LCB
=
φB asin
L sin φA( )⋅
5 L
⎛⎜⎝
⎞
⎠:= φB 18.435 deg=
From the triangle CA'B:
φ'B φB ∆θ+:= φ'B 20.435 deg=
sin φ'B( )
L
sin 180deg φA'−( )
LCB
=
φA' 180deg asin
5 L⋅ sin φ'B( )⋅
L
⎛⎜⎝
⎞
⎠−:=
φA' 128.674 deg=
φ'C 180deg φ'B− φA'−:=
φ'C 30.891 deg=
sin φ'B( )
L
sin φ'C( )
LA'B
= LA'B
sin φ'C( )
sin φ'B( ) L⋅:= LAB 2 L:=
εAB
LA'B LAB−
LAB
:= εAB 0.03977= Ans
Problem 2-11
If a load applied to bar AC causes point A to be displaced to the left by an amount ∆L, determine the
normal strain in wire AB. Originally, θ = 45°.
Given: θ 45deg:=
Solution: εAC
∆L
L
=
LAB L
2 L2+= LAB 2 L=
LCB 2L( )
2 L2+= LCB 5 L=
From the triangle A'AB:
φA 180deg θ−:= φA 135.00 deg=
LA'B ∆L2 LAB2+ 2 ∆L( )⋅ LAB( )⋅ cos φA( )⋅−=
LA'B ∆L2 2 L2⋅+ 2 ∆L( )⋅ L⋅+=
εAB
LA'B LAB−
LAB
=
εAB
∆L2 2 L2⋅+ 2 ∆L( )⋅ L⋅+ 2 L−
2 L
=
εAB
1
2
∆L
L
⎛⎜⎝
⎞
⎠⋅
⎡⎢⎣
⎤⎥⎦
2
1+ ∆L
L
⎛⎜⎝
⎞
⎠+ 1−=
Neglecting the higher-order terms,
εAB 1
∆L
L
+⎛⎜⎝
⎞
⎠
0.5
1−=
εAB 1
1
2
∆L
L
⎛⎜⎝
⎞
⎠⋅+ .....+
⎡⎢⎣
⎤⎥⎦ 1−= (Binomial expansion)
εAB
1
2
∆L
L
⎛⎜⎝
⎞
⎠⋅= Ans
Alternatively,
εAB
LA'B LAB−
LAB
= εAB
∆L sin θ( )⋅
2 L
=
εAB
1
2
∆L
L
⎛⎜⎝
⎞
⎠⋅= Ans
Problem 2-12
The piece of plastic is originally rectangular. Determine the shear strain γxy at corners A and B if the
plastic distorts as shown by the dashed lines.
Given:
a 400mm:= b 300mm:=
∆Ax 3mm:= ∆Ay 2mm:=
∆Cx 2mm:= ∆Cy 2mm:=
∆Bx 5mm:= ∆By 4mm:=
Solution:
Geometry : For small angles,
α
∆Cx
b ∆Cy+
:= α 0.00662252 rad=
β
∆By ∆Cy−
a ∆Bx ∆Cx−( )+:= β 0.00496278 rad=
ψ
∆Bx ∆Ax−
b ∆By ∆Ay−( )+:= ψ 0.00662252 rad=
θ
∆Ay
a ∆Ax+
:= θ 0.00496278 rad=
Shear Strain : 
γxy_B β ψ+:= γxy_B 11.585 10 3−× rad= Ans
γxy_A θ ψ+( )−:= γxy_A 11.585− 10 3−× rad= Ans
Problem 2-13
The piece of plastic is originally rectangular. Determine the shear strain γxy at corners D and C if the
plastic distorts as shown by the dashed lines.
Given:
a 400mm:= b 300mm:=
∆Ax 3mm:= ∆Ay 2mm:=
∆Cx 2mm:= ∆Cy 2mm:=
∆Bx 5mm:= ∆By 4mm:=
Solution:
Geometry : For small angles,
α
∆Cx
b ∆Cy+
:= α 0.00662252 rad=
β
∆By ∆Cy−
a ∆Bx ∆Cx−( )+:= β 0.00496278 rad=
ψ
∆Bx ∆Ax−
b ∆By ∆Ay−( )+:= ψ 0.00662252 rad=
θ
∆Ay
a ∆Ax+
:= θ 0.00496278 rad=
Shear Strain : 
γxy_D α θ+:= γxy_D 11.585 10 3−× rad= Ans
γxy_C α β+( )−:= γxy_C 11.585− 10 3−× rad= Ans
Problem 2-14
The piece of plastic is originally rectangular. Determine the average normal strain that occurs along th
diagonals AC and DB.
