PARALLEL AXIS THEOREM AND IT'S USES IN MOI

Moment Of Inertia of an Area.

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MOI or MOMENTS OF INERTIA is a physical quantity which represents the inertia or resistances shown by the body against the tendency to rotate under the action external forces on the body. It is a rotational axis dependent function as its magnitude depends upon our selection of rotational axis. Although for any axis, we can derive the expression for MOI with the help of calculus, but still it is a cumbersome process.


Now suppose we take a different issue. We know MOI of an area about its centroidal axis is easily be obtained by integral calculus, but can we find a general formula by which we can calculate MOI of an area about any axis if we know its CENTROIDAL MOI.

We shall here find that we can indeed derive an expression by which MOI of any area (A) can be calculated about any Axis, if we know its centroidal MOI and the distance of the axis from it's Centroid G.


If I_GX be the centroidal moment of inertia of an area (A) about X axis, then we can calculate MOI of the Area about a parallel axis (here X axis passing through the point P) at a distance Ŷ-Y'=Y from the centroid if we know the value of I_GX and Y, then I_PX will be
I_PX = I_GX + A.Y² where Y=Ŷ-Y'


I_XX = I_OX = I_GX + A.Ŷ²

<sup>Where I_XX is the moment of inertia of the area about the co-ordinate axis parallel to X axis and passing through origin O, hence we can say,

I_XX = I_OX

 IMPORTANT: The notation of Moment of Inertia

MOI of an area about an axis passing through a point B will be written as I_BX</sup>

Q: Find the Centroidal Moment of Inertia of the figure given above. Each small division represents 50 mm.

To find out Centroidal MOI

QUESTION BANK : ENGINEERING MECHANICS PART-2

TOPICS: NUMERICALS ON FORCE SYSTEM- UNIT-1

5) A bar of AB 12 m long rests in horizontal position on two smooth planes as shown in the figure. Find the distance X at which 100 kN is to be placed to keep the bar in equilibrium.

 

6) 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  

 

7) Find reactions at all the contact points if weight of P is 200 N & diameter is 100 mm, where as weight of Q is 500 N and diameter is 180 mm.

 

8) Determine the force P required to begin rolling the uniform cylinder of mass (m) over the obstacle of height (h) as shown in the figure.  

 

9) A roller of weight 500 N has a radius of 120 mm and is pulled over a step of height 60 mm by a horizontal force P. Find magnitudes of P to just start the roller over the step.

 

10) Two identical rollers each of weight 100 N are supported by an inclined plane of 30° with horizontal and a vertical wall as shown in the figure. Find all the reactions at each contact point.

 

11) A smooth cylinder of radius 500 mm rests on a horizontal plane and is kept from rolling by a rope OA of 1000 mm length. A bar AB of length 1500 mm and weight 1000 N is hinged at point A and placed against the cylinder of negligible weight. Determine the tension in the rope.

 

12)      A flat belt connects pulley B, which drives a pulley A; attached to an electric motor. μ_s =  0.25 and μ_k = 0.2 between both the pulleys and the belt. If maximum allowable tension in the belt is 600 N, determine the largest torque which can be exerted by belt on pulley B.

      

13)       Two blocks of mass MA & MB are kept at equilibrium as shown in the figure. The friction between the block B & the floor is 0.35 and between the blocks is 0.3, then find the minimum force P to just move the block B.

CENTROIDS OF LINES

CENTROID OF A STRAIGHT LINE

If we take a straight line of length (L), then its midpoint will be at a distance (L/2) from either end of the line. Let us denote the centroid as the point G(X_g,Y_g).

Suppose we have a straight line AB of length (L) that makes an angle θ with X axis. Let the coordinate of point A is (X_o,Y_o). Let the centroid be at G(X_g,Y_g), then

X_g = X_o + (Lcos θ)/2
Y_g = Y_o + (Lsin θ)/2

                                                                                                                                                             

     Again, suppose the coordinate of B is given instead of point A. Let it is given as B(Xi,Yi). Then, it will be                                                       

Xg = X₁ - (Lcos θ)/2
Yg = Y₁ - (Lsin θ)/2

For Horizontal lines θ = 0° and for Vertical lines θ = 90°

CENTROID OF A CURVED LINE

The steps to derive the centroid of a quarter circular arc of radius R. 

