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katovenus [111]
3 years ago
6

What is a beam on a bridge? what does it do?

Engineering
1 answer:
kolbaska11 [484]3 years ago
4 0

Answer:

A beam carries vertical loads

Explanation:

In bridge: Beam. The beam bridge is the most common bridge form. A beam carries vertical loads by bending. As the beam bridge bends, it undergoes horizontal compression on the top.

I hope this answered your question, if not let me know :)

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How nany degrées is the included angle of General Purpose Acme threads? A. 60 B. 29 c. 14.5 D. 10
horrorfan [7]

Answer:

B.29

Explanation:

In general purpose acme thread:

  Nominal depth of thread=0.5ltimes Pitch

  Included angle =29 degrees

Generally Acme thread are following types

  1.General purpose(G) Acme

  2.Centralizing(C) Acme

  3. Stub Acme  

Centralizing(C) Acme threads have tighter tolerance during manufacturing as compare to General purpose(G) Acme  threads.

8 0
3 years ago
What is 1000J in Btu?
dybincka [34]

Answer:

0.948 Btu

Explanation:

1 Btu = 1055 J so \frac{1000}{1055} = 0.948 Btu

5 0
4 years ago
Read 2 more answers
A single crystal of a metal that has the BCC crystal structure is oriented such that a tensile stress is applied in the [100] di
VARVARA [1.3K]

Answer:

For [1 1 0] and  [1 0 1] plane, σₓ = 6.05 MPa

For [0 1 1] plane, σ = 0; slip will not occur

Explanation:

compute the resolved shear stress in [111] direction on each of the [110], [011] and on the [101] plane.

Given;

Stress direction: [1 0 0] ⇒ A

Slip direction: [1 1 1]

Normal to slip direction: [1 1 1] ⇒ B

∅ is the angle between A & B

Step 1: cos∅ = A·B/|A| |B| = \frac{[100][111]}{\sqrt{1}.\sqrt{3}  } ⇒ cos∅ = 1/\sqrt{3}

σₓ = τ/cos ∅·cosλ

where τ is the critical resolved shear stress given as 2.47MPa

Step 2: Solve for the slip along each plane

(a) [1 1 0]

cosλ = [1 1 0]·[1 0 0]/(\sqrt{2}·\sqrt{1})        

note: cosλ = slip D·stress D/|slip D||stress D|

cosλ = 1/\sqrt{2}

∵ σₓ = τ/\frac{1}{\sqrt{2} } ·\frac{1}{\sqrt{3} } = \sqrt{6} * 2.47MPa = 6.05MPa

Hence, stress necessary to cause slip on [1 1 0] is 6.05MPa

(b) [0 1 1]

cosλ = [0 1 1]·[1 0 0]/(\sqrt{2}·\sqrt{1}) = 0

∵ σₓ = 2.47MPa/0, which is not defined

Hence, for stress along [1 0 0], slip will not occur along [0 1 1]

(c) [1 0 1]

cosλ = [0 1 1]·[1 0 0]/(\sqrt{2}·\sqrt{1})

cosλ = 1/\sqrt{2}

∵ σₓ = τ/\frac{1}{\sqrt{2} } ·\frac{1}{\sqrt{3} } = \sqrt{6} * 2.47MPa = 6.05MPa

See attachment for the space diagram

3 0
3 years ago
Select the correct answer. Felix aspires to be an engineer working for the government. What credentials will Felix require to ap
Aneli [31]

Answer:

OB

Explanation:

5 0
3 years ago
It is safe to keep a cylinder next to the actual welding or cutting operation, true or false?
Alinara [238K]

Answer:

Explanation:

Cylinders shall be kept far enough away from the actual welding or cutting operation so that sparks, hot s lag, or flame will not reach them. When this is impractical, fire resistant shields shall be provided. Cylinders shall be placed where they cannot become part of an electrical circuit.

7 0
3 years ago
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