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2 years ago

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A cylindrical metal specimen of initial diameter d0 =14 mm, initial length L0=53 mm, strain hardening exponent n=0.31, strength

coefficient K=665 MPa and reduction of area at fracture Arf=0.53 - Determine the true strain at maximum load and true strain at fracture - Determine the length of specimen at maximum load and diameter at fracture - Determine the maximum load force - Determine the Load (F) on the specimen when a true strain Ɛt=0.25 is applied during the tension test

1 answer:

Marrrta [24]2 years ago

3 0

Answer:

a) Ef = 0.755

b) length of specimen( Lf )= 72.26mm

diameter at fracture = 9.598 mm

c) max load ( Fmax ) = 52223.24 N

d) Ft = 51874.67 N

Explanation:

a) Determine the true strain at maximum load and true strain at fracture

True strain at maximum load

Df = 9.598 mm

True strain at fracture

Ef = 0.755

b) determine the length of specimen at maximum load and diameter at fracture

Length of specimen at max load

Lf = 72.26 mm

Diameter at fracture

= 9.598 mm

c) Determine max load force

Fmax = 52223.24 N

d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test

F = 51874.67 N

attached below is a detailed solution of the question above

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Let $U_B$ be the initial velocity of rod B

Let $V_A$ be the final velocity of rod A

Let $V_B$ be the final velocity of rod B

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$M_AU_A + M_BU_B = M_A(V_B - e(U_A - U_B)) + M_BV_B$ ..............(3)

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$V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}$ ....................(4)

From equation (2):

$V_B = V_A + e(U_A -U_B)$ ......(5)

Substitute equation (5) into (1)

$M_AU_A + M_BU_B = M_AV_A + M_B(V_A + e(U_A -U_B))$ ..........(6)

Solving for $V_A$ in equation (6) above:

$V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}$ .........(7)

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$M_A = 2 kg\\M_B = 1 kg\\U_B = -3 m/s( negative x-axis)\\e = 0.65\\U_A = ?$

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Substitute these parameters into equation (7)

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To calculate the final velocity, $V_B$ , substitute the given parameters into (4):

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% $\triangle KE = -47.85 \%$

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