 

Centroid of a curved line can be derived with the help of calculus.

i) Draw the figure in a X-Y coordinate system. Let the curved line has been represented by a function θ

ii) Take an arbitrary point P(X,Y) on the curve. Join the line OP, where O is the origin of the coordinate system. Let OP has a length L and makes an angle (θ) with X axis. Therefore, we can write

                  

                                   X = Rcosθ ----- (a)
                             Y = Rsinθ ----- (b)

iii) Let another point Q, such that PQ = dL where dL is very very small. Let the angle subtended by (dL) be (dθ). So we can write

                                   dL = Rdθ ------ (c)

                             X_g = (1/L) ∫(XdL)
                                  = (1/L) ∫ Rcosθ.Rdθ
                                  = (1/L).R²  ∫ sinθ.dθ ------- (d)

                             Y_g = (1/L) ∫ YdL
                                  = (1/L) ∫ Rsinθ.Rdθ
                                  = (1/L).R²  ∫ sinθ.dθ -------- (e)

CENTROID OF A QUARTER CIRCULAR ARC OF RADIUS R

Suppose we have a quarter circular arc in a co-ordinate system as shown in the figure. Total length of the arc AB = (πR)/2 . We take an arbitrarily small length of the arc CD and denote it as dL.

So,

           dL = Rdθ  ------ (iii)    [ as s=Rθ ]

where R = Radius of the quarter circular arc.

Let the co-ordinate of the point D be D(x,y) where

   
               X = Rcosθ -----(iv) and

            Y = Rsinθ -----(v)

Hence   X_g = (1/L)∫x.dL  ;  here  L = (πR)/2  ;        

                                           X = Rcosθ      

                                          dL = Rdθ

    

             X_g = (2/πR)   ₀∫^π/2Rcosθ.Rdθ 

     =    (2/πR) R²  ₀∫^π/2cosθ.dθ

 =      2R/π
   

   Y_g = (2/πR)   ₀∫^π/2Rsinθ.Rdθ 

     

 =      (2/πR) R²  ₀∫^π/2sinθ.dθ

 =      2R/π

Hence, for a quarter circular arc of radius R will be G(2R/π,2R/π)                                                                

                                                                                      

CENTROID OF A COMPOSITE LINE

In the figure, a composite line A-B-C-D is made of three straight lines AB, BC, CD as shown in the figure.So, a composite line is consisted of several straight or curved lines.


Let a composite line is made of n number of lines, which may straight or curved lines.

STEP-ONE:

Draw the composite line and break it into n number of parts. Let the lengths of these lines are L_1, L_2,  L_{3 ........} L_n. Let the centroids of these lines are G₁(X₁,Y₁),G₂(X₂,Y₂), G₃(X₃,Y₃) ........ G_n(X_n,Y_n).

Calculate length (L_i), and coordinates (X_i,Y_i) for each and every parts.

 

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

X_g = (∑L_iX_i)/(∑L_i) 

    => (L₁X₁ + L₂X₂ + L₃X₃)/(L₁ + L₂ + L₃)
   

Y_g = (∑L_iY_i)/(∑L_i)

    => (L₁Y₁ + L₂Y₂ + L₃Y₃)/(L₁ + L₂ + L₃)
   

CENTROID OF A COMPOSITE LINE

In the figure, a composite line A-B-C-D is made of three straight lines AB, BC, CD as shown in the figure. If the coordinate of point A is (5,5), find the centroid of the composite line.


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








Part -1 : The line AB : Let the centroid of the line be G₁(X₁,Y₁)

length, L₁ = 40 mm;                  


X₁ = 4 + (40*cos 60⁰)/2 = 14  
Y₁ = 4 + (40*sin 60⁰)/2 = 21.32








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


length, L₂ = 15 mm; 


X₂ = 4 + (40*cos 60⁰) + 15/2 = 31.5 
Y₂ = 4 + (40*sin 60⁰) = 38.64




Part -3 : The line CD : Let the centroid of the line be G₃(X₃,Y₃)


length, L₃ = 20 mm; 

X₃ = 4 + (40*cos 60⁰) + 15 = 39 
Y₃ = 4 + (40*sin 60⁰) - 20/2 = 28.64



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

X_g = (∑L_iX_i)/(∑L_i) 

    = (L₁X₁ + L₂X₂ + L₃X₃)/(L₁ + L₂ + L₃)
    = (40 x 14 + 15 x 31.5 + 20 x 39)/(40 + 15 + 20) 
    = 24.17
     

