Torsion in Shafts (Torque)

Hi,
Torsion is a very important to run the vehicles with the wheels. Torsion is required almost in every day to day activities.
For Mechanical and Civil Engineers, the concept of torsion is very important to understand.

Shaft applied with Torsion

Torsion is the amount of couple/moment which is required to rotate a given section through the specified degrees about its polar axis.

In the figure below T is the amount of torque required to rotate a circular section at distance equal to 'L'through an specified angle.
It depends upon Polar moment of Inertia of the section "J". If we analyse the mechanism we get a relationship between, shear stress(t), radial distance(r), Torsion(T), Polar moment of Inertia(J), Modulus of rigidity(G), angle of rotation(@) of the circular section and length(L) as follows:
t/r= T/J = G.@/ L
J is the moment of inertia about the normal axis.

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Several Point Loads acting on a Girder- max Bending Moment and Shear Forces

There are some live examples which you see everyday when a several point loads act on a single girder. On a bridge there are a numbers of wheel loads which act simultaneously and results in the induction of the bending and shear stresses, for design of such structures, it becomes important to find out the magnitude of location of stresses induced.
MAX. B.M. On a Girder when several point loads move from one end of the girder to another end of it.
There are following two important propositions regarding the B.M. on a girder where several point loads act on the girder:
Proposition 1. :  The maximum bending moment(B.M.) for a given point load among the several point loads occurs at the point of action of the load when the distance between the line of action of the load and the equivalent load of the whole load assembly is divided into equal half by the center of the span of the girder.
Proposition 2.  : The maximum bending moment for an assembly of several loads occur at the section which divided the load assembly in such a manner that the average load on either side of the section is same or equal.

Stress Transformation- Analytical and Mohr's Circle Method.

Hi,

• Stress transformation is the process of transforming the stresses from one co-ordinate axis to another which is having some obliquity to the original axes. It is important because in practical civil Engineering problems, a structural element is subjected to combined stress, from more than one directions, so there resultant is going to be in some another axis. To find out the resultant we have to transform the given stresses along the required co-ordinate axis. We have two famous methods of transformation of stresses:

(a) Analytical method   (b) Graphical Method- Mohr's Circle method

• Analytical Method: In this method we analyse the given structural element for the given stress conditions and then we have to do the required mathematical calculations to find out the stresses along the required axis. Suppose n1, n2  are the normal stresses along principle axis and s1 is the applied shearing stress. You are asked to find normal stress(n) and shear stress(s) along a plane which is oblique to the normal with an angle o, then you can use the following formula, which is found after analyzing the given element for the above stresses

             n = [(n1+n2)/2] + [(n1-n2)/2]. (Cos2o) - (s1.Sin2o).
              s = (n1-n2)/2.Sin2o + s1. Cos20

• Graphical Method/Mohr's Circle Method:

Mohr'c Circle method is very simple method to find out the stresses along an plane which is oblique to the normal axis through some angle o. It is drawn on the two rectangular axis, n and s. n is drawn on the horizontal axis and s is generally drawn on the vertical axis, and rightmost value of n on the circumference of the circle and left most value of n gives the major and minor principle stress values. Radius of the circle gives us the maximum shearing stress.

For details please leave a comment....

Rolling Loads- maximum shear force, bending moments

Hello,
Hello, How you doing?

• Rolling Loads- 

• Rolling loads are those loads which roll over the given structural element from one end to the another. You can see many live examples of rolling loads, like a train on the railway track, vehicles on the bridges or roads are rolling loads.

Now to analyse the given structural element for the rolling loads, which classify these rolling loads into the following classes:
(1) Single point rolling loads
(2) Uniformly distributed rolling loads, - a) shorter than span  b) Longer than span.
(3) Two point loads at a fixed distance apart.
(4) Several point loads at fixed distance apart.

•  Single point load: If a single point concentrated load moves from one end of a girder to another end of it, it becomes necessary to find out the maximum values of the shear forces and bending moments at every section of the girder to produce an economical and safe design.

 Now if we analyse the girder then you will find that at any section on the girder maximum 
negative shear force induces when the point load is just on the left of that section, and maximum

positive shear force is induced if the load is just on the right of the section.

Absolute maximum negative shear force is produced at the right end of the girder and absolute maximum shear force is produced at the left end of the section.

