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mylen [45]
3 years ago
9

Assuming the torsional yield strength of a compression spring is 0.43Sut and the maximum shear stress is equal to 434MPa. What i

s the factor of safety if the spring is a music wire (A=2170MPa and m=0.146) with a wire diameter of 4mm?
Engineering
1 answer:
pav-90 [236]3 years ago
4 0

Answer:

factor of safety is 1.756

Explanation:

given data

torsional yield strength = 0.43 sut  

maximum shear stress = 434 MPa

A = 2170 MPa

m = 0.146

diameter = 4 mm

to find out

factor of safety

solution

we know 1 sut is equal to 1772.39 MPa

so 0.43 sut = 0.43 ×1772.39 =  762.128 MPa

so here

factor of safety formula is

factor of safety = yield strength / shear stress   ............1

put here value

factor of safety = 762.128 / 434

factor of safety is 1.756

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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_2 \ e_2 \ f_2] = slip direction

replacing their values;

i.e d_1 = 0 ,e_1 = 0 f_1 =  1 & d_2 = -1 , e_2 = 0 , f_2 = 1

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]

cos \ \lambda = \dfrac{1}{\sqrt{2}}

Also, to find the angle \phi between the stress [001] & normal slip plane [111]

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;

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∴

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

\mathbf{\tau_c =5.675 \ MPa}

3 0
3 years ago
When removing a diesel engine from a truck, Technician A says it is OK to disconnect an air con­ditioning hose, but the refriger
agasfer [191]

Answer:

Technician B is right.

Explanation:

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4 0
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8 0
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