LOADING IN BEAMS

BEAMS & CLASSIFICATION OF BEAMS

BEAM: A beam is a structure generally a horizontal structure on rigid supports and it carries mainly vertical loads. Therefore, beams are a kind of load bearing structures.

Depending upon the types of supports beams can be classified into different catagories.

CANTI-LEVER BEAMS: 

A beam can be at stable equilibrium with a single fixed support at one end and the other end remains free, which is called as the free end while the other end is known as fixed end. This kind of beam is known as Canti lever beam. The fixed joint at the fixed end produces a horizontal, a vertical reactions and a reaction moment at the fixed end.

SIMPLE SUPPORTED BEAM: 

A beam supported as just resting freely on the walls or columns at its both ends is known as simply supported beam.

There will be two vertically upward reactions at the ends of a simply supported beam. A simply supported beam can not resist any horizontal load component.

OVER HANGING BEAM: 

A beam having its end portion or both the end portions extended in the form of a canti-lever beyond the support or supports is called as over hanging beam.

Above those beams are statically determinate. It means that those beams can be analysed applying the conditions of equilibrium. We can determine the values of the unknown reactions.

There are beams which can not be analysed applying the conditions of equilibrium of coplanar forces. These beams are also known as Statically indeterminate structures.

Those types of beams can be classified as,

Fixed beams and Continuous beams.

Fixed Beam: A beam having two fixed joints at the both ends is called fixed beam.

Continuous Beam: The beam which is at rest on more than two supports is called as continuous beam.

What are different types of supports? 

There are four types of supports,

• (i) Simple Supports, 
• (ii) Roller Supports, 
• (iii) Hinged Supports 
• (iv) Fixed Supports.

SOLUTION OF EME-102; TRUSS ANALYSIS

SOLVE THE TRUSS GIVEN BELLOW WITH THE HELP OF METHODS OF JOINT

________________________________________________________________________________

a)      REPLACE JOINTS WITH REACTIONS at A and at B

              

       

b)      Draw FBD of the TRUSS

 

Applying the conditions of Equilibrium of Coplanar Non-concurrent Force System,

 

∑ F_X = 0;        R_b – R_ah = 0  ------ (i)

(-) ← ● → (+)

∑ F_Y = 0;        R_av – 10 – 5 – 15 = 0 => R_av = 30 kN ----- (ii)

∑ M_A = 0;       10 x 4 + 5 x 4 + 15 x 2 – R_b x 3 = 0  ----- (iii)

                        R_b = 30 kN

                        Hence R_ah = R_b = 30 kN

Calculation of Angle θ

The angle θ = tan^-1(3/2) = 56.3°

All the unknown forces will be taken as Tensile, if their magnitudes is found negative, then they will be treated as compressive forces.

First we shall choose a joint having only two unknown forces, either we shall choose joint D or joint A

Let us choose joint D first.

We shall consider point D first, as it has only two unknown force. FBD of the point D is drawn.

∑ F_X = 0;      F₂ = 0

∑ F_Y = 0;      F₁ – 5 = 0

                      F₁ = 5 kN

 

Our next joint will be point E. FBD of the joint E is drawn. As F₁ = 5 kN, hence unknown forces are two. F₃ and F₄

∑ F_X = 0;     – F₃ – F₄ cos 56.3° = 0

∑ F_Y = 0;     – F₁ – 10 –  F₄ sin 56.3° = 0  [ as F₁ = 5 kN]

                           F₄ = –15/ sin 56.3° = – 18.02 kN

                    F₃ = – F₄ cos 56.3° = 10 kN

 

Our next joint is C

 F₅ and F₉ are unknown where as F₄ = – 18.02 kN

 F₂ = 0

 ∑ F_X = 0;     – F₉ + F₄ cos 56.3° = 0

            F₉ = F₄ cos 56.3° = 10 kN

 ∑ F_Y = 0;     F₅ + F₄ sin 56.3° – 15 = 0

                            F₅ = – F₄ sin 56.3° + 15 = 30 kN

 