Given:
a 400mm:= b 300mm:=
∆Ax 3mm:= ∆Ay 2mm:=
∆Cx 2mm:= ∆Cy 2mm:=
∆Bx 5mm:= ∆By 4mm:=
Solution:
Geometry : 
LAC a
2 b2+:= LAC 500 mm=
LDB a
2 b2+:= LDB 500 mm=
LA'C' a ∆Ax+ ∆Cx−( )2 b ∆Cy+ ∆Ay−( )2+:=
LA'C' 500.8 mm=
LDB' a ∆Bx+( )2 b ∆By+( )2+:=
LDB' 506.4 mm=
Average Normal Strain : 
εAC
LA'C' LAC−
LAC
:= εAC 1.601 10 3−×
mm
mm
= Ans
εBD
LDB' LDB−
LDB
:= εBD 12.800 10 3−×
mm
mm
= Ans
Problem 2-15
The guy wire AB of a building frame is originally unstretched. Due to an earthquake, the two columns
of the frame tilt θ = 2°. Determine the approximate normal strain in the wire when the frame isin thi
position. Assume the columns are rigid and rotate about their lower supports.
Given:
a 4m:= b 3m:= c 1m:= θ 2deg:=
Solution:
θ θ
180
⎛⎜⎝
⎞
⎠ π⋅= θ 0.03490659 rad=
Geometry : The vertical dosplacement is negligible. 
∆Ax c θ⋅:= ∆Ax 34.907 mm=
∆Bx b c+( ) θ⋅:= ∆Bx 139.626 mm=
LAB a
2 b2+:= LAB 5000 mm=
LA'B' a ∆Bx+ ∆Ax−( )2 b2+:=
LA'B' 5084.16 mm=
Average Normal Strain : 
εAB
LA'B' LAB−
LAB
:=
εAB 16.833 10 3−×
mm
mm
= Ans
Problem 2-16
The corners of the square plate are given the displacements indicated. Determine the shear strain along
the edges of the plate at A and B.
Given: ax 250mm:= ay 250mm:=
∆v 5mm:= ∆h 7.5mm:=
Solution:
At A : 
tan
θ'A
2
⎛⎜⎝
⎞
⎠
ax ∆h−
ay ∆v+
=
θ'A 2 atan
ax ∆h−
ay ∆v+
⎛⎜⎝
⎞
⎠
⋅:= θ'A 1.52056 rad=
γnt_A
π
2
⎛⎜⎝
⎞
⎠ θ'A−:= γnt_A 0.05024 rad= Ans
At B : 
tan
φ'B
2
⎛⎜⎝
⎞
⎠
ay ∆v+
ax ∆h−
=
φ'B 2 atan
ay ∆v+
ax ∆h−
⎛⎜⎝
⎞
⎠
⋅:= φ'B 1.62104 rad=
γnt_B φ'B
π
2
−:= γnt_B 0.05024 rad= Ans
Problem 2-17
The corners of the square plate are given the displacements indicated. Determine the average normal
strains along side AB and diagonals AC and DB.
Given: ax 250mm:= ay 250mm:=
∆v 5mm:= ∆h 7.5mm:=
Solution:
For AB : 
LAB ax
2 ay
2+:=
LA'B' ax ∆h−( )2 ay ∆v+( )2+:=
εAB
LA'B' LAB−
LAB
:=
LAB 353.55339 mm=
LA'B' 351.89665 mm=
εAB 4.686− 10 3−×
mm
mm
= Ans
For AC : 
LAC 2 ay( ):= LAC 500 mm=
LA'C' 2 ay ∆v+( )⋅:= LA'C' 510 mm=
εAC
LA'C' LAC−
LAC
:= εAC 20.000 10 3−×
mm
mm
= Ans
For DB : 
LDB 2 ax( ):= LDB 500 mm=
LD'B' 2 ax ∆h−( )⋅:= LD'B' 485 mm=
εDB
LD'B' LDB−
LDB
:= εDB 30.000− 10 3−×
mm
mm
= Ans
Problem 2-18
The square deforms into the position shown by the dashed lines. Determine the average normal strain
along each diagonal, AB and CD. Side D'B' remains horizontal. 
