Two-Hinged and Three-Hinged Arches

Introduction:

Arches are the structures, which look somewhat different from the columns and beam. They have the curved shape, of an arch, which can be circular or parabolic. In Civil Engineering, you have to study the analysis of the arches.

In engineering terms, there are three types of arches. They are listed below.

1. Two hinged arches
2. Three hinged arches.
3. Fixed Arches

  

Three hinged arches are the determinate structures, because there are four unknown support reactions, and again there are four numbers of equations of equilibrium, to get the values of these unknowns.

Three Hinged Arches:

See above in fig.2, there are three hinges in the arch, A, B and C. Generally there are three numbers of equilibrium equation, but the fourth equation is derived from the fact the algebraic sum of all the moments at the hing C is 0.  So there are four numbers of equilibrium equations, and we can determine all the four support reactions, Va, Vb, Ha, and Hb.

Three Hinged Arch Image

Two Hinged Arches:

In the fig.1 in the above figure there are two hinges A and B, and there are four support reactions. There are only three numbers of equations of static equilibrium. Hence, the two hinged arches are indeterminate to a degree of 1.

 If we have to find out all the four unknown reactions of the two hinged arch, then, we need one more equilibrium equation. So the indeterminacy of a two hinged arch is equal to 1. 

We can easily find out the Va and Vb, by taking algebraic sum of all the moments about A or B equal to 0.  To find out the horizontal reactions Ha and Hb, many books advise to use the Castigliano's first theorem.

The relative displacement of the either hinge with respect to other is zero, so the partial derivative of the strain energy of the beam with respect to the horizontal reaction will be zero.

 So first we have to find the equation of the strain energy of the whole beam, and then partially differentiate it w.r.t. to the horizontal reactions, and then equate it to zero. 

It becomes the fourth equation, and we can get the value of the horizontal reaction. Now as all the support reactions are found, we can easily plot the bending moment diagram, for the arch. 

Now at any cross of the arch the vertical and the horizontal forces can be resolved along two directions, one is tangent to the cross sectional surface of the arch and another is normal to the cross sectional surface of the arch. It gives rise to another two terms:

1. Radial Shear
2. Normal thrust. 

Radial Shear and Normal Thrust

Due to the curved shape of the arch,  unlike shear force and axial thrust in a straight  beam, the direction of the normal thrust and the radial shear change constantly from one end of the arch to the other.
To find out the magnitude of the radial shear and the normal thrust, first we have to find out the direction of these two.

To find out the direction, we simply find out the slope of the arch(angle w.r.t. horizontal), at that section which, can be found by writing the equation of the circle or the parabola in terms of y and x co-ordinates, depending upon the shape of the arch and  then differentiating the y w.r.t. x gives us the slope of the equation.

Putting the corresponding value of the x at the given cross section, gives us the slope at the given cross section. After finding the angle, it becomes easy to resolve the forces into two directions, of the radial shear and the normal thrust. In the figure shown below, you have to resolve the forces at the section along the R and N, to find out the Radial Shear and the Normal Thrust.

Rib Shortening: 

As we can see that there is a normal thrusting force, which applies the compressing action to the rib/arch. If the magnitude is high it may result in the change in the length of the arch, due to corresponding strain, which can be found out by the Hooke's law. Final result will be the shortening of the arch. This effect is known as the Rib shortening. 

Thank you for your visit!

  

column analogy method

Hi,

Column analogy method is a method of structure analysis, used to analyse the indeterminate structures, specifically the fixed beams, arches etc. 

   

It is based on the analogy between the fixed end moments induced in a fixed beam to the pressure induced at the base of a column subjected to loading of the bending moment diagram of the basic determinate structure of the given fixed beam.

It means that first we have to make the indeterminate fixed beams determinate beams- mostly  simple supported beams, and then we have to find out the bending moment diagram for that structure. This Bending Moment diagram is used as the loading diagram on the top of a column having the cross section as that of the beam-  length equal to the span of the beam and width is kept as one.

 The pressures induced at the two ends of the column base are calculated and they are equal to the fixed end moments induced at the ends of the fixed beams.They are superimposed with the simple supported bending moments and that gives the final bending moment diagram.

Thanks!
Check these relevant books.

   

Moment distribution method - An approximate method

Moment distribution method of structure analysis is the method which was introduced after the slope deflection method by prof. Hardy Cross. This method uses the following properties of the beam for 

(1) one end hinged and far end fixed: The moment required to apply to produce the unit slope at hinged end of the beam when the far end is fixed is equal to 4EI/L. It is known as the stiffness of the beam.  At the same time half the moment applied at A is carried over to the far fixed end B, known as the carry over moment. 

(2) One end hinged and far end also hinged:  

When the far end is also hinged, the stiffness of the beam to produce the unit slope at end A by applying the moment at a is 3EI/L. The moment is not carried over to the far end because it is free.

Procedure to follow the Moment distribution analysis:

 All the joints are initially fixed  so that each span between joints behaves as the independent fixed beams.
Calculate the fixed end moments on each span after adding algebraically moments due to  the external loading, sinking of the supports. Now release the joints which are not fixed one at one time. There will be some unbalanced moment at joints (where two or more than two spans meet), and the ends with hinged or simple supports. balance the joints by distributing the moment at the joints among the member meeting at the joint proportional to their relative stiffness. The carry over moment to the far ends becomes the unbalanced moment for the next cycle of balancing.
Note: It is brief introduction of the method, you can write in if you need any further support in this method.