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Romashka [77]
3 years ago
10

A tensile-testing machine is to test specimens of diameter 15 mm which have ultimate tensile strengths of up to 820 N/mm2. What

is the maximum tensile force the machine must be capable of?
Engineering
1 answer:
trasher [3.6K]3 years ago
8 0

Answer:

14.48 N

Explanation:

We have given the diameter of the specimens d = 15 mm ,so radius r=\frac{15}{2}=7.5 cm

Area of the specimen A=\pi r^2=3.14\times 0.075^2=0.01766m^2

The maximum tensile strength = 820 N/m^2

We know that maximu\ tensile\ strength =\frac{maximum\ tensile\ force}{Area}

820=\frac{maximu\ tensile\ force}{0.01766}=14.48N

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A concentrated load P is applied to the upper end of a 1.47-m-long pipe. The outside diameter of the pipe is D = 112 mm and the
myrzilka [38]

Answer:

Pmax = 38251.73 N

Explanation:

Given info

L = 1.47 m

D = 112 mm ⇒ R = D/2 = 112/2 mm = 56 mm

d = 101 mm  ⇒ r = D/2 = 101/2 mm = 50.5 mm

a) We can apply the following equation in order to get Q (First Moment of Area):

Q = 2*(A₁*y₁-A₂*y₂)

where

A₁ = π*R² = π*(56 mm)² = 3136 π mm²  

y₁ = 4*R/(3*π) = 4*56/(3*π) mm = 224/(3*π) mm

A₂ = π*r² = π*(50.5 mm)² = 2550.25 π mm²

y₂ = 4*r/(3*π) = 4*50.5/(3*π) mm = 202/(3*π) mm

then

Q = 2*(3136 π mm²*224/(3*π) mm-2550.25 π mm²*202/(3*π) mm)

⇒ Q = 62437.833 mm³

b) If  τallow = 83 MPa = 83 N/mm²

P = ?

We can use the equation

τ = V*Q / (t*I)   ⇒  V = τ*t*I / Q

where

t = D - d = 112 mm - 101 mm = 11 mm

I = (π/64)*(D⁴-d⁴) = (π/64)*((112 mm)⁴-(101 mm)⁴) = 2615942.11 mm⁴

Q = 62437.833 mm³

we could also use this equation in order to get Q:

Q = (4/3)*(R³-r³)

⇒  Q = (4/3)*((56 mm)³-(50.5 mm)³) = 62437.833 mm³

then we have

V = (83 N/mm²)*(11 mm)*(2615942.11 mm⁴) / (62437.833 mm³)

⇒ V = 2942.255 N

Finally Pmax = V = 38251.73 N

6 0
3 years ago
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Alexxx [7]

Given :

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Solution :

Displacement vector between point A and B is :

\vec{d} = (3-2)\hat{i} + (5-3)\hat{j}\\\\\vec{d} = \hat{i} + 2\hat{j}

Now, we know work done is given by :

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barxatty [35]

Answer:

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