Monday, 23 February 2015


Bar in torsion
Consider a bar to be rigidly attached at one end and twisted at the other end by a torque or twisting moment T equivalent to F × d, which is applied perpendicular to the axis of the bar, as shown in the figure. Such a bar is said to be in torsion.

For solid or hollow shafts of uniform circular cross-section and constant wall thickness, the torsion relations are:
 \frac{T}{J} = \frac{\tau}{R} = \frac{G\varphi}{\ell}
  • R is the outer radius of the shaft.
  • τ is the maximum shear stress at the outer surface.
  • φ is the angle of twist in radians.
  • T is the torque (N·m ).
  • is the length of the object the torque is being applied to or over.
  • G (OR) C is the shear modulus or more commonly the modulus of rigidity and is usually given in gigapascals (GPa),or N/mm2
  • J is the polar moment of inertia for a round shaft or concentric tube only.
  • the product GJ is called the torsional rigidity.
The shear stress at a point within a shaft is:
 \tau_{\varphi_{z}} = {T r \over J}
  • r is the distance from the center of rotation
Note that the highest shear stress is at the point where the radius is maximum, the surface of the shaft. High stresses at the surface may be compounded by stress concentrations such as rough spots. Thus, shafts for use in high torsion are polished to a fine surface finish to reduce the maximum stress in the shaft and increase its service life.
The angle of twist can be found by using:
 \varphi_{} = {T \ell \over JG}.
Polar moment of inertia : J = Ixx + Iyy
The polar moment of inertia for a solid shaft is:
 J = {\pi \over 2} r^4
where r is the radius of the object.
The polar moment of inertia for a pipe is:
 J = {\pi \over 2} (r_{o}^4 - r_{i}^4)
where the o and i subscripts stand for the outer and inner radius of the pipe.
For a thin cylinder
J = 2π R3 t
where R is the average of the outer and inner radius and t is the wall thickness. 


A shaft rotating with a constant revolutions for minute (r.p.m) N is being acted by a twisting moment T. The power (P) transmitted by the shaft is
P= 2πNT/60 

where T is the torque in N·m, f is the number of revolutions per second, and P is the power in watts.


An aluminum shaft with a constant diameter of 50 mm is loaded by torques applied to gears attached to it as shown in Fig. Using G = 28 GPa, determine the relative angle of twist of gear D relative to gear A.


Three segments aluminum shaft

Monday, 9 February 2015

Mohr's Circle of stress

The transformation equations for plane stress can 
be represented in graphical form by a plot known 
as Mohr’s Circle.
To establish Mohr's Circle, we first recall the stress transformation formulas for plane stress at a given location,

sq = (sx+sy) / 2 + (sx-sy)/2cos2q +txysin2q
tq = (sx-sy)/2sin2q -txycos2q 
Using a basic trigonometric relation (cos22q + sin22q = 1) to combine the two above equations we have,

(sq- save)2  + t2q= R2
save = (sx +  sy)/2  

This is the equation of a circle, plotted on a graph where the abscissa is the normal stress and the ordinate is the shear stress.This circle is with radius of  R2m and centered at  C = (save,0) as shown in the Fig. below, 

  Mohr's circle for plane stress

  In the above Fig., one reads  that  the point

 X = (sx , - txy ) , 
and the point   
Y = (sy txy )

A = (s1 , 0 ) , B = (s2 , 0 )

The points A and B gives the principal stress (s1 s2) , To this end, one can pick the maxium normal stressess as
smax = max(ss2),  smin = min(s1 s2)

Besides, finally one can also read the maxium shear stress as
tmax = 
Construction of Mohr's circle of stress:
Given an initial stress element, or the stress values σx, σy, and τxy, Mohr’s circle can be constructed. 
The following sign conventions may be used:
1. Tensile stresses (positive) are to the right.
2. Compressive stresses (negative) are to the left.
3. Clockwise shear stresses are plotted upward.
4. Counterclockwise shear stresses are plotted downward.
In order to draw Mohr’s circle on a Cartesian coordinate system, the X-axis and Y-axis are identified as the σ-axis and τ-axis, respectively. 
The first two points of the circle to plot are point 1 (σx, τxy) and point 2 (σy, τxy). The line connecting these two points is the diameter of the circle, knowing this the circle can now be drawn.
The center of the circle is located where the diameter intersects the σ-axis, it is also the average normal stress (σavg). The average normal stress can be read from the coordinate system if it was drawn to scale or it can be calculated using the formula
\sigma_{avg}=\frac{(\sigma_x+ \sigma_y)}{2}  

The line from the center of the circle that passes through point 1 will represent the x-axis, this will be used later to relate the direction of the Mohr’s circle stresses to the normal x and y directions.

