Thursday, 19 January 2017

TOTAL ELONGATION OF THE BAR DUE TO ITS OWN WEIGHT


Now we are going further to start our discussion to understand the concept of “Elongation of bar due to its self weight, with the help of this post.

Let us see here the of elongation of bar due to its self weight

As we have already discussed that when an external force will be applied on a body, there will be deformation or changes in the size and shape of the body. But molecular forces which will be acting between the molecules of the body will resist the deformation in the body due to the external force and this is termed as internal resistance developed by the body in order to resist the deformation in the body due to the external force.

This internal resistance force per unit area will be termed as stress.

Same concept will also be applied here when we will analyze stress and strain for a bar of uniform section due to its self weight.

If a bar of uniform cross-section will be fixed at one end and hanging freely under its self weight, there will also be induced stress and strain due to self weight of the bar.

Let us consider one bar AB as shown in following figure. Let us assume that bar is fixed at one end and hanging freely under its self weight. Let us assume that length of the bar is L, cross sectional area of the bar is A, young’s modulus of elasticity of the bar is E and specific weight of the bar i.e. weight per unit volume is ω.
Let us consider one small strip of thickness dy and at a distance y from the lower end as displayed in above figure.

Now we will consider the weight of the bar for a length of y in order to understand the load acting over the small strip of thickness dy towards downward direction.

Therefore weight of the bar for a length of y = ω x A x y

As weight of the bar for a length of y is acting on the small strip of thickness dy towards downward direction, there will be some stress and some changes in length of the bar due to self weight of the bar.

Therefore let us see, stress induced in the bar due to the load acting over the small strip of thickness dy towards downward direction.

Stress, σy = (ω x A x y)/A
Stress, σy = ω x y

And from here we can easily conclude that stress induced in the bar due to the self weight of the bar will be directionally proportional to the y.

Now we will determine the strain and elongation in small strip

Strain = Stress/E
Strain = ω x y /E

Elongation in small strip = Strain x length of small strip
Elongation in small strip = (ω x y /E) x dy

Elongation of the bar due to its self weight will be determined by integrating the above equation from 0 to L.

Elongation of the bar, δL will be determined as displayed here
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Tuesday, 17 January 2017

THERMAL STRESS IN COMPOSITE BARS


Now we are going further to start our discussion to understand the concept of “Thermal stress in composite bar”, in subject of strength of material, with the help of this post. We have already discussed the basics of thermal stress and strain and we will use that concept here too.

Let us see here the concept of thermal stress in composite bar

First we will understand here, what is a composite bar?

Composite bar is basically defined as the bar made by two or more than two bars of similar length but different materials and rigidly fixed with each other in such a way that it behaves as one unit and strain together against external load i.e. it behaves as single unit for compression and extension against compressive and tensile load.

We can say here from the definition of composite bar that strains will be same for each bar of composite bar and hence actual change in length will be similar for each bar or we can say that actual strain will be same for each bar of composite bar and we will use this concept during thermal stress analysis for composite bars.

Let us consider that we have one composite bar consisting two bars of different materials i.e. one bar of brass and other bar of steel and it is displayed here in following figure. Let us assume that we are now going to heat the composite bars up to some temperature, so what will be expected to occur?
As we know that if we are going to heat any material, there will be increase in temperature of the material and hence there will be increase in dimensions of the material. Similarly if we are going to cool the material, there will be decrease in temperature of the material and hence there will be decrease in dimensions of the material.

We can simply say that there will be free expansion or free contraction in the material according to the rising or lowering of temperature of the material. If free expansion or free contraction of the material due to change in temperature is restricted partially or completely, there will be stress induced in the material and this stress will be termed as thermal stress.

We must note it here that if free expansion or free contraction of the material due to change in temperature is not restricted i.e. expansion or contraction of the material is allowed, there will no stress developed in the material.

Now suppose if brass and steel both are not in composite in nature and we want figuring out the change in dimensions separately for steel and brass due to change in temperature, we will be able to say easily that change in length of the brass bar will be more than the change in length of the steel bar for similar rise in temperature of the bar and this is basically due to different co-efficient of linear expansion i.e. α for brass and steel. 

Co-efficient of linear expansion i.e. α for brass will be more than as compared to steel and therefore change in length of the brass bar will be more than the change in length of the steel bar for similar rise in temperature of the bar.

But in this situation, brass and steel are in composite state and therefore both members of composite bars will not be able to expand freely. Hence the expansion of the composite bar, as a whole, will be less than that for brass but more than that for steel.

We must have to note it here that brass will be subjected with compressive load or compressive stress because steel will restrict the brass to expand up to the limit up to which brass could be expanded.

And similarly, steel will be subjected with tensile load or tensile stress because brass will force the steel to expand beyond the limit up to which steel could be expanded. In simple words, we can say that both members of composite bars will be under stress but one will be in tensile stress and other will be in compressive stress i.e. steel will be under tensile stress and brass will be under compressive stress.

