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Sati [7]
3 years ago
5

Line of Action of the Axial Forces for a Uniform Stress Distribution

Engineering
1 answer:
nikklg [1K]3 years ago
4 0

Answer:

Line of action of axial force for a uniform stress distribution should pass through the centroid of the cross sectional area.

Explanation:

If the line of action of the force is along the centroidal axis of the cross sectional area there is no eccentricity in the line of application of force hence no moment is generated in the cross sectional area hence we get a uniform stress distribution as theorized by hookes law.

In case of eccentric force there is an additional moment in addition of the force. This induced moment induces bending in the section thus giving a non uniform stress distribution in the section.

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Explain the conductivity results observed for ionic compounds in the solid state and in aqueous solution.i.Solid ionic compounds
Svetach [21]

Answer:

Aqueous solution of ionic compounds conduct electricity while solid ionic compounds don't.

Explanation:

Ionic compound conduct electricity when liquid or in aqueous solution that is resolved in water because the ionic bonds of the compound become weak and the ions are free to move from place to place.

Ionic compounds don't conduct electricity while in solid state because the ionic bonds are to strong and ions cannot move around with lack of space for movement which makes the electric conductivity zero.

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Whats viruses c liver?
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Hepatitis C is a viral infection that causes liver inflammation, sometimes leading to serious liver damage. The hepatitis C virus (HCV) spreads through contaminated blood.

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2 years ago
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List and explain 4 factors you need to observe while stick welding to make a good “consistent” bead
ch4aika [34]

Answer:I don’t know this one

Explanation:

5 0
3 years ago
Sandwich materials typically use a high density core with non-structural cover plates. a)True b)- False
Lorico [155]

Answer: False

Explanation: Sandwich materials are usually in composite material form which has a fabrication of two thin layers which are stiff in nature and have  light weighing and thick core .The construction is based on the ratio that is of stiffness to the weight .Therefore, the density of the material in the core is not high and are only connected with the skin layer through adhesive .So the given statement is false that sandwich materials typically use a high density core with non- structural cover plates.

4 0
3 years ago
A stainless steel (AISI 304) tube used to transport a chilled pharmaceutical has an inner diameter of 36 mm and a wall thickness
Bingel [31]

Answer:

a) heat gain per unit tube length = \frac{23-6}{1.35} = 12.6W/m

b) heat gain per unit tube length = \frac{23-6}{2.20} = 7.7W/m

Explanation:

Assumptions:

  1. Constant properties
  2. Steady state conditions
  3. Negligible effect of radiation
  4. Negligible constant resistance between tube and insulation
  5. one dimensional radial conduction

a) What is the heat gain per unit tube length

R_{conv,i}'=\frac{1}{2\pi r_1h_i}

d_1=36mm Therefore r_1=\frac{d_1}{2} =36/2=18mm=18*10^{-3}

r_2=2mm=2*10^{-3}m

k_{st}=14.2W/m.k

h_o=6W/m^2

h_i=400W/m^2

R_{conv,i}'=\frac{1}{2\pi * 1.8*10^{-3}*400}= 0.221m.K/W

R_{cond,st}'=\frac{ln(r_2/r_1)}{2\pi k_{st}} =\frac{ln(20/18)}{2\pi *14.2} =1.18*10^{-3}m.K/W

R_{conv,o}'=\frac{1}{2\pi r_2h_0}=\frac{1}{2\pi *2*10^{-3}*6}=1.33m.K/W

R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.33=1.35m.K/W

heat gain per unit tube length = \frac{23-6}{1.35} = 12.6W/m

b) What is the heat gain per unit length if a 10-mm-thick layer of calcium silicate insulation (k_ins = 0.050 W/m.K) is applied to the tube

r_3=r_1+r_2+10mm=30mm=0.03m

R_{conv,i}' and R_{cond,st}' are the same, but R_{conv,o}' changes.

Therefore:

R_{conv,o}'=\frac{1}{2\pi r_3h_0} = \frac{1}{2\pi *0.03*6}=0.88m.K/W

R_{conv,ins}'=\frac{ln(r_3/r_)}{2\pi k_{ins}} =\frac{ln(30/20)}{2\pi *0.05} =1.29m.K/W

The total resistance R_{tot}'=R_{conv,i}'+R_{cond,st}'+R_{conv,ins}'+R_{conv,o}'=0.221+(1.18*10^{-3})+1.29+0.88=2.20m.K/W

heat gain per unit tube length = \frac{23-6}{2.20} = 7.7W/m

8 0
3 years ago
Read 2 more answers
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