F₃ = 10 kN;   F₅ = 30 kN

        ∑ F_X = 0;     F₃ = F₆ + F₇ cos 56.3°

        ∑ F_Y = 0;     – F₅ – F₇ sin 56.3° =0

       F₇ = – F₅/ sin 56.3° = – 36.05 kN

F₆ = F₃ – F₇ cos 56.3° = 10 + 20 = 30 kN

R_av = 30 kN;  R_ah = 30 kN

 ∑ F_X = 0;      F₆  = R_ah = 30 kN

 ∑ F_Y = 0;     F₈ = R_av = 30 kN

Sl no	Link	Force	Magnitude	Nature
01	ED	F1	5 kN	T
02	CD	F2	0
03	FE	F3	10 kN	T
04	CE	F4	18.02 kN	C
05	FC	F5	30 kN	T
06	AF	F6	30 kN	T
07	BF	F7	36.05 kN	C
08	AB	F8	30 kN	T
09	BC	F9	10 kN	T

SOLUTION OF EME-102; CENTROID 2

HOW TO FIND THE CENTROID OF A COMPOSITE AREA

(a composite area consists of several straight or curved lines.)

(i) Draw the figure in a coordinate system. Draw the dimensions too. Every dimensions will be measured with respect to origin of the coordinate system

(ii) Divide the composite area into several parts of basic geometric areas. Lebel them as part-1, part-2, part-3, .......part-n. Let the corresponding areas are A₁, A₂, A₃, .... A_n. Let the centroids are G₁(X₁,Y₁), G₂(X₂,Y₂), G₃(X₃,Y₃), ...... G_n(X_n,Y_n).

(iii) Let the centroid of the composite area be G(Xg,Yg). Hence,
Xg = (A₁X₁ + A₂X₂ +A₃X₃)/(A₁ + A₂ + A₃)

Yg = (A₁Y₁ + A₂Y₂ +A₃Y₃)/(A₁ + A₂ + A₃)

 

(a) Suppose we have certain area of magnitude (A) in a coordinate system. The centroid of the area will be at its mid-point. A centroid is denoted by G.

                       In the figure we have a complex geometrical area composed of three basic geometrical areas. Three rectangles. Let us denote the centroids as G₁, G₂, G₃ for the given areas in the figure.

We shall have to find the Centroid of the entire area composed of  A₁, A₂, A₃.

At first, the composite line is divided into three parts.

Part -1 : The middle rectangle : Let the centroid of the area A₁ be G₁(X₁,Y₁)

Area, A₁ = (100-15-15)x15 mm² = 70x15  mm²  = 1050  mm²                 
          X₁ = 7.5 mm
          Y₁ =0


Part -2 : The lower Rectangle : Let the centroid of the A₂ be G₂(X₂,Y₂)


Area, A₂ = 50 x 15  mm² = 750 mm²                
          X₂ = 25 mm
          Y₂ = - 42.5 mm


Part -3 : The upper Rectangle : Let the centroid of the area Area, A₃ be G₃(X₃,Y₃)


Area, A₃ = 50 x 15  mm² = 750 mm²                
          X₃ = 25 mm
          Y₃ = 42.5 mm






If the centroid of the composite line be G  (X_g,Y_g)

X_g = (∑A_iX_i)/(∑A_i) 

    = (A₁X₁ + A₂X₂ +A₃X₃)/(A₁ + A₂ + A₃)
    = (1050 x 7.5 + 750 x 25 + 750 x 25)/( 1050 + 750 + 750)
    = 17.79 mm
     

Y_g = (∑A_iY_i)/(∑A_i) 

    = (A₁Y₁ + A₂Y₂ +A₃Y₃)/(A₁ + A₂ + A₃)
    = {1050 x 0 + 750 x (-42.5) + 750 x 42.5}/( 1050 + 750 + 750)
    = 0

As the figure is symmetrical about X axis, hence when we take the line of symmetry as our X axis (as we have taken here), we can directly write, Y_g =0

SOLUTION OF EME-102; CENTROID

HOW TO FIND THE CENTROID OF A COMPOSITE AREA

(a composite area consists of several straight or curved lines.)

(i) Draw the figure in a coordinate system. Draw the dimensions too. Every dimensions will be measured with respect to origin of the coordinate system

(ii) Divide the composite area into several parts of basic geometric areas. Lebel them as part-1, part-2, part-3, .......part-n. Let the corresponding areas are A₁, A₂, A₃, .... A_n. Let the centroids are G₁(X₁,Y₁), G₂(X₂,Y₂), G₃(X₃,Y₃), ...... G_n(X_n,Y_n).

(iii) Let the centroid of the composite area be G(Xg,Yg). Hence,
Xg = (A₁X₁ + A₂X₂ +A₃X₃)/(A₁ + A₂ + A₃)

Yg = (A₁Y₁ + A₂Y₂ +A₃Y₃)/(A₁ + A₂ + A₃)

 

(a) Suppose we have certain area of magnitude (A) in a coordinate system. The centroid of the area will be at its mid-point. A centroid is denoted by G.