Given:
a 50mm:= b 50mm:=
∆Bx 3− mm:=
∆Cx 8mm:= ∆Cy 0mm:=
θ'A 91.5deg:= LAD' 53mm:=
Solution:
For AB : 
∆By LAD' cos θ'A 90deg−( )⋅ b−:= ∆By 2.9818 mm=
LAB a
2 b2+:= LAB 70.7107 mm=
LAB' a ∆Bx+( )2 b ∆By+( )2+:= LAB' 70.8243 mm=
εAB
LAB' LAB−
LAB
:=
εAB 1.606 10 3−×
mm
mm
= Ans
For CD : 
∆Dy ∆By:= ∆Dy 2.9818 mm=
LCD a
2 b2+:= LCD 70.7107 mm=
LC'D' a ∆Cx+( )2 b ∆Dy+( )2+ 2 a ∆Cx+( )⋅ b ∆Dy+( )⋅ cos θ'A( )⋅−:=
LC'D' 79.5736 mm=
εCD
LC'D' LCD−
LCD
:=
εCD 125.340 10 3−×
mm
mm
= Ans
Problem 2-19
The square deforms into the position shown by the dashed lines. Determine the shear strain at each of
its corners, A, B, C, and D. Side D'B' remains horizontal. 
Given: a 50mm:= b 50mm:=
∆Bx 3− mm:=
∆Cx 8mm:= ∆Cy 0mm:=
θ'A 91.5deg:= LAD' 53mm:=
Solution: θ'A 1.597 rad=
Geometry : 
∆By LAD' cos θ'A 90deg−( )⋅ b−:= ∆By 2.9818 mm=
∆Dy ∆By:= ∆Dy 2.9818 mm=
∆Dx LAD'− sin θ'A 90deg−( )⋅:= ∆Dx 1.3874− mm=
In triangle C'B'D' : 
LC'B' ∆Cx ∆Bx−( )2 b ∆By+( )2+:=
LC'B' 54.1117 mm=
LD'B' a ∆Bx+ ∆Dx−:= LD'B' 48.3874 mm=
LC'D' a ∆Cx+ ∆Dx−( )2 b ∆Dy+( )2+:=
LC'D' 79.586 mm=
cos θB'( ) LC'B'
2 LD'B'
2+ LC'D'2−
2 LC'B'( )⋅ LD'B'( )⋅=
θB' acos
LC'B'
2 LD'B'
2+ LC'D'2−
2 LC'B'( )⋅ LD'B'( )⋅
⎡⎢⎢⎣
⎤⎥⎥⎦
:= θB' 101.729 deg= θB' 1.7755 rad=
θD' 180deg θ'A−:= θD' 88.500 deg= θD' 1.5446 rad=
θC' 180deg θB'−:= θC' 78.271 deg= θC' 1.3661 rad=
Shear Strain : 
γxy_A 0.5π θ'A( )−:= γxy_A 26.180− 10 3−× rad= Ans
γxy_B 0.5π θB'( )−:= γxy_B 204.710− 10 3−× rad= Ans
γxy_C 0.5π θC'( )−:= γxy_C 204.710 10 3−× rad= Ans
γxy_D 0.5π θD'( )−:= γxy_D 26.180 10 3−× rad= Ans
Problem 2-20
The block is deformed into the position shown by the dashed lines. Determine the average normal
strain along line AB.
Given: ∆xBA 70 30−( )mm:= ∆yBA 100mm:=
∆xB'A 55 30−( )mm:=
∆yB'A 1102 152−( )mm:=
Solution:
For AB : 
LAB ∆xBA2 ∆yBA2+:= LAB 107.7033 mm=
LAB' ∆xB'A2 ∆yB'A2+:= LAB' 111.8034 mm=
εAB
LAB' LAB−
LAB
:=
εAB 38.068 10 3−×
mm
mm
= Ans
Problem 2-21
A thin wire, lying along the x axis, is strained such that each point on the wire is displaced ∆x = k x2
along the x axis. If k is constant, what is the normal strain at any point P along the wire?
Given: ∆x k x2⋅=
Solution:
ε
x
∆xd
d
=
ε 2 k⋅ x⋅= Ans
Problem 2-22
The rectangular plate is subjected to the deformation shown by the dashed line. Determine the averag
shear strain γxy of the plate.