Y_g = (∑L_iY_i)/(∑L_i) 

    = (L₁Y₁ + L₂Y₂ + L₃Y₃)/(L₁ + L₂ + L₃)
    = (40 x 21.32 + 15 x 38.64 + 20 x 28.64)/(40 + 15 + 20) 
    = 26.74

TRUSS ANALYSIS: THEORY OF TRUSS:

TRUSS ANALYSIS: THEORY OF TRUSS:

TRUSS :

Truss is a kind of framed structure made of entirely by rigid metallic rods joined by pin. The rods are called as Links or Linkages and the pins are called as joints. Their primary goal is to support the applied loads or we can say they are primarily load bearing structures. We often encounter trusses in our daily life as trusses are used to support roofs of various kinds of industrial sheds. Trusses are used as poles carrying high tension electricity.

LINK/ LINKAGES :

A link is a rigid rod which can bear any external load applied on it. A link can bear two types of forces.

COMPRESSIVE FORCES :

When the external forces applied on the link or rod tries to decrease the length of the rod, then they are called as External Compressive Forces. A truss in equilibrium counters this compressive force by inducing an internal force, equal and opposite the externally applied force. The internal force thus induced balancing the external compressive force is named as Internal Compressive Forces. Generally Compressive Forces are considered as negative in truss analysis.

TENSILE FORCES :

When external loads applied on a link try to increase the length of the link, we call them External Tensile Loads. To neutral the tensile load applied on a link, an equal but opposite internal force is generated named as Internal Tensile forces. Tensile forces are generally considered as positive internal forces.

THE SIMPLEST TRUSS:

A triangular shaped truss made of three linkages and three joints is the simplest type of truss. As it is the simplest geometric shape where there is no change in shape with the application of forces at the joints if the length of rods/ linkages remain unchanged / constant.

MAXWELL'S TRUSS EQUATION:

To distinguish between "statically determinate structure" and "statically indeterminate structure" Maxwell formulated an equation involving the number of linkages (m) and number of joints (j).

The trusses which satisfies the equation,
m = 2j - 3
are statically determinate structures and named as "Perfect Trusses".

If m > 2j - 3, then the number of linkages are more than required, hence, called as "Redundant Trusses".

Where as if m < 2j - 3 for any truss, then the number of linkages are less than that of a perfect truss. These kinds of trusses are called as "Deficient Trusses".



ASSUMPTIONS CONSIDERED WHILE ANALYZING TRUSSES :

While analyzing trusses, to simplify the analysis we often consider certain assumptions. The purpose of these assumptions are the simplification of a complex problems. The assumptions are

(i) The links are perfectly rigid bodies, ie there occurs no change in the dimensions of the links.

(ii) The pin joints are perfectly smooth, ie there is no friction in the each and every joints.

(iii) The mass and weights of the links are so small compare to the magnitudes of the applied forces, that for truss analysis we shall neglect them. It means the links are massless as well as weightless.

(iv) The cross-sections and material of the links are uniform by nature.

(v) The external loads are only applied on a joint in the truss, whenever we shall place any external load, we must place it one of the joints in the truss.

(vi) Stress in each member is constant along its length.

The objective of analyzing the trusses is to determine the reactions and member forces. The methods used for carrying out the truss analysis with the equations of equilibrium and by considering only parts of the structure through analyzing its free body diagram to solve the unknowns.

Method of Joints

 

 

The first to analyze a truss by assuming all members are in tension reaction. A tension member is when a member experiences pull forces at both ends of the bar and usually denoted as positive (+ve) sign. When a member experiencing a push force at both ends, then the bar was said to be in compression mode and designated as negative (-ve) sign.
In the joints method, a virtual cut is made around a joint and the cut portion is isolated as a Free Body Diagram (FBD). Using the equilibrium equations of ∑ F_x = 0 and ∑ F_y = 0, the unknown member forces could be solve. It is assumed that all members are joined together in the form of an ideal pin, and that all forces are in tension (+ve) of reactions.
An imaginary section may be completely passed around a joint in the truss. The joint has become a free body in equilibrium under the forces applied to it. The equations ∑ H = 0 and ∑ V = 0 may be applied to the joint to determine the unknown forces in members meeting there. It is evident that no more than two unknowns can be determined at a joint with these two equations.
 
