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balandron [24]
3 years ago
5

For a bronze alloy, the stress at which plastic deformation begins is 294 MPa and the modulus of elasticity is 121 GPa. (a) What

is the maximum load that can be applied to a specimen having a cross-sectional area of 327 mm2 without plastic deformation? (b) If the original specimen length is 129 mm, what is the maximum length to which it may be stretched without causing plastic deformation?
Physics
1 answer:
FromTheMoon [43]3 years ago
7 0

Answer:

a) load in Newton is 96,138 b) 129.314mm

Explanation:

Stress = force/ area (cross sectional area of the bronze)

Force(load) = 294*10^6*327*10^-6 = 96138N

b) modulus e = stress/ strain

Strain = stress/ e = (294*10^6)/ (121*10^ 9) = 2.34* 10^ -3

Strain = change in length/ original length = DL/ 129

Change in length DL = 129 * 2.34*10^ -3 = 0.31347

Maximum length = change in length + original length = 129.314mm

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Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

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coefficient of static friction μ =  0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

\sum f_y =0

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\mu_s N_A + \mu_s N_B  = P

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replacing our value from equation (1)

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b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for  the cabinet is not to tip over is calculated by determining the limiting condition  of the unbalanced torque whose effect is canceled by the normal reaction at N_A and it is shifted to N_B:  

Then:

\sum M _B = 0

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