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4 years ago

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An 18-in.-long titanium alloy rod is subjected to a tensile load of 24,000 lb. If the allowable tensile stress is 60 ksi and the

allowable total elongation is not to exceed 0.05 in., compute the required rod diameter. Assume the proportional limit to be 120 ksi.

1 answer:

Vilka [71]4 years ago

3 0

Answer:

Required Diameter = 302.65 inches

Explanation:

We are given;

Allowable tensile stress = 60 ksi

Weight of tensile load = 24,000 lb

Elongation = 0.05 in

Original length = 18 in

We'll need to check the diameters under stress and strain.

Now, we know that the formula for stress is;

Stress = Force/Area

Thus,

Area = Force/stress

So for this stress, area required is;

A_req = 24000/60 = 4000 in²

So let's find the required diameter here.

Area = πd²/4

So, 4000 = πd²/4

(4000 x 4)/π = d²

d² = 5092.96

Required diameter here is;

d = √5092.96

d = 71.36 in

For Strain;

Formula for strain is;

Strain = stress/E

We are given E = 120 ksi

stress = P/A = 24,000/A

strain = elongation/original length = 0.05/18 = 0.00278

Thus;

0.00278 = P/(A•E)

0.00278 = 24000/(120 x A)

Making A the subject to obtain;

A = 24000/(120 x 0.00278)

A_required = 71942 in²

Area = πd²/4

So, 71942 = πd²/4

(71942 x 4)/π = d²

d² = 91599.4

Required diameter here is;

d = √91599.4

d = 302.65 in

The larger diameter is 302.65 inchesand it's therefore the required one.

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3 years ago

Lumber jacks use cranes and giant tongs to hoist their goods into trucks for transport. Fortunately, smaller versions of these d

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Complete Question

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 $P= 210 N$

and the moment M is $M = -48.3N \cdot m$

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

$\sum M_A = 0$

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

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

=> $10.29 +F_{BC} (0.02763+0.03552) =0$

=> $F_{BC} =\frac{10.29}{0.06315}$

=> $F_{BC} = - 162.925 N$

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

At Equilibrium the moment about D is

$\sum M_D = 0$

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

=> $\frac{210}{2} * 0.230 -F_T (0.345 +0.035) - (-162.925)cos(52.125^o) *0.035\\-(-162.925)sin(52.125^o)(0.010 +0.040) =0$

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

=> $F_T =89.67N$

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4 years ago

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Answer:

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4 years ago

2. Using an RLC circuit to design a band-reject filter, with the two cut off frequencies 40k and 90k Hz. Design the values of R,

nadya68 [22]

Answer:

C = 59.17 nF

Q = 2.6

Explanation:

given data

frequencies = 40k Hz

frequencies = 90k Hz

solution

we take here R, L C take in series

so cut off frequency is express as

Wc1 = $-\frac{R}{2L}+ \sqrt{(\frac{R}{2L})^2+\frac{1}{LC}}$ = 40000

wc2 = $\frac{R}{2L}+ \sqrt{(\frac{R}{2L})^2+\frac{1}{LC}}$ = 90000

so here

wc2 - wc1 will be

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so

$\frac{R}{L}$ = 50000

we consider here R is 500

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and here total cut off frequency is

total cut off frequency = 40000 + 90000 = 130000

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so C = 59.17 nF

quality factor Q will be

Q = $\frac{130000}{50000}$

Q = 2.6

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3 years ago