Figure 1: A simple truss model supported by pinned and roller support at its end. Each triangle has the same length, L and it is equilateral where degree of angle, θ is 60° on every angle. The support reactions, R_a and R_c can be determine by taking a point of moment either at point A or point C, whereas H_a = 0 (no other horizontal force).
Here are some simple guidelines for this method of truss analysis:

1. Firstly draw the Free Body Diagram (FBD),
2. Solve the reactions of the given structure,
3. Select a joint with a minimum number of unknown (not more than 2) and analyze it with ∑ F_x = 0 and ∑ F_y = 0,
4. Proceed to the rest of the joints and again concentrating on joints that have very minimal of unknowns,
5. Check member forces at unused joints with ∑ F_x = 0 and ∑ F_y = 0,
6. Tabulate the member forces whether it is in tension (+ve) or compression (-ve) reaction.

 
Figure 2: The figure showing 3 selected joints, at B, C, and E. The forces in each member can be determine from any joint or point. The best way to start by selecting the easiest joint like joint C where the reaction R_c is already obtained and with only 2 unknown, forces of F_CB and F_CD. Both can be evaluate with ∑ F_x = 0 and ∑ F_y = 0 rules. At joint E, there are 3 unknown, forces of F_EA, F_EB and F_ED, which may lead to more complex solution compare to 2 unknown values. For checking purposes, joint B is selected to shown that the equation of ∑ F_x is equal to ∑ F_y which leads to zero value, ∑ F_x = ∑ F_y = 0. Each value of the member’s condition should be indicate clearly as whether it is in tension (+ve) or in compression (-ve) state.

* (Trigonometric Functions:
Taking an angle between member x and z…

• Cos θ = x / z
• Sin θ = y / z
• Tan θ = y / x )

Method of Sections

 

 

The section method is an effective method when the forces in all members of a truss are being able to determine. Often we need to know the force in just one member with greatest force in it, and the method of section will yield the force in that particular member without the labor of working out the rest of the forces within the truss analysis.
If only a few member forces of a truss are needed, the quickest way to find these forces is by the method of sections. In this method, an imaginary cutting line called a section is drawn through a stable and determinate truss. Thus, a section subdivides the truss into two separate parts. Since the entire truss is in equilibrium, any part of it must also be in equilibrium. Either of the two parts of the truss can be considered and the three equations of equilibrium ∑ F_x = 0, ∑ F_y = 0, and ∑ M = 0 can be applied to solve for member forces.
 

Figure 3: Using the same model of simple truss, the details would be the same as previous figure with 2 different supports profile. Unlike the joint method, here we only interested in finding the value of forces for member BC, EC, and ED.
Few simple guidelines of section truss analysis:

1. Pass a section through a maximum of 3 members of the truss, 1 of which is the desired member where it is dividing the truss into 2 completely separate parts,
2. At 1 part of the truss, take moments about the point (at a joint) where the 2 members intersect and solve for the member force, using ∑ M = 0,
3. Solve the other 2 unknowns by using the equilibrium equation for forces, using ∑ F_x = 0 and ∑ F_y = 0.

Note: The 3 forces cannot be concurrent, or else it cannot be solve.



 

Figure 4: A virtual cut is introduce through the only required members which is along member BC, EC, and ED. Firstly, the support reactions of R_a and R_d should be determine. Again a good judgment is require to solve this problem where the easiest part would be consider either on the left hand side or the right hand side. Taking moment at joint E (virtual pint) on clockwise for the whole RHS part would be much easier compare to joint C (the LHS part). Then, either joint D or C can be consider as point of moment, or else using the joint method to find the member forces for F_CB, F_CE, and F_DE. Note: Each value of the member’s condition should be indicate clearly as whether it is in tension (+ve) or in compression (-ve) state.

CENTROID OF COMPLEX GEOMETRIC FIGURES:

So in this articles, we are going to discuss the concepts of centroid for one dimensional as well as two dimensional objects. 

Let's first discuss about 1D and 2D objects, one by one, an 1D object is a line, practically a metallic rod will be considered as a linear, 1D object. Where as any thin plate of negligible thickness can be considered a 2D body. Suppose we have a thin metallic rectangular plate. If it is considered as a 2D rectangular area of b X h.

The concept of centroid has been developed on the basis of resultant of several areas. We know that an area can be represented as the cross product of two vectors, hence it is also an vector. Suppose we have an area A, in a cartesian 2D coordinate system. We just divide the area into n parts, and represent them as a₁, a₂, a₃, ..... a_n.

