Saturday, 23 September 2017

WHAT IS POLAR MODULUS OF SECTION?

We were discussing the concept of section modulus of beams, derivation of mass moment of inertia and also the derivation of area moment of inertia in our previous posts. 

Now we are going further to start a new topic i.e. Polar modulus with the help of this post. We will also see here the polar modulus for various sections in this post.

So, what is polar modulus?

 
Before going ahead, we must have to understand here first the basics of polar moment of inertia and after that it will be easy to understand the concept of Polar modulus.

Polar moment of inertia of a plane area is basically defined as the area moment of inertia about an axis perpendicular to the plane of figure and passing through the center of gravity of the area.
Polar moment of inertia will be displayed by J.

Polar modulus

Polar modulus is basically defined as the ratio of the polar moment of inertia (J) to the radius of the shaft. Polar modulus will be displayed by ZP.

Polar modulus is also known as torsional section modulus. We can write here the expression for polar modulus as displayed here.
Polar Modulus = Polar moment of inertia/Radius of the shaft
ZP = J/R

Polar modulus for various sections

 
Following figure, displayed here, indicates the polar modulus for various sections

Do you have suggestions? Please write in comment box.

 
We will now derive the torsional equation, in the category of strength of material, in our next post.

Reference:

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

Also read

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Thursday, 21 September 2017

DERIVATION OF TORSIONAL EQUATION


Now we are going further to start a new topic i.e. Derivation of torsional equation with the help of this post.

Before going ahead, let us recall the basic definition of twisting moment or torsion.

 
A shaft will said to be in torsion, if it will be subjected with two equal and opposite torques applied at its two ends.

When a shaft will be subjected to torsion or twisting moment, there will be developed shear stress and shear strain in the shaft material.

We will consider here one case of circular shaft which will be subjected to torsion and we will derive here the torsion equation for circular shaft.

We have following information from above figure

R = Radius of the circular shaft
D = Diameter of the circular shaft
dr = Thickness of small elementary circular ring
r = Radius of the small elementary of circular ring
q = Shear stress at a radius r from the centre of the circular shaft
τ = Shear stress at outer surface of shaft
dA = Area of the small elementary of circular ring
dA = 2П x r x dr

Shear stress, at a radius r from the centre, could be determined as mentioned here
q/r = τ /R
q = τ x r/R

Turning force due to shear stress at a radius r from the centre could be determined as mentioned here
dF = q x dA
dF = τ x r/R x 2П x r x dr
dF = τ/R x 2П r2dr

Twisting moment at the circular elementary ring could be determined as mentioned here
dT = Turning force x r
dT = τ/R x 2П r3dr
dT = τ/R x r2 x (2П x r x dr)
dT = τ/R x r2 x dA

 
Total torque could be easily determined by integrating the above equation between limits 0 and R
Therefore total torque transmitted by a circular solid shaft could be given in following way as displayed here in following figure.
Let us recall here the basic concept of Polar moment of inertia and we can write here the formula for polar moment inertia. Further, we will use this formula of polar moment of inertia in above equation.
Polar moment of inertia
Therefore total torque transmitted by a circular solid shaft could be given by following equation as mentioned here.
We have already derived the expression for shear stress produced in a circular shaft subjected to torsion and therefore we have following result from that expression
Considering above two equations, we can write here the expression for torsion equation for circular shaft as displayed here.
Where, 
C = Modulus of rigidity
L = Length of shaft
θ = Angle of twist (Radian)

 
Do you have suggestions? Please write in comment box.

We will now discuss another topic i.e. Power transmitted by a circular solid shaft, in the category of strength of material, in our next post.

Reference:

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

Also read

Continue Reading

Wednesday, 20 September 2017

POLAR MOMENT OF INERTIA FOR VARIOUS SECTIONS

Now we are going further to start a new topic i.e. Polar moment of inertia with the help of this post. we will also see here the polar moment of inertia for various sections in this post.

 

So, what is polar moment of inertia?

Polar moment of inertia of a plane area is basically defined as the area moment of inertia about an axis perpendicular to the plane of figure and passing through the center of gravity of the area.
Polar moment of inertia will be displayed by J.

Mathematically, we can write polar moment of inertia i.e. J as mentioned here.
We can also say from above equation of polar moment of inertia that, Polar moment of inertia of an element will be basically the resultant of the product of element area and square of its distance from the axis.