Given: a 150mm:= b 200mm:=
∆a 0mm:= ∆b 3− mm:=
Solution:
∆θ atan ∆b
a
⎛⎜⎝
⎞
⎠:=
∆θ 1.146− deg=
∆θ 19.9973− 10 3−× rad=
Shear Strain : 
γxy ∆θ:=
γxy 19.997− 10 3−× rad= Ans
Problem 2-23
The rectangular plate is subjected to the deformation shown by the dashed lines. Determine the averag
shear strain γxy of the plate.
Given: a 200mm:= ∆a 3mm:=
b 150mm:= ∆b 0mm:=
Solution:
∆θ atan ∆a
b
⎛⎜⎝
⎞
⎠:=
∆θ 1.146 deg=
∆θ 19.9973 10 3−× rad=
Shear Strain : 
γxy ∆θ:=
γxy 19.997 10 3−× rad= Ans
Problem 2-24
 The rectangular plate is subjected to the deformation shown by the dashed lines. Determine the
average normal strains along the diagonal AC and side AB.
Given: a 200mm:= b 150mm:=
∆Ax 3− mm:= ∆Ay 0mm:=
∆Bx 0mm:= ∆By 0mm:=
∆Cx 0mm:= ∆Cy 0mm:=
∆Dx 3− mm:= ∆Dy 0mm:=
Solution:
Geometry : 
LAC a
2 b2+:= LAC 250 mm=
LAB b:= LAB 150 mm=
LA'C a ∆Cx+ ∆Ax−( )2 b ∆Cy+ ∆Ay−( )2+:=
LA'C 252.41 mm=
LA'B ∆Ax2 b2+:=
LA'B 150.03 mm=
Average Normal Strain : 
εAC
LA'C LAC−
LAC
:= εAC 9.626 10 3−×
mm
mm
= Ans
εAB
LA'B LAB−
LAB
:= εAB 199.980 10 6−×
mm
mm
= Ans
Problem 2-25
The piece of rubber is originally rectangular. Determine the average shear strain γxy if the corners B
and D are subjected to the displacements that cause the rubber to distort as shown by the dashed lines
Given: a 300mm:= b 400mm:=
∆Ax 0mm:= ∆Ay 0mm:=
∆Bx 0mm:= ∆By 2mm:=
∆Dx 3mm:= ∆Dy 0mm:=
Solution:
∆θAB atan
∆By
a
⎛⎜⎝
⎞
⎠:=
∆θAB 0.38197 deg=
∆θAB 6.6666 10 3−× rad=
∆θAD atan
∆Dx
b
⎛⎜⎝
⎞
⎠:=
∆θAD 0.42971 deg=
∆θAD 7.4999 10 3−× rad=
Shear Strain : 
γxy_A ∆θAB ∆θAD+:=
γxy_A 14.166 10 3−× rad= Ans
Problem 2-26
The piece of rubber is originally rectangular and subjected to the deformation shown by the dashed
lines. Determine the average normal strain along the diagonal DB and side AD.
Given: a 300mm:= b 400mm:=
∆Ax 0mm:= ∆Ay 0mm:=
∆Bx 0mm:= ∆By 2mm:=
∆Dx 3mm:= ∆Dy 0mm:=
Solution:
Geometry : 
LDB a
2 b2+:= LDB 500 mm=
LAD b:= LAD 400 mm=
LD'B' a ∆Bx+ ∆Dx−( )2 b ∆Dy+ ∆By−( )2+:= LD'B' 496.6 mm=
LAD' ∆Dx2 b ∆Dy+( )2+:= LAD' 400.01 mm=
Average Normal Strain : 
εBD
LD'B' LDB−
LDB
:= εBD 6.797− 10 3−×
mm
mm
= Ans
εAD
LAD' LAD−
LAD
:= εAD 28.125 10 6−×
mm
mm
= Ans
Problem 2-27
The material distorts into the dashed position shown. Determine (a) the average normal strains εx 
and εy , the shear strain γxy at A, and (b) the average normal strain along line BE.