Let the elemental areas are at a distance x₁, x₂, x₃, ..... x_n, from Y axis and y₁, y₂, y₃, ...y_n from X axis.

The total moments produced about Y axis will be equal to the summation of all the individual moments produced by n elemental areas. 

Now moment is a vector quantity and we know vectors of same kind can be added together, therefore, all the n moment vectors can be added to get a single value of Resultant Moment. 

We also know this resultant moment's position vector. Let the resultant moment passes through a point G. The point through which resultant moment passes through is called Center of the Area or Centroid.

How can we find out the point G, whose coordinates are (X_g,Y_g)?

As moment of an area also obeys VARIGNON'S THEOREM OF MOMENT, then sum of all the moments produced by individual elemental areas will exactly be equal to the moment produced by the total area, i.e. the resultant of all those elemental areas. Now if all the areas are added to have the resultant area which will pass through the centroid G such that it produces a moment of X_gA about Y axis and Y_gA about X axis.

But Varignon's theorem states us that, for a vector system, resultant vector produces the moment about a point, is exactly equal to the sum of all the moments produced by all elemental areas about the same point and in the same plane. Hence, we can write now that,

Sum(a₁x₁+ a₂x₂+ + +a_nx_n) = AX_g
we can use summation sign ∑ to represent these equations,
∑a_ix_i = (∑a_i)X_g
=> X_g = (∑a_ix_i)/((∑a_i)

Sum(a₁y₁+ a₂y₂+ + +a_ny_n) = AY_g
∑a_iy_i = (∑a_i)Y_g
=> Y_g = (∑a_iy_i)/((∑a_i)

Algorithm to find out the Centroid G(X_g, Y_g) of a Complex Geometric Figure.

Step1:
Take a complex 2D figure like an Area or Lamina.

Step2:
Try to identify the basic figures whose algebraic combination produces our problem figure, whose centroid we shall find out.

Step3:
Choose a coordinate system, and make it as our frame of reference. All the distances and coordinate must be define with respect to our frame of reference.

Step4:
Compute the area (a_i), coordinates of their own centroid G_i (x_i, y_i) for each and every elemental areas. While measuring the centroids, all the measurements will be based on according to our chosen Axes.

Step5:
If any particular area has to subtracted to get the complex figure, the area will be negative, where as any area addition will be positive area.

Step6:
If the Centroid of the complex figure be G(X_g,Y_g)then,

=> X_g = (∑a_ix_i)/((∑a_i)

=> Y_g = (∑a_iy_i)/((∑a_i)

Here G₁ is the centroid of the part one where G₂ is the centroid of the circular area that has to be removed where as G₃ is the centroid of the triangular area that has to be removed also.

If we are asked to find moment of inertia of an area, which is nothing but the "second moment of area" then we shall have to find the centroidal moment of inertia first. Then we shall transfer the Moment of Inertia to another axis ie we shall apply parallel axis theorem to transfer moment of inertia from one axis (here centroidal axis) to another parallel axis.

INTELLIGENT OBJECTIVE QUESTIONS IN MECHANICS

1) A cantilever beam of square cross-section (100 mm X 100 mm) and length 2 m carries a concentrated load of 5 kN at its free end. What is the maximum normal bending stress at its mid-length cross-section?

(a) 10 N/mm²
(b) 20 N/mm²
(c) 30 N/mm²
(d) 40 N/mm²

2) A hollow shaft of outside diameter 40 mm and inside diameter 20 mm is to replaced by a solid shaft of 30 mm diameter. If the maximum shear stresses induced in the two shafts are to be equal, what is the ratio of the maximum resistible torque in the hollow to that of solid shaft?

(a) 10/9
(b) 20/9
(c) 30/9
(d) 40/9

(3) A cannonball is fired from a tower 80 m above the ground with a horizontal velocity of 100 m/s. Determine the horizontal distance at which the ball will hit the ground. (take g=10 m/s²)

(a) 400 m,
(b) 280 m,
(c) 200 m,
(d) 100 m.

(4) Water drops from a tap at the rate of four droplets per second. Determine the vertical separation between two consecutive drops after the lower drop attained a velocity of 4 m/s. Take g=10 m/s².