The two end points of the horizontal diameter are σ1and σ2
The point σ1 represents the maximum normal stress (σmax) and the point σ2 is the minimum normal stress (σmin).
The equations for finding these values are

\sigma_{max} &= \left(\frac{\sigma_x + \sigma_y}{2} \right )+\sqrt{(\frac{\sigma_x- \sigma_y}{2})^2+ {\tau_{xy}}^2}

\sigma_{min} &= \left(\frac{\sigma_x + \sigma_y}{2} \right )-\sqrt{(\frac{\sigma_x- \sigma_y}{2})^2+ {\tau_{xy}}^2}

The vertical diameter passes through σavg and goes up to positive τmax and down to negative τmin. The equation to find the value of τmax,minis;

\tau_{max,min}= \pm \frac{\sqrt{(\sigma_x- \sigma_y )^2+ {4\tau_{xy}}^2 }}{2}
The next value to determine is the angle that the normal stress is acting at. This is the angle between the positive σ-axis and the x-axis. The measure of the angle is found by:

                                            2\theta =  \arctan\left[\frac{2\tau_{xy}}{\sigma_x- \sigma_y }\right]

To find the angle that the maximum stress is acting at the following equation is used
                                            2\theta' =  \arctan\left[\frac{\sigma_x- \sigma_y }{2\tau_{xy}}\right]

It is important to pay attention to the use of these two equations as they look similar.
The last thing to do is to draw the initial stress element as if it was not given, the normal stress element at an angle θ, and the maximum shear stress element at an angle of θ'.


Problem no 1
For a given loading conditions the state of stress in the wall of a cylinder is subjected to normal stresses 85 MN/m2 tensile,in x- direction and  25 MN/m2 tensile in y- direction,along with  Shear stresses of 60 MN/m2 on the planes on which the stresses are act.  the sheer couple acting on planes carrying the 25 MN/m2 stress is clockwise.
a)Calculate the principal stresses and the planes on which they act. 
b)What would be the effect on these results if owing to a change of loading if the 85MN/m2 becomes compressive while other stresses are remain unchanged

a)The principle stresses are given by the formula
\sigma_{max} &= \left(\frac{\sigma_x + \sigma_y}{2} \right )+\sqrt{(\frac{\sigma_x- \sigma_y}{2})^2+ {\tau_{xy}}^2}

For finding out the planes on which the principle stresses act us the equation
The solution of this equation will yeild two values q i.e they q1 and q2 giving,
                                       q1= 31071' and q2= 121071'

(b) In this case only the loading (a) is changed i.e. its direction had been changed. While the other stresses remains unchanged hence now the block diagram becomes.
Again the principal stresses would be given by the equation.

q1= -23.74 and q2= 113.740


Principal Plane is the Plane in which  no Shear Stresses are acting. The normal stresses which are acting on Principal Plane are called as Principal Stresses.
Stresses on Oblique Plane :

The normal stresses (sx and sy) and the shear stress (txy) are acting vary smoothly on a body, the normal and tangential stresses or shear stress acting on a Oblique plane making a rotation of an angle q to the vertical face are given by,

 sq = (sx+sy) / 2 + (sx-sy)/2cos2q +txysin2q
tq = (s-sy)/2sin2q -txycos2q 
In above equations, there exist a couple of particular angles where the stresses take on special values.
First, there exists an angle qp where the shear stress txy becomes zero. That angle is found by setting txy to zero in the above shear transformation equation and solving for q (set equal to qp). The result is,
The angle qp defines the principal directions where the only stresses are normal stresses. These stresses are called principal stresses and are found from the original stresses (expressed in the x,y directions) via,

The transformation to the principal directions can be illustrated as:

Maximum Shear Stress Direction:
Another important angle, qs, is where the maximum shear stress occurs. This is found by finding the maximum of the shear stress transformation equation, and solving for q. The result is,
The maximum shear stress is equal to one-half the difference between the two principal stresses,

The transformation to the maximum shear stress direction can be illustrated as:

 Transformation of stresses in two dimensions,showing the planes of action of principal stresses, and maximum and minimum shear stresses.

