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3 years ago

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For a bronze alloy, the stress at which plastic deformation begins is 297 MPa and the modulus of elasticity is 113 GPa. (a) What

is the maximum load that can be applied to a specimen having a cross-sectional area of 316 mm2 without plastic deformation? (b) If the original specimen length is 128 mm, what is the maximum length to which it may be stretched without causing plastic deformation?

1 answer:

Alenkinab [10]3 years ago

5 0

Answer:

a) 93.852 kN

b) 128.043 mm

Explanation:

Stress is load over section:

σ = P / A

If plastic deformation begins with a stress of 297 MPa, the maximum load before plastic deformation will be:

P = σ * A

316 mm^2 = 3.16*10^-4

P = 297*10^6 * 3.16*10^-4 = 93852 N = 93.852 kN

The stiffness of the specimen is:

k = E * A / l

k = 113*10^9 * 3.16*10^-4 / 0.128 = 279 MN/m

Hooke's law:

x' = x0 * (1 + P/k)

x' = 0.128 * (1 + 93.852*10^3 / 279*10^6) = 0.128043 m = 128.043 mm

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Lumber jacks use cranes and giant tongs to hoist their goods into trucks for transport. Fortunately, smaller versions of these devises are available for weekend warriors who want to play with their chain saws. Let us model the illustrated tongs as a planar mechanism that carries a log of weight 210 N. Given the following dimensions: 35 mm 10 mm 40 mm 230 mm 85 mm 45 mm 10 mm 35 mm 345 mm determine the force in N and moment in Nm that our worker exerts on the tongs. Also determine the pinching force magnitude in N that the tongs exert on the log; i.e. determine the horizontal force that the tong's teeth exert on the log. Assume the point E is centered between the tong's teeth.

The diagram for this question is shown on the first uploaded image

Answer:

The force P is $P= 210 N$

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The horizontal force that the tong teeth exerts is $F_T =89.67N$

Explanation:

First let denote the dimension to corresponding to the diagram

$a=35mm , b= 10mm, c= 40mm, d= 230mm, e= 85mm,f= 45mm,\\g= 10mm,h=35mm,i=345mm.$

Next looking at the diagram let us consider the vertical direction

At equilibrium

$\sum F_{vertical} =0$

This mean that

$P+ W = 0$

Since they are acting in opposite direction the equation becomes

$P - W = 0$

=> $P= W$

=> $P= 210 N$

At Equilibrium Moment about F gives

$\sum M_f = 0$

=> $F_T * (e +f + g+ h+ i) - F_T * (e+ f+g+ h+i) - W *d -M =0$

=> $M = -W *d$

=> $M = -210 * 0.230$

=> $M = -48.3N \cdot m$

Here $F_T$ is the horizontal force that the tong teeth exerts

Now let consider the part BAF of the system as shown on the second uploaded image

Now the angle $\theta$ is mathematically given as

$tan \theta = \frac{g+h}{a}$

=> $\theta = arctan \frac{g+h}{a}$

$= arctan (\frac{10+35}{35} )$

$=52.125^o$

Now at equilibrium the moment about A is

$\sum M_A = 0$

=> $P * (c+d) +M + F_{BC} cos \theta *f+F_{BC}sin\theta *(a+b) =0$

$210 * (0.040 + 0.230)-48.3+F_{BC} cos (52.125^o)*0.045+ F_{BC}sin(52.125^o)$ $* (0.035 +0.010) =0$

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Looking at the forces acting on the teeth as shown on the third uploaded image

At Equilibrium the moment about D is

$\sum M_D = 0$

=> $\frac{W}{2} *d - F_T *(i+h) -F_{BC} cos \theta *h -F_{BC} sin \theta * (b+c) =0$

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=> $34.081 = F_T(0.345 +0.035)$

=> $F_T =89.67N$

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