Do you have any suggestions? Please write in comment box

Reference:

Strength of material, By R. K. Bansal
Image Courtesy: Google

We will see another important topic, in the category of strength of material, in our next post.

Also read

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Monday, 16 January 2017

WHAT IS THERMAL STRESS AND STRAIN?


Now we are going further to start our discussion to understand the concept of “Thermal stress and strain”, in subject of strength of material, with the help of this post.

Let us see here the concept of thermal stress and strain

First we will understand here the concept of thermal stress and after that we will see here, in this post, thermal strain.

As we know that if we are going to heat any material, there will be increase in temperature of the material and hence there will be increase in dimensions of the material. Similarly if we are going to cool the material, there will be decrease in temperature of the material and hence there will be decrease in dimensions of the material.

Let us consider that we are going to heat or cool any material, as we have discussed, there will be changes in temperature of the material and therefore there will be changes in dimensions of the material.

We can simply say that there will be free expansion or free contraction in the material according to the rising or lowering of temperature of the material. If free expansion or free contraction of the material due to change in temperature is restricted partially or completely, there will be stress induced in the material and this stress will be termed as thermal stress.

We must note it here that if free expansion or free contraction of the material due to change in temperature is not restricted i.e. expansion or contraction of the material is allowed, there will no stress developed in the material.

So what we have concluded for thermal stress and thermal strain?

Thermal stress will be basically defined as the stress developed in the material due to change in temperature and free expansion or free contraction of the material, due to change in temperature, will be restricted.

Respective strain developed in the material will be termed as the thermal strain.

Derivation for thermal stress and thermal strain formula

Let us consider one metal bar AB as shown in following figure. Let us assume that initial temperature of the metal bar is T1 and we are heating the metal bar, as shown in figure, at one end to achieve the final temperature of the metal bar T2. Let us assume that initial length of the metal bar is L.
As we have discussed above that if we are going to heat the material, there will be increase in temperature of the material and therefore there will be increase in dimensions of the material too.

Increase in temperature of the metal bar, ΔT = T2-T1
Let us think that free expansion of the material, due to rise in temperature, is not restricted i.e. free expansion of the material is allowed and it is displayed in figure as BC.

Let us assume that free expansion, as a result of increase in temperature of the metal bar due to heating, is δL. Free expansion will be given by the following formula as displayed here.

δL = α. L. ΔT
Thermal strain will be determined, as written here, with the help of following result
Thermal strain, Ԑ = δL/L

Thermal strain, Ԑ = α. ΔT

Let us think that if one load P will be exerted at B as shown in figure and hence as we have discussed above, free expansion or free contraction of the material due to change in temperature will be restricted and hence there will be stress induced in the material i.e. thermal stress will be developed in the material.

Thermal stress, σ = thermal strain x Young’s modulus of elasticity

Thermal stress, σ = α. ΔT. E

We must have to note it here that thermal stress is also known as temperature stress and similarly thermal strain is also termed as temperature strain.
Do you have any suggestions? Please write in comment box

Reference:

Strength of material, By R. K. Bansal
Image Courtesy: Google

We will see another important topic i.e. Thermal stresses in composite bars, in the category of strength of material, in our next post.

Also read


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Sunday, 15 January 2017

STRESS AND ELONGATION OF UNIFORMLY TAPERING RECTANGULAR ROD

We were discussing the concept of stress and strain and also we have discussed the different types of stress and also different types of strain as well as concept of Poisson ratio in our previous posts. We have also seen the concept of Hook’s Law , types of modulus of elasticity in our recent post.

Now we are going further to start our discussion to understand “Elongation of uniformly tapering rectangular rod”, in subject of strength of material, with the help of this post.

Let us see here the elongation of uniformly tapering rectangular rod

First we will understand here, what is a uniformly tapering rectangular rod?

A rectangular rod is basically taper uniformly from one end to another end throughout the length and therefore its one end will be of larger width and other end will be of smaller width. However thickness of the rectangular rod will be constant throughout the length of rod.

Let us consider the uniformly tapering rectangular rod as shown in figure, length of the uniformly tapering rectangular rod is L. Width of larger end of the rod is a and as we have discussed that rectangular rod will be uniformly tapered and hence other end width of the rectangular rod will be smaller and let us assume that width of other end is b. Let us assume that thickness of the rectangular rod is t.

Let us consider that uniformly tapering rectangular rod is subjected with an axial tensile load P and it is displayed in following figure.
Let us consider one infinitesimal smaller element of length dx and its width will be at a distance x from its larger diameter end as displayed in above figure.