                       In the figure we have a complex geometrical area composed of three basic geometrical areas. A rectangle, a semi circle and a isosceles triangle. Let us denote the centroids as G₁, G₂, G₃ for the given areas in the figure.

We shall have to find the Centroid of the entire area composed of  A₁, A₂, A₃.

At first, the composite line is divided into three parts.

Part -1 : The semi-circle : Let the centroid of the area A₁ be G₁(X₁,Y₁)

Area, A₁ = (π/2)x(25)² mm² = 981.74 mm²                  
          X₁ = { 25 -  (4x25)/(3xπ)} mm = 14.39 mm
          Y₁ = 25 mm


Part -2 : The Rectangle : Let the centroid of the A₂ be G₂(X₂,Y₂)


Area, A₂ = 100 x 50  mm² = 5000 mm²             
          X₂ = 25 + (100/2) = 75 mm
          Y₂ = 25 mm
Part -3 : The Triangle : Let the centroid of the area Area, A₃ be G₃(X₃,Y₃)


Area, A₃ = (1/2) x 50 x 50 mm² = 1250 mm²             
          X₃ = 25 + 50 + 25 = 100 mm
          Y₃ = 50 + (50/3) =  66.67 mm






If the centroid of the composite line be G  (X_g,Y_g)

X_g = (∑A_iX_i)/(∑A_i) 

    = (A₁X₁ + A₂X₂ +A₃X₃)/(A₁ + A₂ + A₃)
    = (981.74 x 14.39 + 5000 x 75 + 1250 x 100)/( 981.74 + 5000 + 1250)
    = 71.09
     

Y_g = (∑A_iY_i)/(∑A_i) 

    = (A₁Y₁ + A₂Y₂ +A₃Y₃)/(A₁ + A₂ + A₃)
    = (981.74 x 25 + 5000 x 25 + 1250 x 66.67)/ ( 981.74 + 5000 + 1250)
    = 32.20

SOLUTION OF EME-102; EQUILIBRIUM OF FORCES

EQUILIBRIUM OF FORCES IN 2D

A light string ABCDE whose extremity A is fixed, has weights W₁ & W₂ attached to it at B & C. It passes round a small smooth pulley at D carrying a weight of 300 N at the free end E as shown in figure. If in the equilibrium position, BC is horizontal and AB & CD make 150° and 120° with BC, find (i) Tensions in the strings and (ii) magnitudes of W₁ & W₂

Although ABCDE is a single string/rope but still the tensions in the string/rope will be different at different segments like in the segment AB the tensions will be T₁ , but in BC segment it will be different as the weight is attached at a fixed point (point B) on the string, hence it will be T₂ here and in CD it will be T₃  there. As at point D the string is not attached rather passes over a smooth pulley hence the tension in DE and CD will be same ie. T₃  again.

. 

To solve for equilibrium of forces in 2D, follow these steps:

1. Draw a free body diagram: Draw a diagram of the object in question and identify all the forces acting on it. This will help you visualize the problem and identify any unknown forces or angles.

2. Break forces into components: Resolve all forces into their x- and y-components. This is done by using trigonometry to determine the horizontal and vertical components of each force.

3. Apply Newton's second law: For the object to be in equilibrium, the sum of the forces in the x-direction and the sum of the forces in the y-direction must be equal to zero. This gives you two equations to solve simultaneously.

4. Solve the equations: Solve the equations for the unknown forces or angles by using algebraic methods. This will give you the values of the unknown forces or angles required for the object to be in equilibrium.

5. Check for consistency: Check that the forces and angles you have calculated are consistent with the problem and the free body diagram. For example, make sure that the direction of the forces makes sense and that the angles are reasonable.

6. Interpret the results: Interpret the results and explain what they mean in the context of the problem. This might involve calculating the tension in a rope, the force required to lift an object, or the angle required for an object to remain stationary.

Physics of Inertial Forces

Force Effects Caused by Inertia and Accelerating Reference Frames

Inertial forces are not real forces, rather they are often called as "pseudo forces". They produce effects that feel like forces but actually arise from Newton's inertial law in a reference frame that is accelerating.