Given: a 80mm:= Bx 0mm:= Ex 80mm:=
b 125mm:= By 100mm:= Ey 50mm:=
∆Ax 0mm:= ∆Cx 10mm:= ∆Dx 15mm:=
∆Ay 0mm:= ∆Cy 0mm:= ∆Dy 0mm:=
Solution:
∆xAC ∆Cx ∆Ax−:= ∆xAC 10.00 mm=
∆yAC ∆Cy ∆Ay−:= ∆yAC 0.00 mm=
∆θAC atan
∆Cx
b
⎛⎜⎝
⎞
⎠:= ∆θAC 79.8300 10
3−× rad=
Since there is no deformation occuring along the y- and x-axis, 
εx_A ∆yAC:= εx_A 0= Ans
εy_A
∆xAC2 b2+ b−
b
:=
εy_A 0.00319= Ans
γxy_A ∆θAC:=
γxy_A 79.830 10 3−× rad= Ans
Geometry : 
∆Bx
∆Cx
By
b
= ∆Bx
By
b
⎛⎜⎝
⎞
⎠ ∆Cx⋅:= ∆Bx 8 mm=
∆By 0mm:=
∆Ex
∆Dx
Ey
b
= ∆Ex
Ey
b
⎛⎜⎝
⎞
⎠ ∆Dx⋅:= ∆Ex 6 mm=
∆Ey 0mm:=
LBE Ex Bx−( )2 Ey By−( )2+:= LBE 94.34 mm=
LB'E' a ∆Ex+ ∆Bx−( )2 Ey ∆Ey+ ∆By−( )2+:= LB'E' 92.65 mm=
εBE
LB'E' LBE−
LBE
:= εBE 17.913− 10 3−×
mm
mm
= Ans
Note: Negative sign indicates shortening of BE.
Problem 2-28
The material distorts into the dashed position shown.Determine the average normal strain that occurs
along the diagonals AD and CF.
Given: a 80mm:= Bx 0mm:= Ex 80mm:=
b 125mm:= By 100mm:= Ey 50mm:=
∆Ax 0mm:= ∆Cx 10mm:= ∆Dx 15mm:=
∆Ay 0mm:= ∆Cy 0mm:= ∆Dy 0mm:=
Solution:
∆xAC ∆Cx ∆Ax−:= ∆xAC 10.00 mm=
∆yAC ∆Cy ∆Ay−:= ∆yAC 0.00 mm=
∆θAC atan
∆Cx
b
⎛⎜⎝
⎞
⎠:= ∆θAC 79.8300 10
3−× rad=
Geometry : 
LAD a
2 b2+:= LAD 148.41 mm=
LCF a
2 b2+:= LCF 148.41 mm=
LA'D' a ∆Dx+ ∆Ax−( )2 b ∆Dy+ ∆Ay−( )2+:=
LA'D' 157.00 mm=
LC'F a ∆Cx−( )2 b ∆Cy−( )2+:=
LC'F 143.27 mm=
Average Normal Strain : 
εAD
LA'D' LAD−
LAD
:= εAD 57.914 10 3−×
mm
mm
= Ans
εCF
LC'F LCF−
LCF
:= εCF 34.653− 10 3−×
mm
mm
= Ans
Problem 2-29
The block is deformed into the position shown by the dashed lines. Determine the shear strain at
corners C and D.
Given: a 100mm:= ∆Ax 15− mm:=
b 100mm:= ∆Bx 15− mm:=
LCA' 110mm:=
Solution:
Geometry : 
∆θC asin
∆Ax
LCA'
⎛⎜⎝
⎞
⎠
:= ∆θC 7.84− deg=
∆θC 0.1368− rad=
∆θD asin
∆Bx
LCA'
⎛⎜⎝
⎞
⎠
:= ∆θD 7.84− deg=
∆θD 0.1368− rad=
Shear Strain : 
γxy_C ∆θC:=
γxy_C 136.790− 10 3−× rad= Ans
γxy_D ∆θD−:=
γxy_D 136.790 10 3−× rad= Ans
Problem 2-30
The bar is originally 30 mm long when it is flat. If it is subjected to a shear strain defined by γxy = 0.0
x, where x is in millimeters, determine the displacement ∆y at the end of its bottom edge. It is distorte
into the shape shown, where no elongation of the bar occurs in the x direction.
Given: L 300mm:= γxy 0.02 x⋅=
unit 1mm:=
Solution:
dy
dx
tan γxy( )=
dy
dx
tan 0.02 x⋅( )=
0
∆y
y1
⌠⎮⌡ d 0
L
xtan 0.02 x⋅( ) unit( )⋅⌠⎮⌡ d=
∆y
0
30
xtan 0.02x( ) unit( )⋅⌠⎮⌡ d:=
∆y 9.60 mm= Ans
Problem 2-31
The curved pipe has an original radius of 0.6 m. If it is heated nonuniformly, so that the normal strain
along its length is ε = 0.05 cos θ, determine the increase in length of the pipe.