Friday, 4 April 2014

Short Answer questions -S.A.II

        1.     What is Slope Deflection Method?(Oct/Nov 2010)
          The Slope Deflection  equations gives the moments at end of a member. These are expressed in terms of  the deformations (ie.slopes & displacements).
This method is used for the analysis statically indeterminate beams and rigid frames.
   2.     What are the assumptions used in the slope deflection method?(Oct/Nov 2010)
     1) The method of the structure is linearly elastic.
    2) The structure is loaded within the elastic limit of its material and the principle of superposition is also for the analysis.
   3) The axial displacements and shear displacements are negligible and they are neglected in the analysis.
     4) The flexural deformations (caused by B.M) are only considered.

    3. Explain  Stiffness Factor?(Oct/Nov 2010)
The term ”Stiffness” is defined as the ratio moment of inertia(I) to the length of the member(L).
 It is denoted by ‘K’.
    4. State the reasons for the side sway?
   In the rigid portal frames, the side sway occurs due to one of the following cases:-
   1) Eccentric (or) the unsymmetrical loading on the rigid frames.
   2) Unsymmetrical geometry of the frames.
   3) Different boundary conditions of the column.
   4) The cross-section of the members of the frame are not uniform.
   5) Application of horizontal load on the frame (wind, seismic forces etc).

   6) Differential settlement of supports of rigid frames.
   5. What is Moment Distribution method?
     Moment Distribution Method is exactly same as the slope deflection method except the difficulty of solving the simultaneous equations. This was proposed by Prof.Hardy Cross in the year 1930.
 6. What are Carry over moments?
    When the moment ‘M’(balancing moment)is distributed among the members in any  joint ‘A’, all the adjacent joints to A are prevented from rotation. The moment needed to prevent the rotations are called “Carry-over Moments”.

7. What is Carry over Factor(CF)?
   The ratio of the carry over moment to the distribution moment for a particular member is called ”Carry-over Factor”.
    For all prismatic members, the value of CF will be same and equal to ½.
8. What is meant by Distribution Factor?( Oct/Nov 2010)

  The fraction of total moment shared by each member at a joint is called as “Distribution Factor” which is equal to ratio of the stiffness of the member to total stiffness of all the members meeting  at a joint.
                                                 Dij=Kij / Kij
9. State Modified Stiffness Factor?
 The stiffness factor whose far end is hinged in a member will be (¾)*(I/L).
                                  MSF= (3/4)*K
10. What is a Cable?
 In the structural application, the term “Cable” means a flexible tension member.
 These are commonly used to support the suspension bridges as permanent guys for transmission towers, chimneys and building roofs. They carry only the tensile forces.

11. What is meant by Dip?
 The vertical distance from tower top to the lowest point on cable is called as the “Central Dip”. This is also called as “Sag”.  The central dip for the cable will be 1/10 to 1/12.    
12. What is Shape of a cable if it is supported to point loads ? (Oct/Nov 2010)
 Under the point loads, the cable will take the shape of “Funicular Polygon”.
13. What is Shape of a cable if it is supported to U.D.L along the span?
Under the uniformly distributed loads, the cable will take the shape of  "Parabola"
14. What is Shape of a cable due to its self weight?
Under the self weight ,the cable take the shape of "Catenary"
15. State eddy's theorem
The bending moment at any section of an arch is proportional to the vertical intercept between the Linear arch and center line of the Actual arch. 
16. What are the  Assumptions in  portal method ?
1. The points of inflection are located at the mid-height of each column above the first floor. If the base of the column is fixed, the point of inflection is assumed at mid height of the ground floor columns as well; otherwise it is assumed at the hinged column base.
2. Points of inflection occur at mid span of beams.
3. Total horizontal shear at any floor is distributed among the columns of that floor such that the exterior columns carry half the force carried by the inner columns. 
17. What are the  Assumptions in cantilever method ?
1. The points of inflection are located at the mid-height of each column above the first floor. If the base of the column is fixed, the point of inflection is assumed at mid height of the ground floor columns as well; otherwise it is assumed at the hinged column base.
2. Points of inflection occur at mid span of beams.
3. The basic assumption of the method can be stated as “the axial force in the column at any floor is linearly proportional to its distance from the centroid of all the columns at that level.