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# Multiple choice question for engineering

## Set 1

1. In the calculation of the radius of gyration, we use intensity of loadings. So whenever the distributed loading acts perpendicular to an area its intensity varies __________
a) Linearly
b) Non-Linearly
c) Parabolically
d) Cubically

Answer: a [Reason:] The load intensity is varying linearly in the structures. Thus the intensity is not varying parabolically nor is it cubically. It cannot be a vector also. Thus the intensity is linearly varied.

2. The radius of gyration is related to moment of the body. So the calculation of the moment of the body due to the loadings involve a quantity called ____________
a) Moment
b) Inertia
c) Moment of Inertia
d) Rotation

Answer: c [Reason:] The calculation of the moment of the body due to the loadings involve a quantity called moment of inertia. This is having much significance in the various fields in the engineering sector. The main types are the ‘I’ section structures which are being much used.

3. Find the moment of the force about the point R.

a) 5000Nm
b) 5550Nm
c) 6000Nm
d) 7000Nm

Answer: a [Reason:] As we know that the moment is the cross product of the distance and the force we will try to apply the same here. We see that the perpendicular distance is 5m. Thus we get the distance. And hence multiply it with the force, 5x2000xsin30. Because the force component perpendicular to the distance need to be taken.

4. How far from the shaft at P a 200N vertical force must act so as to create the same moment as produced by the 75N, at P?

a) 1.2m
b) 1.125m
c) 0.6m
d) 0m

Answer: b [Reason:] As we know that the moment is the cross product of the distance and the force we will try to apply the same here. So let the distance be y, then the force of 200N will act downward. The distance perpendicular is xcos60. Thus equating the 112.5 to 200ycos60.

5. Moment of Inertia is the integration of the square of the distance of the centroid and the del area along the whole area of the structure and this moment is at a distance from the rotating axis, known as radius of gyration.
a) True
b) False

Answer: a [Reason:] The moment of inertia of the section is the integration of the square of the distance of the centroid and the del area along the whole area of the structure. This is having much significance in the various fields in the engineering sector. The main types are the ‘I’ section structures which are being much used.

6. There is perpendicular axis theorem for the calculation of radius of gyration.
a) True
b) False

Answer: b [Reason:] There is no perpendicular axis theorem for the radius of gyration. In spite there is the theorem as parallel axis for any area. Thus we have the theorem which is used to add the two mutually perpendicular moment of inertias.

7. Radius of gyration is indirectly to the parallel axis theorem. The parallel axis theorem gives the moment of inertia ______________ to the surface of considesrance.
a) Linear
b) Non-Linear
c) Perpendicular
d) Parallel

Answer: c [Reason:] Parallel axis for any area is used to add the two mutually perpendicular moment of inertias for areas. It gives a moment of inertia perpendicular to the surface of the body. That is the moment of inertia perpendicular to the surface in considerance.

8. The parallel axis theorem can add any angle varied moment of inertias to give the perpendicular moment of inertia and the radius of gyration can be calculated.
a) True
b) False

Answer: b [Reason:] Parallel axis for any area is used to add the two mutually perpendicular moment of inertias for areas. It gives a moment of inertia perpendicular to the surface of the body. That is the moment of inertia perpendicular to the surface in considerance.

9. As radius of gyration is indirectly to the parallel axis theorem, parallel axis theorem uses the ____________ of the distance.
a) Square root
b) Square
c) Cube root
d) Cube

Answer: b [Reason:] Parallel axis for any area is used to add the two mutually perpendicular moment of inertias for areas. It gives a moment of inertia perpendicular to the surface of the body. And uses the square of the distance from the axis of rotation.

10. The distance in the parallel axis theorem is multiplied by:
a) Area
b) Volume
c) Linear distance
d) Area/Volume

Answer: a [Reason:] Parallel axis for any area is used to add the two mutually perpendicular moment of inertias for areas. It gives a moment of inertia perpendicular to the surface of the body. And uses the square of the distance from the axis of rotation multiplied by the area.

11. One of the use of the centre of mass or centroid is as in the radius of gyration is that the net force acts at the ___________ of the loading body.
a) Centroid
b) The centre axis
c) The corner
d) The base

Answer: a [Reason:] In the moment of inertia calculations we see that the net force acts at the centroid of the loading body. That is if the loading system is in the form of the triangle then the at the distance 2 by 3 of the base the net force of the loading will act. And the load will be half the area of the loading.