Let us consider that width of infinitesimal smaller element is Cx
Cx = a-[(a-b)/L] X
Cx = a- KX
Where we have assumed that K= (a-b)/L

Let us consider that area of cross section of rectangular bar at a distance x from its larger diameter end is Ax and we will determine area as mentioned here.
Ax = Width x Thickness
Ax = (a- KX) t

Stress

Let us consider that stress induced in rectangular bar at a distance x from its bigger width end is σx and we will determine stress as mentioned here.
σx = P/ Ax
σx = P/ [(a- KX) t]

Strain

Let us consider that strain induced in rectangular bar at a distance x from its bigger width end is Ԑx and we will determine strain as mentioned here.
Strain = Stress / Young’s modulus of elasticity
Ԑx = P/ [(a- KX) t. E]

Change in length of infinitesimal smaller element

Change in length of infinitesimal smaller element will be determined by recalling the concept of strain.

Δ dx = Ԑx. dx
Where, Ԑx = P/ [E (a- KX) t]

Now we will determine the total change in length of the uniformly tapering rectangular rod by integrating the above equation from 0 to L.

And we can say that elongation of uniformly tapering rectangular rod will be calculated with the help of following result.
Do you have any suggestions? Please write in comment box

Reference:

Strength of material, By R. K. Bansal
Image Courtesy: Google

We will see another important topic i.e. Elongation of uniformly tapering circular rod in the category of strength of material.

Also read


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Saturday, 14 January 2017

ELONGATION OF UNIFORMLY TAPERING CIRCULAR ROD

We were discussing the concept of stress and strain and also we have discussed the different types of stress and also different types of strain as well as concept of Poisson ratio in our previous posts. We have also seen the concept of Hook’s Law , types of modulus of elasticity in our recent post.

Now we are going further to start our discussion to understand “Elongation of uniformly tapering circular rod”, in subject of strength of material, with the help of this post.

Let us see here the elongation of uniformly tapering circular rod

First we will understand here, what is a uniformly tapering circular rod?
A circular rod is basically taper uniformly from one end to another end throughout the length and therefore its one end will be of larger diameter and other end will be of smaller diameter.

Let us consider the uniformly tapering circular rod as shown in figure, length of the uniformly tapering circular rod is L and larger diameter of the rod is D1 at one end and as we have discussed that circular rod will be uniformly tapered and hence other end diameter of the circular rod will be smaller and let us assume that diameter of other end is D2.
Let us consider that uniformly tapering circular rod is subjected with an axial tensile load P and it is displayed in above figure.

Let us consider one infinitesimal smaller element of length dx and its diameter will be at a distance x from its larger diameter end as displayed in above figure.

Let us consider that diameter of infinitesimal smaller element is Dx
Dx = D1-[(D1-D2)/L] X
Dx = D1- KX
Where we have assumed that K= (D1-D2)/L

Let us consider that area of cross section of circular bar at a distance x from its larger diameter end is Ax and we will determine area as mentioned here.
Ax = (П/4) Dx2
Ax = (П/4) (D1- KX) 2

Stress

Let us consider that stress induced in circular bar at a distance x from its larger diameter end is σx and we will determine stress as mentioned here.
σx = P/ Ax
σx = P/ [(П/4) (D1- KX) 2]
σx = 4P/ [П (D1- KX) 2]

Strain

Let us consider that strain induced in circular bar at a distance x from its larger diameter end is Ԑx and we will determine strain as mentioned here.

Strain = Stress / Young’s modulus of elasticity
Ԑx = σx /E
Ԑx = 4P/ [П E (D1- KX) 2]

Change in length of infinitesimal smaller element

Change in length of infinitesimal smaller element will be determined by recalling the concept of strain.

Δ dx = Ԑx. dx
Where, Ԑx = 4P/ [П E (D1- KX) 2]

Now we will determine the total change in length of the uniformly tapering circular rod by integrating the above equation from 0 to L.
And we can say that elongation of uniformly tapering circular rod will be calculated with the help of following result.

Do you have any suggestions? Please write in comment box

Reference:

Strength of material, By R. K. Bansal
Image Courtesy: Google
We will see another important topic i.e. Stress analysis of bars of varying sections in the category of strength of material.

Also read

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Friday, 13 January 2017

STRESS ANALYSIS OF BARS OF VARYING SECTIONS

We were discussing the concept of stress and strain and also we have discussed the different types of stress and also different types of strain as well as concept of Poisson ratio in our previous posts. We have also seen the concept of Hook’s Law , types of modulus of elasticity in our recent post.

Now we are going further to start our discussion to understand “Stress analysis of bars of varying sections”, in subject of strength of material, with the help of this post.

Let us see here the stress analysis of bars of varying sections

Let us see the following figure, where we can see a bar of having different length and different cross-sectional area and bar is subjected with an axial load P. As we can see here that length and cross-sectional areas of each section of bar is different and therefore stress induced, strain and change in length too will be different for each section of bar.
Young’s modulus of elasticity of each section might be same or different depending on the material of the each section of bar.