When a car accelerates forward rapidly, a person inside the car feel pushed back into their seats. When the car turns around a curve, the person feels pulled to the outside of the curve. If the car suddenly comes to a stop, the persons inside the car not wearing seatbelts fly forward, possibly hitting the windshield. The car’s occupants may feel like some force is pushing them around, but in reality there are no forces shoving them in the directions they move inside the car. They feel shoved around the car because the car is accelerating. The occupants, however, follow Newton’s first law, the inertial law, and continue their original motion, as the car accelerates.


Newton’s First Law

Also called the inertial law, Newton’s first law requires that any object with no outside forces acting on it continues to move at a constant velocity. A constant velocity is a constant speed in a straight line because in physics the velocity includes direction. Any change in an object’s velocity (increasing speed, decreasing speed, or changing direction) is called an acceleration and requires an external force to act on the object. This tendency for objects to continue to move at a constant velocity is called inertia.


Inertial Forces

Despite the name, inertial forces are not real forces. Rather they are effects caused by an object’s inertia when the object is in or on something that is accelerating, what physicists call an accelerating reference frame.

For example, consider the occupants of a car rapidly increasing its speed. They feel pushed back into their seats, but not because some force is shoving them in the chest. The only real force acting on them is the back of the car seat accelerating them forward. Because of Newton’s first law, however, these occupants have inertia that tends to keep them at rest. They feel squeezed back into the car seat because the car is accelerating forward while their inertia would tend to keep them at rest.

When the car goes around a curve, it is also accelerating because the direction of the car’s velocity is changing. The occupants’ inertia tends to keep them moving in a straight line. Hence they feel pulled sideways in the car because they experience an inertial force (or effect) caused by the car’s acceleration as it changes direction.

Flying forward into the windshield when a car stops suddenly (Always wear seatbelts!) is a similar inertial effect. The car accelerates to a stop, and the occupants continue moving forward until their seatbelts (or the windshield) exerts a stopping force on the occupants. There is no real force pushing them forward, they just continue their forward motion, as required by Newton’s first law, while the car accelerates (slowing) to a stop.
Circular Motion

To move in a circular path, an object must have a centripetal force acting on it. The centripetal force points inward towards the center of the circle. The outward centrifugal effect is the tendency of the object to continue in a straight line motion. Hence, the centrifugal effect is an example of an inertial force and is not a real force acting on an object moving in a circular path.

Inertial forces feel like forces, but they are not real forces. They are effects caused by an object’s inertia when it is in or on something that is accelerating.


 D'Alembert's principle of inertial forces

D'Alembert showed that one can transform an accelerating rigid body into an equivalent static system by adding the so-called "inertial force" and "inertial torque" or moment. The inertial force must act through the center of mass and the inertial torque can act anywhere. The system can then be analyzed exactly as a static system subjected to this "inertial force and moment" and the external forces. The advantage is that, in the equivalent static system' one can take moments about any point (not just the center of mass). This often leads to simpler calculations because any force (in turn) can be eliminated from the moment equations by choosing the appropriate point about which to apply the moment equation (sum of moments = zero). Even in the course of Fundamentals of Dynamics and Kinematics of machines, this principle helps in analyzing the forces that act on a link of a mechanism when it is in motion. In textbooks of engineering dynamics this is sometimes referred to as d'Alembert's principle.

d’Alembert’s principle,  alternative form of Newton’s second law of motion, stated by the 18th-century French polymath Jean le Rond d’Alembert. In effect, the principle reduces a problem in dynamics to a problem in statics. The second law states that the force F acting on a body is equal to the product of the mass m and acceleration a of the body, or F = ma; in d’Alembert’s form, the force F plus the negative of the mass m times acceleration a of the body is equal to zero: F - ma = 0. In other words, the body is in equilibrium under the action of the real force F and the fictitious force -ma. The fictitious force is also called an inertial force and a reversed effective force.

QUESTION BANKS: Analyse the following Trusses:

 Analyse the following Trusses:

(1)      A cantilever truss has been as shown in the figure. Find the value of W which will produces a force of magnitudes 15 kN  in the member AB.

(2)        A cantilever truss is loaded as shown in the figure. Find the nature and magnitudes of the forces in each link.

(3)      A cantilever truss has been as shown in the figure. Find the value of W which will produces a force of magnitudes 15 kN  in the member AB.

(4)      A truss has been loaded as shown in the figure. Find the nature and the magnitudes of the forces in the links BC, CH and GH by the methods of sections.

(5) A truss has been loaded as shown in the figure. Find the internal forces in each of the beam.

(6)      A truss has been loaded as shown in the figure. Find the forces in each member and tabulate them by any methods.

compiled by Subhankar Karmakar

more content: click the following links for more questions.

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