12. If the non-Uniform loading is of the type of parabola then for calculating the radius of gyration?
a) The net load will not be formed as all the forces will be cancelled
b) The net force will act the centre of the parabola
c) The net force will act on the base of the loading horizontally
d) The net force will act at the centroid of the parabola

Answer: d [Reason:] The net force will act at the centroid of the parabola. Whether it be a parabola or the cubic curve the centroid is the only point at which the net force act. Force can’t be acted horizontally if the loading is vertical. Hence whatever be the shape of the loading, the centroid is the point of action of net force. Thus the use of centroid.

13. If any external force also is applied on the structure and we are determining radius of gyration then what should we consider?
a) The net force will act at the centroid of the structure only
b) The net load will not be formed as all the forces will be cancelled
c) The net force will act on the base of the loading horizontally
d) The net force will not to be considered, there would be a net force of the distribution, rest will be the external forces

Answer: d [Reason:] The external forces are treated differently. They are not added by the force of the distributed loading. That is the force not only acts at the centroid always. It can be shifted also. Depending on the external forces. Thus the use of centroid or centre of mass.

14. The body is sometimes acted by two or three force members and we need to find radius of gyration for the same. The difference between the two and the three force members is:
a) The former is collinear and the latter is parallel
b) The former is parallel and the latter is perpendicular
c) The former is perpendicular and the latter is collinear
d) The former is acting on two points in the body while the latter is on three points

Answer: d [Reason:] The definition of the two force member only defines that the forces are being acted on the two points on the body. So does is the definition of the three forces members. The points of action of the three forces are three.

## Set 2

1. The uniformly distributed load is having two different values of load per unit length on the ends of the distribution.
a) True
b) False

Answer: b [Reason:] The answer is no. The uniform distributed load is not having two different values of the load per unit meter on the corner. Rather the distribution is uniform. That is it is having the same value of the same on both of the corners.

2. There are two types of loading. The uniformly distributed and the non-uniformly distributed that is the one having two different values at corners.
a) The first part of the statement is false and other part is true
b) The first part of the statement is false and other part is false too
c) The first part of the statement is true and other part is false
d) The first part of the statement is true and other part is true too

Answer: d [Reason:] The two forms of the loading are the uniformly distributed and the non-uniformly distributed that is the one having two different values at corners. The uniform distributed load is not having two different values of the load per unit meter on the corner.

3. The type of reduction of the loading is different for the uniformly distributed and the non-uniformly.
a) True
b) False

Answer: a [Reason:] The methods for the reduction of the loadings is different for both uniformly distributed and the non-uniformly. In the latter the simplification uses the quadratic equation. The latter also uses the integration for the same purpose.

4. Determine the magnitude of the resultant force acting on the shaft shown from left.

a) 640N
a) 675N
a) 620N
a) 610N

Answer: b [Reason:] The net force will act at the centroid of the parabola. Whether it be a parabola or the cubic curve the centroid is the only point at which the net force act. Force can’t be acted horizontally if the loading is vertical. Hence whatever be the shape of the loading, the centroid is the point of action of net force.

a) Centroid
b) The centre axis
c) The corner
d) The base

a) Tangent
b) Sine
c) Cosine
d) Sine inverse

7. Determine the magnitude of the resultant force acting on the shaft shown. The length is 2m.

a) 160N
b) 16N
c) 169N
d) 111N

Answer: a [Reason:] The net force will act at the centroid of the parabola. Whether it be a parabola or the cubic curve the centroid is the only point at which the net force act. Force can’t be acted horizontally if the loading is vertical. Hence whatever be the shape of the loading, the centroid is the point of action of net force.

8. If the non-Uniform loading is of the type of parabola then?
a) The net load will not be formed as all the forces will be cancelled
b) The net force will act the centre of the parabola
c) The net force will act on the base of the loading horizontally
d) The net force will act at the centroid of the parabola

Answer: d [Reason:] The net force will act at the centroid of the parabola. Whether it be a parabola or the cubic curve the centroid is the only point at which the net force act. Force can’t be acted horizontally if the loading is vertical. Hence whatever be the shape of the loading, the centroid is the point of action of net force.

9. If any external force also is applied on the distributed loading then?
a) The net force will act at the centroid of the structure only
b) The net load will not be formed as all the forces will be cancelled
c) The net force will act on the base of the loading horizontally
d) The net force will not to be considered, there would be a net force of the distribution, rest will be the external forces

Answer: d [Reason:] The external forces are treated differently. They are not added by the force of the distributed loading. That is the force not only acts at the centroid always. It can be shifted also. Depending on the external forces.