Axial load for each section will be same i.e. P. When we will go to determine the total change in length of the bar of varying sections, we will have to add change in length of each section of bar.

P = Bar is subjected here with an axial Load
A1, A2 and A3 = Area of cross section of section 1, section 2 and section 3 respectively
L1, L2 and L3 = Length of section 1, section 2 and section 3 respectively
σ 1, σ 2 and σ 3 = Stress induced for the section 1, section 2 and section 3 respectively
ε1, ε 2 and ε 3 = Strain developed for the section 1, section 2 and section 3 respectively
E= Young’s Modulus of the bar

Let us see here stress and strain produced for the section 1

Stress, σ1= P / A1
Strain, ε1 = σ1/E
Strain, ε1 = P / A1E

Similarly stress and strain produced for the section 2

Stress, σ2= P / A2
Strain, ε2 = σ2/E
Strain, ε2 = P / A2E

Similarly stress and strain produced for the section 3

Stress, σ3= P / A3
Strain, ε3 = σ3/E
Strain, ε3 = P / A3E
Now we will determine change in length for each section with the help of definition of strain
ΔL1= ε1L1
ΔL2= ε2L2
ΔL3= ε3L3

Total change in length of the bar

As we have already discussed above that when we will go to determine the total change in length of the bar of varying sections, we will have to add change in length of each section of the bar.
ΔL= ΔL1+ ΔL2+ΔL3
Let us consider that Young’s modulus of elasticity of each section is different, in this situation we will have following equation to determine the total change in length of the bar.
Do you have any suggestions? Please write in comment box

Reference:

Strength of material, By R. K. Bansal
Image Courtesy: Google

We will see another important topic i.e. stress analysis of bars of composite sections in our next post.

Also read

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Thursday, 12 January 2017

STRESS ANALYSIS OF BARS OF COMPOSITE SECTIONS

We were discussing the concept of stress and strain and also we have discussed the different types of stress and also different types of strain as well as concept of Poisson ratio in our previous posts. We have also seen the concept of Hook’s Law and types of modulus of elasticity in our recent post.

Now we are going further to start our discussion to understand the “Stress analysis of bars of composite sections”, in subject of strength of material, with the help of this post.

Let us see here the stress analysis of bars of composite sections

First we will understand here, what is a composite bar?

Composite bar is basically defined as a bar made by two or more than two bars of similar length but different materials and rigidly fixed with each other in such a way that it behaves as one unit and strain together against external load i.e. it behaves as single unit for compression and extension against compressive and tensile load.

We can say here from the definition of composite bar that strains will be same for each bar of composite bar and hence stress produced in each bar of composite bar will be dependent over the young modulus of elasticity of respective material.

So let us see here the important points in respect of composite bar, those are very important to keep in mind during stress analysis of bars of composite sections.

As we have seen above that composite bar will have two or more than two bars of similar length and these bars will be rigidly fixed with each other and therefore change in length will be similar for each bar or we can say that strains will be same for each bar of composite bar.

Total external load which will be acting over the composite bar will be shared by the each bar of composite bar and hence we can say that total external load on composite bar will be equal to the addition of the load shared by each bar of composite bar.

Let us see here the following figure which represents two bar of similar length but different materials and rigidly fixed with each other to form one composite bar.

A1 and A2 = Area of cross section of bar 1 and bar 2 respectively
E1 and E2 = Young’s modulus of elasticity for material of bar 1 and material of bar 2 respectively
P1 and P2 = Load shared by bar 1 and bar 2 respectively
σ1 and σ2 = Stress induced in bar 1 and bar 2 respectively

As we have already discussed that total external load which will be acting over the composite bar will be shared by the each bar of composite bar and therefore we will have following equation.

P = P1 + P2
Stress induced in bar 1, σ1= P1 / A1
Stress induced in bar 1, σ2= P2 / A2
P = σ1A1 + σ2 A2

We have also discussed above that composite bar will have two or more than two bars of similar length and these bars will be rigidly fixed with each other and therefore change in length will be similar for each bar or we can say that strains will be same for each bar of composite bar.

Strain in bar 1, Ԑ1= σ1/ E1
Strain in bar 2, Ԑ2 = σ2/ E2

From above statement that strains will be same for each bar of composite bar, we will have following equation.

σ1/ E1 = σ2/ E2

Conclusion

We have secured here two very important equations as mentioned below

σ1/ E1 = σ2/ E2
P = σ1A1 + σ2 A2

These equations will be required for determining the value of stresses in case of bars of composite sections.

Do you have any suggestions? Please write in comment box

Reference:

Strength of material, By R. K. Bansal
Image Courtesy: Google

We will see another important topic i.e. Thermal stress in strength of material in our next post.

Also read

Continue Reading

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