Polar moment of inertia is basically a quantity which is used to specify the body resistance against twisting and it also suggests the strength of body against torsion loading.

 
Polar moment of inertia is quite similar to area moment of inertia. We must have to note it here that both are used in design analysis but area moment of inertia will be under consideration when structure will be subjected with bending or deflection, while polar moment of inertia will be under consideration when structure will be subjected with torsional loading.

Polar moment of inertia for various sections

Do you have suggestions? Please write in comment box.

 
We will now derive the torsional equation, in the category of strength of material, in our next post.

Reference:

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Sunday, 17 September 2017

POWER TRANSMITTED BY A CIRCULAR HOLLOW SHAFT


Now we are going further to start a new topic i.e. Power transmitted by a circular hollow shaft with the help of this post. We recommend to first review the recent post “Torsion or twisting moment” for having basic information about the concept of torsion or twisting moment.

So, what is torsion or twisting moment?

 
A shaft will said to be in torsion, if it will be subjected with two equal and opposite torques applied at its two ends.

When a shaft will be subjected to torsion or twisting moment, there will be developed shear stress and shear strain in the shaft material.

We will discuss here one case of a hollow circular shaft which will be subjected to torsion and we will secure here the expression for maximum torque transmitted by a hollow circular shaft.
We have following information from above figure
Ro = Outer radius of the hollow circular shaft
Ri = Inner radius of the hollow circular shaft
Do = Outer diameter of the hollow circular shaft
Di = Inner diameter of the hollow circular shaft
dr = Thickness of small elementary circular ring
r = Radius of the small elementary of circular ring
q = Shear stress at a radius r from the centre of the hollow circular shaft
τ = Maximum shear stress at outer surface of shaft
dA = Area of the small elementary of circular ring
dA = 2П x r x dr

Shear stress, at a radius r from the centre, could be determined as mentioned here
q/r = τ / Ro
q = τ x r/ Ro

Turning force due to shear stress at a radius r from the centre could be determined as mentioned here
dF = q x dA
dF = τ x r/ Ro x 2П x r x dr
dF = τ/ Ro x 2П r2dr

 
Twisting moment at the circular elementary ring could be determined as mentioned here
dT = Turning force x r
dT = τ/ Ro x 2П r3dr

Total torque could be easily determined by integrating the above equation between limits Ri and Ro.
Therefore total torque transmitted by a hollow circular shaft will be given by following formula
Let us consider that N is the R.P.M of the shaft and ω is the angular velocity of the shaft. We have already secured here the torque transmitted by circular solid shaft and it is mentioned above.  
Now we will find here the expression for power transmitted by solid circular shaft as mentioned here.
Do you have suggestions? Please write in comment box.

We will now discuss the Torque in terms of polar moment of inertia, in the category of strength of material, in our next post.

Reference:

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

Also read

Continue Reading

Saturday, 16 September 2017

POWER TRANSMITTED BY SOLID CIRCULAR SHAFT


Now we are going further to start a new topic i.e. Power transmitted by a circular solid shaft with the help of this post. We recommend to first review the recent post “Torsion or twisting moment” for having basic information about the concept of torsion or twisting moment.

So, what is torsion or twisting moment?

 
A shaft will said to be in torsion, if it will be subjected with two equal and opposite torques applied at its two ends.

When a shaft will be subjected to torsion or twisting moment, there will be developed shear stress and shear strain in the shaft material.

We will discuss here one case of circular shaft which will be subjected to torsion and we will secure here the expression for power transmitted by a circular solid shaft.
We have following information from above figure

R = Radius of the circular shaft
D = Diameter of the circular shaft
dr = Thickness of small elementary circular ring
r = Radius of the small elementary of circular ring
q = Shear stress at a radius r from the centre of the circular shaft
τ = Shear stress at outer surface of shaft
dA = Area of the small elementary of circular ring
dA = 2П x r x dr

Shear stress, at a radius r from the centre, could be determined as mentioned here
q/r = τ /R
q = τ x r/R

Turning force due to shear stress at a radius r from the centre could be determined as mentioned here
dF = q x dA
dF = τ x r/R x 2П x r x dr
dF = τ/R x 2П r2dr

 
Twisting moment at the circular elementary ring could be determined as mentioned here
dT = Turning force x r
dT = τ/R x 2П r3dr

Total torque could be easily determined by integrating the above equation between limits 0 and R
Therefore total torque transmitted by a circular solid shaft

T = (П/16) τ D3

Let us consider that N is the R.P.M of the shaft and ω is the angular velocity of the shaft. We have already secured here the torque transmitted by circular solid shaft and it is mentioned above. Now we will find here the expression for power transmitted by solid circular shaft as mentioned here.