10. The resultant force acting, of the uniformly distributed loading is dependent on:
a) Area
b) Vertical distance
c) Length of the supports
d) The distance of the supports between them

Answer: a [Reason:] The resultant force acting, of the uniformly distributed loading is dependent on the area that the distribution is covering. The more the area the more is the force. That is the more is the tension created over the structure on which loading is kept. Hence the answer.

a) The latter has the involvement of the integration for the calculation of the net force
b) The former has the involvement of the integration for the calculation of the net force
c) The latter has the involvement of the differentiation for the calculation of the net force
d) The former has the involvement of the differentiation for the calculation of the net force

Answer: a [Reason:] The non-uniformly distributed loading is using the integration for getting the net force given by it. And the former only need the area or the linear distance of the loading. That is when the distance is multiplied by the uniformly distributed load the net force is obtained. But that is not so with the non-uniform one.

12. Determine the location of the resultant force acting on the shaft shown from left. The length is 2m.

a) 1.5m
b) 0.5m
c) 0.7m
d) 1.8m

Answer: a [Reason:] The net force will act at the centroid of the parabola. Whether it be a parabola or the cubic curve the centroid is the only point at which the net force act. Force can’t be acted horizontally if the loading is vertical. Hence whatever be the shape of the loading, the centroid is the point of action of net force.

13. Determine the location of the resultant force acting on the shaft shown from left.

a) 4.5m
b) 4m
c) 3.5m
d) 2m

Answer: b [Reason:] The net force will act at the centroid of the parabola. Whether it be a parabola or the cubic curve the centroid is the only point at which the net force act. Force can’t be acted horizontally if the loading is vertical. Hence whatever be the shape of the loading, the centroid is the point of action of net force.

15. Determine the magnitude of the resultant force acting on the shaft shown from left.

a) 1650N
b) 150N
c) 1250N
d) 1450N

Answer: a [Reason:] The net force will act at the centroid of the parabola. Whether it be a parabola or the cubic curve the centroid is the only point at which the net force act. Force can’t be acted horizontally if the loading is vertical. Hence whatever be the shape of the loading, the centroid is the point of action of net force.

## Set 3

1. What does the moment of the force measure in the rolling of the body?
a) The tendency of rotation of the body along any axis
b) The moment of inertia of the body about any axis
c) The couple moment produce by the single force acting on the body
d) The total work done on the body by the force

Answer: a [Reason:] The moment of the force measures the tendency of the rotation of the body along any axis, whether it be the centroid axis of the body, or any of the outside axis. The couple moment is produced by two forces, not by a single force. The total work done is the dot product of force and distance not the cross.

2. Determine the vertical force acting in the given figure.

a) 236N
b) 600n
c) 403N
d) 830N

Answer: a [Reason:] The net forces acting on the body is shown by the help of the resultant forces. There are two types, first the frictional and the second is the normal. This is because the resultant forces have the sum of all the forces which are acting on the direction which is same.

3. If a car is moving forward, what is the direction of the moment of the moment caused by the rolling of the tires, assume non slippery surface?
a) It is heading inwards, i.e. the direction is towards inside of the car
b) It is heading outwards, i.e. the direction is towards outside of the car
c) It is heading forward, i.e. the direction is towards the forward direction of the motion of the car
d) It is heading backward, i.e. the direction is towards back side of the motion of the car

Answer: a [Reason:] When you curl your wrist in the direction in which the tires are moving then you will find that the thumb is pointing outwards. That is outwards the body of the car. This phenomena is also observed in rainy seasons. When cars travel on the roads, the water is thrown outside from the tires, due to moment.

4. The moment of the force is the product of the force and the perpendicular distance of the axis and the point of action of the force. Is this also true for rolling?
a) True
b) False

Answer: a [Reason:] The moment is the product of the force applied to the body and the perpendicular distance of the point of action of the force to the axis about which the body is being rotated. That is the moment is the cross product of the force and the distance between the axis and the point of action. Yes this is also true for rolling.

5. For the rolling of the body the calculation of the moment of the force about the axis of rolling, the cross product table, i.e. the 3X3 matrix which is made for doing the cross product having 3 rows, contains three elements. Which are they from top to bottom?
a) Axis coordinates, point coordinates and the force coordinates
b) Point coordinates, axis coordinates, and the force coordinates
c) Axis coordinates, force coordinates and the point coordinates
d) Force coordinates, point coordinates and the axis coordinates

Answer: a [Reason:] The 3X3 matrix which is being made is having axis coordinates, point coordinates and the force coordinates. They are from top to bottom placed. The order cannot be changed. Or if changed then one needs to apply the negative sign appropriately. Negative because the directions gets reversed.