(a) 0.49 m
(b) 0.31 m
(c) 0.50 m
(d) 0.30 m

MULTIPLE CHOICE QUESTIONS: sub: engg. mechanics. Sub: Engineering Mechanics, Sub Code: EME-202, Semester: 2nd Sem, Course: B.Tech

Q.1) The example of Statically indeterminate structures are,
a. continuous beam,
b. cantilever beam,
c. over-hanging beam,
d. both cantilever and fixed beam.

Q.2) A redundant truss is defined by the truss satisfying the equation,
a. m = 2j - 3,
b. m < 2j + 3, 

c. m > 2j - 3,
d. m > 2j + 3

Q.3) The property of a material to withstand a sudden impact or shock is called,
a. hardness 
b. ductility, 
c. toughness, 
d. elasticity of the material

Q.4) The stress generated by a dynamic loading is approximately _____ times of the stress developed by the gradually applying the same load.

Q.5) The ratio between the volumetric stress to the volumetric strain is called as
a. young's modulus
b. modulus of elasticity
c. rigidity modulus,
d. bulk modulus

Q.6) In a Cantilever beam, the maximum bending moment is induced at
a. at the free end
b. at the fixed end
c. at the mid span of the beam
d. none of the above

Q.7) The forces which meet at a point are called
a. collinear forces
b. concurrent forces
c. coplanar forces
d. parallel forces

Q.8) The coefficients of friction depends upon
a. nature of the surface
b. shape of the surface
c. area of the contact surface
d. weight of the body

Q.9) The variation of shear force due to a triangular load on simply supported beam is
a. uniform 
b. linear 
c. parabolic 
d. cubic

Q.10) A body is on the point of sliding down an inclined plane under its own weight. If the inclination of the plane is 30 degree, then the coefficient of friction between the planes will be

a. 1/√3

b. √3

c. 1

d. 0

11. A force F of 10 N is applied on a mass of 2 kg. What is the acceleration of the mass?

A. 2 m/s²

B. 5 m/s²

C. 10 m/s²

D. 20 m/s²

Answer: B

12. What is the moment of a force of 50 N applied at a distance of 2 meters from a fixed point?

A. 25 Nm

B. 50 Nm

C. 100 Nm

D. 200 Nm

Answer: C

13. A 2000 kg car traveling at 20 m/s collides with a 500 kg car traveling at 10 m/s in the opposite direction. What is the velocity of the cars after the collision?

A. 6.7 m/s

B. 10 m/s

C. 13.3 m/s

D. 16.7 m/s

Answer: A

14. A 500 N force is applied to a 100 kg object on a flat surface. What is the coefficient of static friction if the object is just about to move?

A. 0.5

B. 0.7

C. 0.8

D. 1.0

Answer: D

15. A beam of length 4 m and moment of inertia of 1000 kg/m² is supported at each end. What is the maximum load that the beam can support if it is uniformly loaded?

A. 500 N

B. 1000 N

C. 2000 N

D. 4000 N

Answer: C

16. A block of mass 2 kg is hanging from a string. What is the tension in the string if the block is stationary?

A. 19.6 N

B. 20 N

C. 29.4 N

D. 30 N

Answer: B

17. A roller coaster car of mass 500 kg is traveling at 20 m/s at the bottom of a  loop-the-loop. What is the minimum radius of the loop required for the car to remain in contact with the track?

A. 40 m

B. 50 m

C. 60 m

D. 70 m

Answer: D

18. A body of mass 10 kg is moving with a velocity of 5 m/s. What is the kinetic energy of the body?

A. 50 J

B. 100 J

C. 125 J

D. 250 J

Answer: B

19. A body of mass 5 kg is placed on an inclined plane which makes an angle of 30° with the horizontal. What is the force acting on the body parallel to the plane?

A. 4.9 N

B. 7.5 N

C. 8.7 N

D. 10 N

Answer: B

20. A force of 100 N is applied on a body of mass 20 kg. What is the work done by the force in moving the body through a distance of 5 meters?

A. 250 J

B. 500 J

C. 1000 J

D. 2000 J

Answer: B

21. What is the principle of moments?

A. The sum of the moments about any point of a system in equilibrium is zero.

B. The sum of the forces acting on a system in equilibrium is zero.

C. The sum of the torques acting on a system in equilibrium is zero.

D. The sum of the accelerations of a system in equilibrium is zero.

Answer: A

22. What is the difference between static and dynamic equilibrium?

A. In static equilibrium, there is no motion, while in dynamic equilibrium, there is motion.

B. In static equilibrium, the forces are balanced, while in dynamic equilibrium, the forces are unbalanced.

C. In static equilibrium, the sum of the forces and moments is zero, while in dynamic equilibrium, the sum of the forces and moments is not zero.