Do you have suggestions? Please write in comment box.

 
We will now discuss the topic, Power transmitted by a hollow circular shaft, in the category of strength of material, in our next post.

Reference:

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

Also read

Continue Reading

TORQUE TRANSMITTED BY A HOLLOW CIRCULAR SHAFT


Now we are going further to start a new topic i.e. Torque transmitted by a circular hollow shaft with the help of this post.

So, what is torsion or twisting moment?

 
A shaft will said to be in torsion, if it will be subjected with two equal and opposite torques applied at its two ends.

When a shaft will be subjected to torsion or twisting moment, there will be developed shear stress and shear strain in the shaft material.
We will discuss here one case of a hollow circular shaft which will be subjected to torsion and we will secure here the expression for maximum torque transmitted by a hollow circular shaft.

We have following information from above figure

Ro = Outer radius of the hollow circular shaft
Ri = Inner radius of the hollow circular shaft
Do = Outer diameter of the hollow circular shaft
Di = Inner diameter of the hollow circular shaft
dr = Thickness of small elementary circular ring
r = Radius of the small elementary of circular ring
q = Shear stress at a radius r from the centre of the hollow circular shaft
τ = Maximum shear stress at outer surface of shaft
dA = Area of the small elementary of circular ring
dA = 2П x r x dr

Shear stress, at a radius r from the centre, could be determined as mentioned here
q/r = τ / Ro
q = τ x r/ Ro

 
Turning force due to shear stress at a radius r from the centre could be determined as mentioned here
dF = q x dA
dF = τ x r/ Ro x 2П x r x dr
dF = τ/ Ro x 2П r2dr

Twisting moment at the circular elementary ring could be determined as mentioned here
dT = Turning force x r
dT = τ/ Ro x 2П r3dr

Total torque could be easily determined by integrating the above equation between limits Ri and Ro.
Therefore total torque transmitted by a hollow circular shaft will be given by following formula
Do you have suggestions? Please write in comment box.

 
We will now discuss the Principal planes and principal stresses, in the category of strength of material, in our next post.

Reference:

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

Also read

Continue Reading

Tuesday, 12 September 2017

TORQUE TRANSMITTED BY A CIRCULAR SOLID SHAFT


Now we are going further to start a new topic i.e. Maximum torque transmitted by a circular solid shaft with the help of this post. We recommend to first review the recent post “Torsion or twisting moment” for having basic information about the concept of torsion or twisting moment.

 

So,what is torsion or twisting moment?

A shaft will said to be in torsion, if it will be subjected with two equal and opposite torques applied at its two ends.

When a shaft will be subjected to torsion or twisting moment, there will be developed shear stress and shear strain in the shaft material.

We will discuss here one case of circular shaft which will be subjected to torsion and we will secure here the expression for maximum torque transmitted by a circular solid shaft.
We have following information from above figure
R = Radius of the circular shaft
D = Diameter of the circular shaft
dr = Thickness of small elementary circular ring
r = Radius of the small elementary of circular ring
q = Shear stress at a radius r from the centre of the circular shaft
τ = Shear stress at outer surface of shaft
dA = Area of the small elementary of circular ring
dA = 2П x r x dr

 
Shear stress, at a radius r from the centre, could be determined as mentioned here
q/r = τ /R
q = τ x r/R

Turning force due to shear stress at a radius r from the centre could be determined as mentioned here
dF = q x dA
dF = τ x r/R x 2П x r x dr
dF = τ/R x 2П r2dr

Twisting moment at the circular elementary ring could be determined as mentioned here
dT = Turning force x r
dT = τ/R x 2П r3dr

Total torque could be easily determined by integrating the above equation between limits 0 and R
Therefore total torque transmitted by a circular solid shaft
T = (П/16) τ D3

Do you have suggestions? Please write in comment box.
 
We will now discuss the another topic, Torque transmitted by a hollow circular shaftin the category of strength of material, in our next post.

Reference:

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

Also read

Continue Reading

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