Given: r 0.6m:= ε 0.05 cos θ( )⋅=
Solution:
∆L Lε⌠⎮⌡ d=
∆L
0
90deg
rθ( )0.05 cos θ( )⋅⌠⎮⌡ d=
∆L
0
90deg
θ0.05 r⋅ cos θ( )⋅⌠⎮⌡ d:=
∆L 30.00 mm= Ans
Problem 2-32
Solve Prob. 2-31 if ε = 0.08 sin θ .
Given: r 0.6m:= ε 0.08 sin θ( )⋅=
Solution:
∆L Lε⌠⎮⌡ d=
∆L
0
90deg
rθ( )0.08 sin θ( )⋅⌠⎮⌡ d=
∆L
0
90deg
θ0.08 r⋅ sin θ( )⋅⌠⎮⌡ d:=
∆L 0.0480 m= Ans
Problem 2-33
A thin wire is wrapped along a surface having the form y = 0.02 x2, where x and y are in mm. Origina
the end B is at x = 250 mm. If the wire undergoes a normal strain along its length of ε = 0.0002x,
determine the change in length of the wire. Hint: For the curve, y = f (x), ds = 1
dy
dx
⎛⎜⎝
⎞
⎠
2
+ dx⋅ .
Given: Bx 250mm:= ε 0.0002 x⋅= unit 1mm:=
Solution:
y 0.02x2=
dy
dx
0.04x=
ds 1
dy
dx
⎛⎜⎝
⎞
⎠
2
+ dx⋅=
ds 1 0.04x( )2+ dx⋅=
∆L sε⌠⎮⌡ d= ∆L 0
Bx
x0.0002x( ) 1 0.04x( )2+⋅⌠⎮⌡ d=
∆L unit( )
0
250
x0.0002x( ) 1 0.04x( )2+⋅⌠⎮⌡ d⋅:=
∆L 42.252 mm= Ans
Problem 2-34
The fiber AB has a length L and orientation If its ends A and B undergo very small displacements uA
 and vB, respectively, determine the normal strain in the fiber when it is in position A'B'.
Solution:
Geometry:
LA'B' L cos θ( )⋅ uA−( )2 L sin θ( )⋅ vB−( )2+=
LA'B' L
2 uA
2+ vB2+ 2L vB sin θ( )⋅ uA cos θ( )⋅−( )⋅+=
Average Normal Strain:
εAB
LA'B' L−⋅
L
=
εAB 1
uA
2 vB
2+
L2
+
2 vB sin θ( )⋅ uA cos θ( )⋅−( )⋅
L
+ 1−=
Neglecting higher-order terms uA
2 and vB
2 ,
εAB 1
2 vB sin θ( )⋅ uA cos θ( )⋅−( )⋅
L
+ 1−=
Using the binomial theorem:
εAB 1
1
2
2 vB sin θ( )⋅ uA cos θ( )⋅−( )⋅
L
⎡⎢⎣
⎤⎥⎦+ ..+ 1−=
εAB
vB sin θ( )⋅
L
uA cos θ( )⋅
L
−= Ans
Problem 2-35
If the normal strain is defined in reference to the final length, that is,
 ε '
n
 = 
p'p
∆s' ∆s−
∆s'
⎛⎜⎝
⎞
⎠lim→
instead of in reference to the original length, Eq.2-2, show that the difference in these strains is
represented as a second-order term, namely, εn - ε'n = εn ε'n.
Solution:
εn
∆S' ∆S−
∆S=
εn ε'n−
∆S' ∆S−
∆S
⎛⎜⎝
⎞
⎠
∆S' ∆S−
∆S'
⎛⎜⎝
⎞
⎠−=
εn ε'n−
∆S'2 2 ∆S( )⋅ ∆S'( )⋅− ∆S2+
∆S( ) ∆S'( )⋅=
εn ε'n−
∆S' ∆S−( )2
∆S( ) ∆S'( )⋅=
εn ε'n−
∆S' ∆S−
∆S
⎛⎜⎝
⎞
⎠
∆S' ∆S−
∆S'
⎛⎜⎝
⎞
⎠⋅=
εn ε'n− εn( ) ε'n( )⋅= (Q.E.D.)
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