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kirza4 [7]
3 years ago
6

Draw the front, side and the top view please hurryyyy

Engineering
1 answer:
Firlakuza [10]3 years ago
4 0
Sorry my drawings aren’t great, but I hope this helps

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Gelneren [198K]

Answer:

C

Explanation:

7 0
3 years ago
I need help with this ASAP pls help
Evgen [1.6K]

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3 years ago
The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change i
ehidna [41]

Answer:

Change in length = 0.1257 mm

Change in diameter= -0.03771mm

Explanation:

Given

Diameter, d = 15 mm

Length of rod, L = 200mm

F = Force= 300N

d = 0.015m

Ep=2.70 GPa, np=0.4.

First, we have to calculate the normal stress using

σ = F/A where F = Force acting on the Cross-sectional area

A = Area

Area is calculated as πd²/4 where d = 0.015m

A = 22/7 * 0.015²/4

A = 0.000176785714285m²

A = 1.768E-4m²

So, stress. σ = 300N/1.768E-4m²

σ = 1696832.579185520Pa

σ = 1.697MPa

Calculating E(long)

E(long) = σ /Ep

E(long) = 1.697E-3/2.70

E(long) = 0.0006285

At this point, we fan now calculate the change in length of the element;

∆L = E(long) * L

∆L = 0.0006285 * 200mm

∆L = 0.1257mm

Calculating E(lat)

E(lat) = -np * E(long)

E(lat) = -4 * 0.0006285

E(lat) = -0.002514

At this point, we can now calculate the change in diameter of the element;

∆D = E(lat) * D

∆L = -0.002514 * 15mm

∆L = -0.03771mm

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A pin-supported structure has unrestrained rotations at the support locations.a) True b) False
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Answer:

a)True

Explanation:

Yes it is true a pin support can not resist the rotation motion . It can resist only lateral or we can say that only linear motion of structure and can not resit angular moment of motion about hinge or pin joint.On the other hand a fixed support can resist linear as well rotation motion of structure.

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3 years ago
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