Week 4  1 slides

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Mechanics of Solids
Axial Load

Thermal Strain
Thermal Deformation
Thermal Strain
Thermal Stress
thermal = E thermal A B

25 mm 20 mm
A B C
300 mm 450 mm
D D=25 mm
a
a
Section aa
0.2 mm
E
Segment AB: EAB = 44.7 GPa; AB = 26 x106/oC
Segment CD: ECD = 68.9 GPa; CD = 24 x 106/oC
A B C D D=25 mm
E FAB FDC
FAB = FCD = F
YAB = 152 MPa
YCD = 255 MPa
= 152 x (252202) = 107.5 kN
= 255 x 12.52 = 125.2 kN
107.5 kN
485 The AM1004T61 magnesium alloy tube AB is capped with a rigid plate. The gap between E and end C of the 6061T6 aluminum alloy solid circular rod CD is 0.2 mm when the temperature is 30oC . Determine the highest temperature to which it can be raised without causing yielding either in the tube or the rod. Neglect the thickness of the rigid cap. (p. 157)
Thermal Strain

25 mm 20 mm
A B C
300 mm 450 mm
D D=25 mm
a
a
Section aa
0.2 mm
E
A B C D D=25 mm
E FAB FDC
FAB = FCD = F 107.5 kN
D=25 mm FDC FDC
AD= 0.2 mm = AB + CD (Compatibility)
T F = 0.2 mm
485 The AM1004T61 magnesium alloy tube AB is capped with a rigid plate. The gap between E and end C of the 6061T6 aluminum alloy solid circular rod CD is 0.2 mm when the temperature is 30oC . Determine the highest temperature to which it can be raised without causing yielding either in the tube or the rod. Neglect the thickness of the rigid cap. (p. 157)
Thermal Strain

485 The AM1004T61 magnesium alloy tube AB is capped with a rigid plate. The gap between E and end C of the 6061T6 aluminum alloy solid circular rod CD is 0.2 mm when the temperature is 30oC . Determine the highest temperature to which it can be raised without causing yielding either in the tube or the rod. Neglect the thickness of the rigid cap. (p. 157)
25 mm 20 mm
A B C
300 mm 450 mm
D D=25 mm
a
a
Section aa
0.2 mm
E
A B C D D=25 mm
E FAB FDC
FAB = FCD = F 107.5 kN
AD = 0.2 mm = AB + CD (Compatibility)
= 0.2 mm
= 1.02 mm = 1.43 mm
=2.65 mm
T = 142 oC Segment AB: EAB = 44.7 GPa; AB = 26 x106/oC
Segment CD: ECD = 68.9 GPa; CD = 24 x 106/oC YAB = 152 MPa
YCD = 255 MPa Th = 30 + T = 172 oC
Thermal Strain

Axial Load: Deformation
= E SaintVenant Principle Hookes Law
15 kN
A D
20 kN
B C
a
a
N [kN]
L
NAB
A B C D 0
NBC
NCD
NBC  NAB =  20 kN
NCD  NBC= + 15 kN

Poissons Ratio
When a deformable body is stretched by a tensile force, not only does it elongate but it also contract laterally, i.e. it would contract in other two dimensions. Likewise, a compressive force acting on a deformable body cause it to contract in the direction of force and yet its sides expand laterally
Lateral strain is the same in all lateral direction;
Poissons ratio is a constant.
Usually 0 v 0.5. For most linearly elastic material v 0.3;

Lateral strain is the same in all lateral direction;
Poissons ratio is a constant.
Usually 0 v 0.5. For most linearly elastic material v 0.3;
z
y
x
o
zz
yy
xx
z
= + + zz
'yy
'xx
'zz
=
'z =zz/E ''z z =  (yy/E) =  (xx/E)
z + z + z = [zz  (yy + xx ) ]/E x = [xx  (yy + zz ) ]/E
y = [yy  (xx + zz ) ]/E
Poissons Ratio

Lateral strain is the same in all lateral direction;
Poissons ratio is a constant.
Usually 0 v 0.5. For most linearly elastic material v 0.3;
z
y
x
o
zz
yy
xx
= [zz  (yy + xx ) ]/E
x = [xx  (yy + zz ) ]/E
y = [yy  (xx + zz ) ]/E
dx dz
dy
dV=dx dy dz
(1+x)dx z (1+z)dz
(1+y)dy
V
V = (1+x) (1+y) (1+z) dx dy dz
= (1+x) (1+y) (1+z) dx dy dz  dx dy dz
= (x+y+z) dx dy dz
e= V dV = x+y+z Volume Strain: =
1  2 E
(zz + yy + xx)
Poissons Ratio

Lateral strain is the same in all lateral direction;
Poissons ratio is a constant.
Usually 0 v 0.5. For most linearly elastic material v 0.3;
z
y
x
o
zz
yy
xx
dx dz
dy
dV=dx dy dz
(1+x)dx
(1+z)dz
(1+y)dy
V = (1+x) (1+y) (1+z) dx dy dz
e= V dV
Volume Strain:
= 1  2 E (zz + yy + xx)
zz = yy = xx = p
= 3(1  2)
E p e k = Volume Modulus of elasticity:
(Bulk modulus)
Poissons Ratio

325 The plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. E=2.70GPa, v=0.4. (p. 111)
200 mm
300 N 300 N
A B
= 0.126/200=0.00063
=  0.00063 x 0.4 =  0.00025
=  0.00025 x 15 =  0.00377 mm
Ans.
Ans.
Poissons Ratio

Lateral strain is the same in all lateral direction;
Poissons ratio is a constant.
Usually 0 v 0.5. For most linearly elastic material v 0.3;
z
y
x
o
zz
yy
xx
dx dz
dy
dV=dx dy dz
(1+x)dx
(1+z)dz
(1+y)dy
V = (1+x) (1+y) (1+z) dx dy dz
e= V dV
Volume Strain:
= 1  2 E (zz + yy + xx)
zz = yy = xx = p
= 3(1  2)
E p e k = Volume Modulus of elasticity:
(Bulk modulus)
Poissons Ratio

Strain Energy
Work of a Force:
F W = F x
P
x
F = f(x)
Axial Load: F
x 0
F = k x
P
(k = P/=EA/L )
W
W
External Work

P
Axial Load: F
x 0
F = k x
P
(k = P/ )
External Work
Internal Work Strain Energy y
y
dx dz dy
dFy = ydA = ydxdz
dz = dz
We = Ui
W
Strain Energy

Internal Work Strain Energy
0
z
Strain Energy Density  Strainenergy per unit volume
i
Modulus of resilience  When stress reaches the proportional limit, strain energy density is referred to Modulus of resilience
pl
pl
x fr
y
y
dx dz dy
Modulus of toughness The higher the modulus of toughness, the more strain energy can be stored (absorbed) inside the material before fracturing.
z
Strain Energy

Internal Work Strain Energy y
y
dx dz dy
F
Total Strain Energy inside deformable body The above equation is the elemental strain energy, by integrating it, the total volume and substituting Hooke's law we have that:
External Work
If the deflection of a structure or member under a single concentrated load P is known, the corresponding Work W may be obtained by writing
W
Strain Energy

1473 Determine the horizontal displacement of join B. Each A36 member has a crosssectional area of 1250 mm2. (p. 760)
A
B
C 1.8 m
2.4 m
1 kN
1.8 m
Strain Energy
Mechanics of SolidsSlide Number 2Slide Number 3Slide Number 4Slide Number 5Slide Number 6Slide Number 7Slide Number 8Slide Number 9Slide Number 10Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17