Bending moment is positive for any position of the load, but maximum bending moment at a section occurs when the load is on the section itself and absolute maximum bending moment is produced at the central section when load is also at the center.

•  Uniformly distributed Loads:

(a) Longer than span: When the given udl(uniformly distributed load) is longer than the given span, it is easy to find out the maximum negative shear force , positive shear force and bending moment values on the section.

Rolling Loads(udl) on bridge

 
Maximum negative shear force is induced when the udl is on the left part of the span and right end of the udl is just on the section itself.

So, maximum negative shear force is induced at the right most end of the span when the end of the load is just to the left of the section.

Similarly, maximum positive shear force is induced at the left end of the girder when left end of the load is just on the right of the left end.

Maximum bending moment at a section is induced when whole of the span is loaded with the udl, and the value of the absolute maximum bending moment is induced at the center of the span and its value is given by (w.l^2)/8.


to be continued...\


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Cement - A Concreting Material

Hi,

Fresh Concrete is a material formed by mixing cementing material, coarse aggregates, fine aggregates and water in some definite proportion and which after curing of the required period forms a hard material like stone known as concrete.

Cement: 
Setting times of cement: 
(a) Initial Setting Time: Initial setting time is the time required by cement to start setting. So it is the time which is utilized for the mixing, transporting and placing of the cement concrete.
 Initial setting time is adjusted by mixing Gypsum to cement.
If gypsum is not added then cement will set as soon as water is added to it. BIS code recommends a initial setting time not less than 30 minutes.

(b)Final Setting time: Final setting time is the time after which cement starts gaining strength. BIS code recommends a final setting time not more than 10 hours.

Image source: Google Images

(c) Progressive hardening and strengthening of cement.

Compounds of cement(Bogue's compounds):
After the burning of the raw material there is a formation of the compounds like tri-calcium silicate, di-calcium silicate, Tri-calcium aluminate and tetra-calcium alumino ferrate.

Tricalcium silicate and dicalcium silicate are compounds which are responsible for the early strength and ultimate strength of the cement respectively. Tricalcium aluminate is responsible for almost all the un-desirable properties of cement.

So, the four Bogue's compounds are :
1)Tri-calcium silicate
2)Di-calcium silicate
3)Tri-calcium aluminate
4)Tetra-calcium alumino ferrate

Thanks!

Reference:
1.GK Publishers - GATE 2013 - Civil Engineering.
2. Google Maharaj

Civil GATE Preparation- Structural Analysis- one liner - part-1

Here are the one liners from structural analysis, which may need further explanation but due to some restrictions I am not able to submit those explanations here, if you really need them, you can leave a comment with email , and I will reply to you for sure.

a. A bar AB of diameter 40 mm and 4 m long is rigidly fixed at its ends. A torque of 600 Nm is applied at a section of the bar 1 m from end A. The fixing couples Ta and Tb at the supports A and B are 450N-m and 150 N-m respectively.

b. A propped cantilever of span 'l' carries a uniformly distributed load of 'w' per unit run over its entire span. The value of prop reaction to keep the beam horizontal is 3/8(w.l).

c. A two hinged semi-circular arch of radius 'R' carries a concentrated load 'W' at the crown. The horizontal thrust is w/pi  (pi  =3.14)

Part-3 GATE 2014 preparation - Structural mechanics- one liners

1. A brittle material of 4 sq. cm cross section carries an axial tensile load of 20 tonnes. The maximum shear stress in the block will be 2500 kg/cm^2.

2. Maximum allowable shear stress in a section is 100 kg.cm^2. If bar is subjected to a tensile force of 5000kg and the section is square shaped, then dimension of sides of the square will be 5 cm.

3. A simply supported beam of span L carries a concentrated load P at mid-span. If d be the width of the beam being constant and its depth varying through out its length of the spans, its mid-span depth when its design stress is 'f' will be [(3PL)/(2df)]^(1/2)

4. If an element of a stressed body is in a state of pure shear with a magnitude of 40 N/mm^2, then magnitude of maximum principal stress at that location is 40 N/mm^2.

5. Dimension of the flexural rigidity of a beam element in mass[M] and length[L] and time[T] is given by [M][L]^3[T]^-2

6. A cylinderical shell made of mild steel plate of 100 cm diameter is to be subjected to an internal pressure of 10 kg/cm^2. If material yields at 2000 kg/cm^2, assuming factor of safety as 4 and using maximum principal stress theory, thickness of the plate will be 1cm.