D. In static equilibrium, the sum of the forces and moments is not zero, while in dynamic equilibrium, the sum of the forces and moments is zero.

Answer: C

23. What is the moment of inertia?

A. The resistance of an object to angular acceleration.

B. The force required to rotate an object.

C. The distance between the center of mass and the axis of rotation.

D. The angular velocity of an object.

Answer: A

24.What is the difference between stress and strain?

A. Stress is the deformation per unit length, while strain is the force per unit area.

B. Stress is the force per unit area, while strain is the deformation per unit length.

C. Stress is the force applied to an object, while strain is the resulting deformation.

D. Stress is the resistance of an object to deformation, while strain is the resistance of an object to stress.

Answer: B

25. What is Hooke's Law?

A. The stress applied to an elastic material is proportional to the strain produced.

B. The strain produced in an elastic material is proportional to the stress applied.

C. The deformation produced in an elastic material is proportional to the force applied.

D. The force applied to an elastic material is proportional to the deformation produced.

Answer: A

26.What is the difference between a beam and a truss?

A. A beam is a one-dimensional structure, while a truss is a two-dimensional structure.

B. A beam is made up of several members connected at their ends, while a truss is made up of several members connected at their joints.

C. A beam is used to support loads that are perpendicular to its axis, while a truss is used to support loads that are parallel to its axis.

D. A beam is a rigid structure, while a truss is a flexible structure.

Answer: B

27. What is the difference between a force and a moment?

A. A force is a vector quantity, while a moment is a scalar quantity.

B. A force is a scalar quantity, while a moment is a vector quantity.

C. A force is a push or a pull, while a moment is a twist or a turn.

D. A force is a linear motion, while a moment is a rotational motion.

Answer: C

28. What is the center of mass?

A. The point where the weight of an object is concentrated.

B. The point where the forces acting on an object are balanced.

C. The point where the moments acting on an object are balanced.

D. The point where the acceleration of an object is zero.

Answer: A

29. What is the method used to determine the forces in a truss?

A. Method of joints

B. Method of sections

C. Both A and B

D. None of the above

Answer: C

30. In a truss, which members are in tension and which members are in compression?

A. All members are in tension.

B. All members are in compression.

C. Members with angled force vectors are in tension, and members with vertical force vectors are in compression.

D. Members with vertical force vectors are in tension, and members with angled force vectors are in compression.

Answer: C

31. What is the difference between a simple truss and a compound truss?

A. A simple truss is made up of one triangle, while a compound truss is made up of two or more triangles.

B. A simple truss is made up of straight members only, while a compound truss may have curved members.

C. A simple truss is statically determinate, while a compound truss may be statically indeterminate.

D. A simple truss is used for short spans, while a compound truss is used for long spans.

Answer: A

32.How many unknown forces are there in a simple truss?

A. 2

B. 3

C. 4

D. It depends on the number of joints in the truss.

Answer: B

33. What is the method used to analyze a truss with multiple loadings?

A. Superposition method

B. Substitution method

C. Iterative method

D. None of the above

Answer: A

34. What is the maximum number of reactions that can be present in a truss?

A. 1

B. 2

C. 3

D. 4

Answer: B

35. What is the difference between a statically determinate and a statically indeterminate truss?

A. A statically determinate truss has only one solution for the unknown forces, while a statically indeterminate truss may have more than one solution.

B. A statically determinate truss has more unknown forces than the number of equations available to solve them, while a statically indeterminate truss has fewer unknown forces than the number of equations available to solve them.

C. A statically determinate truss is easier to analyze, while a statically indeterminate truss requires more advanced techniques.

D. A statically determinate truss is always more efficient than a statically indeterminate truss.

Answer: C

36. What is the difference between a pinned support and a roller support?

A. A pinned support allows rotation but not translation, while a roller support allows translation but not rotation.

B. A pinned support allows both rotation and translation, while a roller support allows neither.

C. A pinned support is used for horizontal loads, while a roller support is used for vertical loads.

D. A pinned support is always more stable than a roller support.

Answer: A

37. What is the maximum number of members that can be present in a simple truss?

A. 2n-2, where n is the number of joints

B. 2n-3, where n is the number of joints

C. n-1, where n is the number of joints

D. n+1, where n is the number of joints

Answer: B

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