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mariarad [96]
3 years ago
10

For a rod of annealed AISI 1018 steel with a cross sectional area of 0.65 in^2?; what is the maximum tensile load Pmax that shou

ld be placed on it given a design factor of 3 to avoid yielding?
Engineering
1 answer:
lawyer [7]3 years ago
7 0

Answer:

maximum tensile load Pmax is 11.91 ksi

Explanation:

given data

area = 0.65 in²

design factor of safety = 3

to find out

what is the maximum tensile load Pmax

solution

we know here area is 0.65 in² and FOS = 3

so by steel table for rod of annealed AISI 1018 steel table we know σy = 55 ksi

so

we use here design factor formula that is

\frac{ \sigma y}{FOS}  = \frac{Pmax}{area}  .............1

put here all these value we get Pmax in equation 1

\frac{55}{3}  = \frac{Pmax}{0.65}

Pmax = 11.91 ksi

so maximum tensile load Pmax is 11.91 ksi

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P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diame
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Answer:

a)  m=0.17kg/s

b)  Ma=0.89

Explanation:

From the question we are told that:

Pressure P=60kPa

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Generally the Power series equation for Mach number is mathematically given by

\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}

\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}

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\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}

\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}

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6 0
2 years ago
Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a
Elena L [17]

Answer:

\mathbf{\tau_c =5.675 \ MPa}

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [\bar 101]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]

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[d_1\ e_1 \ f_1] = directional indices for tensile stress

[d_2 \ e_2 \ f_2] = slip direction

replacing their values;

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cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]

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Then;

cos \  \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]

replacing their values;

i.e d_1 = 0 ,e_1 = 0 f_1 =  1 & d_3 = 1 , e_3 = 1 , f_3 = 1

cos \  \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]

cos \phi= \dfrac{1} {\sqrt{3} }

However, the critical resolved SS(shear stress) \mathbf{\tau_c} can be computed using the formula:

\tau_c = (\sigma )(cos  \phi )(cos \lambda)

where;

applied tensile stress \sigma = 13.9 MPa

∴

\tau_c =13.9\times (  \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})

\mathbf{\tau_c =5.675 \ MPa}

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