Reference: 
 1. GATE 2013 : GK Publishers

To find out the EI(rigidity) of a beam by using an overhanging beam

AIM: To find out the rigidity(EI) of a beam by using an overhanging beam apparatus.

Overhanging beam apparatus

APPARATUS:  One can use a overhanging beam apparatus along with the other weights which are used to be hanged and the dial gauge to note down the deflection.
Theory:

We know that if we apply a load equal to W at the two far ends of the overhangs of length 'a' each of an overhanging beam with the central span equal to 'L' then the amount of central deflection is 

given  by   y = W.a.L^2/ 8.E.I  

Now to find out EI, you have to use the following formula

EI= W.a.L^2/ 8.y

W, L and a are known from the above apparatus dimensions and the loads which are to be used are also known to you, and you have to note down the central up-rise(deflection) i.e. y with the help of the dial gauge and then put the values in the above formula.

So you get the value of the rigidity of the given overhanging beam.

Compare the values obtained for different loads and for different deflections and use the average.

Thank you!

part 2 - GATE preparation- Structural Mechanics- one liners

1. At a point in a steel member, major principal stress is 1000 kg/cm^2, and minor principal stress is compressive. If uni-axial tensile yield stress is 1500 kg/cm^2, then magnitude of minor principal stress at which yielding will commence, according to maximum shearing stress theory is, 500 kg/cm2 (Sy/2 = (S1-S2)/2.

2. If an element is subjected to pure shearing stress T, then maximum principal stress T is equal to T.

3. A brittle material of 4 sq. cm cross section carries an axial tensile load of 20 T. What will be the maximum shear stress in the block?  Ans : 2.5 T/cm^2 = 2500 kg/cm^2.

4. Given EI = (1/100)wL^3, where w is central load as well as total udl load acting on the span.  What will be the deflection when a udl equal to w and a concentrated load w is acting centrally to a simply supported beam?     Ans: 1300/384

GATE 2014- one liners - Structural Mechanics- Part 1

1. The relation between E(Young's modulus of Elasticity) and bulk modulus K, when U(Poisson's ratio) = 0.25 is   E=1.5K

2. If K and G are bulk modulus and modulus of rigidity the Poisson's ratio will be (3K-2G)/(6K+2G).

3. An axial tensile force is acting on a body and normal strain in axial direction is 1.25 mm/m. If Poisson's ratio is 0.3, the Volumetric strain in the body is 1.7*10^-4 (=logitudinal strain*lateral strain^2)

4. A point is a strained body is subjected to a tensile stress of 100 MPa on one plane and a tensile stress of 50 MPa on a plane at right angle to it. If these planes are carrying shear stresses of 50 MPa, then principal stresses inclines to the layer normal stress at an angle = 1/2[(tan)^-1(2)]

5. A simply supported beam of span l carries over its full span a load varying linearly from zero at either end to w /unit length at other end, then maximum bending moment occurs at mid span and is equal to wl^2/12.

6. Slope at the end of simply supported beam of span 2 m and load 5 kg/ unit length over the entire span, will be 5/(3EI)

7. If a simply supported beam of circular cross section with diameter d and length l carries a concentrated load W at the center of the beam, then strength of the beam is proportional to (z =I/y) D^3.

8. When both ends of the column are fixed, the cripling load is F. If one end of the column is made free, then value of cripling load will be changed to F/16.

9. From Rankine's hypothesis Rankine's criteria for failure of brittle material is maximum principal stress.

10. If a circular shaft is subjected to a torque T which is half of the bending moment applies, then the ratio of maximum bending stress and maximum shear stress is 4


Reference: GK publishers

Rules for drawing Mohr's Circle to find out Principal Stresses

Rules for drawing Mohr's circle for normal and shear stresses:
1. On rectangular S - T (S- normal stress, T - shear stress) axis, plot the points having co-ordinates A(Sx, Tyx) and B ( Sy, Tyx). Assume tension as positive/plus, compression as minus, and shear stress as plus when its moment about center is clockwise.
2. Join these points, with a straight line AB which gives us the diameter of the circle whose center is on S-axis.
3. The radius of the circle to any point on its circumference represents the axis directed normal to the plane whose stress components are given by the co-ordinates of that point.
4. The angle between the radii to selected points on Mohr's circle is twice the angle between the normals to the actual planes represented by these points.