I bh 3 12 b width of the specimen h thickness of the specimen I steel b steel h

I bh 3 12 b width of the specimen h thickness of the

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I = bh 3 /12 b = width of the specimen h = thickness of the specimen I steel = b steel h 3 steel /12 = (0.0221m) (0.0063m) 3 /12 = 4.61 X 10 -10 m 4 I aluminum = b aluminum h 3 aluminum /12 = (0.0232m) (0.0065m) 3 /12 = 5.31 X 10 -10 m 4 I brass = b brass h 3 brass /12 = (0.0214m) (0.00613m) 3 /12 = 4.11 X 10 -10 m 4 Material Load, F (N) Deflection, δ (m) Inertia, I (m^4) Aluminum 5 1.87E-04 5.31E-10 Aluminum 10 3.53E-04 5.31E-10 Aluminum 15 5.20E-04 5.31E-10 Aluminum 20 7.00E-04 5.31E-10 Steel 5 6.33E-05 4.61E-10 Steel 10 1.40E-04 4.61E-10 Steel 15 2.20E-04 4.61E-10 Steel 20 2.70E-04 4.61E-10 Brass 5 2.37E-04 4.11E-10 Brass 10 3.33E-04 4.11E-10 Brass 15 7.87E-04 4.11E-10 Brass 20 1.03E-03 4.11E-10 Elasticity, E (Part III, A) E= (F/ δ) (3.5L 3 /384I)
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F = Weight of the load δ = Deflection of the specimen L = Clamping Length I = Moment of Inertia E = Modulus of Elasticity The modulus of elasticity can determine by plotting the graph deflection of the specimen against the load by rearranged the equation: δ = (3.5L 3 /384IE) F This equation is similar to the directly proportional equation, y=mx. Let y= δ, x=F, m = 3.5L 3 /384IE Then, the modulus of elasticity is determined using the gradient in the graph: m = 3.5L 3 /384IE E = 3.5L 3 /384Im Elasticity of Aluminum (Part3A)
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0 5 10 15 20 0 0 0 0 0 0 0 0 0 f(x) = 0x Load vs Deflection of Aluminum Load(N) Deflection (m) m = 3.50x10 -5 , L = 0.5m, I = 5.31X10 -10 m 4 E = 3.5L 3 /384Im E = 3.5(0.5m) 3 / [384(5.31X10 -10 m 4 ) (3.50x10 -5 )] E = 61.30 GPa Elasticity of Steel (Part3A)
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0 5 10 15 20 0 0 0 0 0 f(x) = 0x Load vs Deflection of Steel Load(N) Deflection (m) m = 1.39x10 -5 , L = 0.5m, I = 4.61X10 -10 m 4 E = 3.5L 3 /384Im E = 3.5(0.5m) 3 / [384(4.61X10 -10 m 4 ) (1.39x10 -5 )] E = 177.80 GPa Elasticity of Brass (Part3A)
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0 5 10 15 20 0 0 0 0 0 0 0 0 0 0 0 f(x) = 0x Load vs Deflection of Brass Load(N) Deflection (m) m = 4.92x10 -5 , L = 0.5m, I = 4.11X10 -10 m 4 E = 3.5L 3 /384Im E = 3.5(0.5m) 3 / [384(4.11X10 -10 m 4 ) (4.92x10 -5 )] E = 56.34 GPa Part 3B: Moment of Inertia, I
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I = bh 3 /12 b = width of the specimen h = thickness of the specimen I steel = b steel h 3 steel /12 = (0.0259m) (0.00608m) 3 /12 = 4.85X 10 -10 m 4 I aluminum = b aluminum h 3 aluminum /12 = (0.0256m) (0.00638m) 3 /12 = 5.55 X 10 -10 m 4 I brass = b brass h 3 brass /12 = (0.0260m) (0.00662m) 3 /12 = 6.29 X 10 -10 m 4 Material Load, F (N) Deflection, δ (m) Inertia, I (m^4) Aluminum 5 1.32E-03 5.55E-10 Aluminum 10 2.64E-03 5.55E-10 Aluminum 15 4.00E-03 5.55E-10 Aluminum 20 5.29E-03 5.55E-10 Steel 5 6.80E-04 4.85E-10 Steel 10 1.20E-03 4.85E-10 Steel 15 1.84E-03 4.85E-10 Steel 20 2.40E-03 4.85E-10 Brass 5 7.17E-04 6.29E-10 Brass 10 1.54E-03 6.29E-10 Brass 15 2.38E-03 6.29E-10 Brass 20 3.12E-03 6.29E-10 Elasticity, E (Part III, B) E= (F/ δ) (3.5L 3 /384I)
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F = Weight of the load δ = Deflection of the specimen L = Clamping Length I = Moment of Inertia E = Modulus of Elasticity The modulus of elasticity can determine by plotting the graph deflection of the specimen against the load by rearranged the equation: δ = (3.5L 3 /384IE) F This equation is similar to the directly proportional equation, y=mx. Let y= δ, x=F, m = 3.5L 3 /384IE Then, the modulus of elasticity is determined using the gradient in the graph: m = 3.5L 3 /384IE E = 3.5L 3 /384Im Elasticity of Brass (Part3B)
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0 5 10 15 20 0 0 0 0 0 0.01 f(x) = 0x Load vs Deflection of Brass Load(N) Deflection (m) m = 1.56x10 -4 , L = 0.8m, I = 6.29X10 -10 m 4 E = 3.5L 3 /384Im E = 3.5(0.8m) 3 / [384(6.29X10 -10 m 4 ) (1.56x10 -4 )] E = 47.56 GPa Elasticity of Steel (Part3B)
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0 5 10 15 20 0 0 0 0 0 0.01 f(x) = 0x Load vs Deflection of Steel Load(N) Deflection (m) m = 1.21x10 -4 , L = 0.5m, I = 4.85X10 -10 m 4 E = 3.5L 3 /384Im E = 3.5(0.8m) 3 / [384(4.85X10 -10 m 4 ) (1.21x10 -4 )] E = 79.52 GPa Elasticity of Aluminum (Part3B)
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0 5 10 15 20 0 0 0 0 0 0.01 0.01 0.01 0.01 f(x) = 0x Load vs Deflection of Aluminum Load(N) Deflection (m) m = 2.65x10 -4 , L = 0.5m, I = 5.55X10 -10 m 4 E = 3.5L 3 /384Im E = 3.5(0.8m) 3 / [384(5.55X10 -10 m 4 ) (2.65x10 -4 )] E = 31.73 GPa Percentage Error
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  • Spring '15
  • Tensile strength, elastic modulus, deflection

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