6. Find the moment of the force about the point R.

a) 5000Nm
b) 5550Nm
c) 6000Nm
d) 7000Nm

Answer: a [Reason:] As we know that the moment is the cross product of the distance and the force we will try to apply the same here. We see that the perpendicular distance is 5m. Thus we get the distance. And hence multiply it with the force, 5x2000xsin30. Because the force component perpendicular to the distance need to be taken.

7. In rolling there is the involvement of the vector math. So for rolling which of the following is correct? (For A representing the vector representation of the axis of rotation, r the radius vector and F the force vector)
a) A.(rxF)
b) Ax(rxF)
c) A.(r.F)
d) Fx(r.F)

Answer: a [Reason:] The correct form of the equation is given by A.(rxF). Where A represents the vector representation of the axis of rotation, r the radius vector and F the force vector. This is usually done for determining the moment of the force about the axis. That is if body is being rotated by the force about an axis.

8. In the equation A.(rxF), the r vector is what?
a) The magnitude of the axis, i.e. the length of the axis
b) The length of the force vector
c) The length of the radius

Answer: d [Reason:] The r in the equation A.(rxF) is the radius vector of the rolling body. That is it is the vector which is having the start point at the axis of rolling and the end point at the point of action of the force on the rolling body. This vector is being crossed by the force vector, which is then followed by the dot product with the axis vector.

9. The ___________ forces do not cause the rotation/rolling of the body if the rotation is considered in about the axis of the body or the centroid axis of the body.
a) Non-concurrent
b) Concurrent
c) Parallel
d) Non-Parallel

Answer: b [Reason:] The concurrent forces are the which are somewhere touching the axis of rotation. If any of the force is touching that axis, that force is not considered, or is insufficient to cause a rotation. If a force is concurrent then the perpendicular distance of the force from the line of axis is zero, thus no rotation. As we know rotation is caused by moment.

10. For the rolling of the body right handed coordinate system means (consider the mentioned axis to be positive)?
a) Thumb is z-axis, fingers curled from x-axis to y-axis
b) Thumb is x-axis, fingers curled from z-axis to y-axis
c) Thumb is y-axis, fingers curled from x-axis to z-axis
d) Thumb is z-axis, fingers curled from y-axis to x-axis

Answer: a [Reason:] As right handed coordinate system means that you are curling your fingers from positive x-axis towards y-axis and the thumb which is projected is pointed to the positive z-axis. Thus visualizing the same and knowing the basic members of axis will not create much problem. The right handed coordinate system is universal throughout.

11. In the equation Wa/r generally used in the rolling frictional calculations, what does each stands for?
a) P is the force, W is weight of the body, r radius and a coefficient of rolling friction all in normal powers
b) P is the force in nano Newton, W is weight of the body, r radius and a coefficient of rolling friction
c) P is the force, W is weight of the body in kg, r radius and a coefficient of rolling friction
d) P is the force, W is weight of the body, r radius cm and a coefficient of rolling friction

Answer: a [Reason:] The rolling of the cylinder is termed as perfect rolling only when the cylinder is rolling at a constant velocity. Also the rolling should be along the rigid surface. The normal forces are acting perpendicular to the surface of the rolling. And for rolling equation Wa/r is generally used.

12. Which statement is correct about the vector F acting parallel to the direction of the motion of the rolling body?
a) F= Fcos β + Fcos α + Fcosγ
b) F= Fsin β + Fcos α + Fcosγ
c) F= Fcos β + Fsin α + Fcosγ
d) F= Fcos β + Fcos α + Fsinγ

Answer: a [Reason:] As we know the α, β and γ are the angles made by the x, y and z-axis respectively. Thus, is the magnitude of the vector is F, the F= Fcos β + Fcos α + Fcosγ. Which means the force is the resultant of all its axis’ components. The force driving the rolling motion of the body can be divided like this.

13. What if the moment of the force calculated about the axis of rolling is negative?
a) It means that the force is applied in the opposite direction as imagined
b) It means that the direction of the motion is in the opposite sense as imagined
c) It means that the radius vector is in the opposite sense as imagined
d) Such calculation means that the calculations are wrongly done

Answer: b [Reason:] It means that the direction of the rolling is in the opposite sense as imagined. We can’t say about the direction of the force or the direction of the radius vector. But yes we can say about the direction of the rolling as it is the thing which is going to be calculated. Rest all the parts are fixed. They can’t be altered.

14. In the equation A.(rxF) the r is heading from ______________ and ending at _____________
a) Axis of rolling, Force vector
b) Axis of rolling, Force vector’s point of action on the body
c) Force vector, Axis of rolling
d) Force vector’s point of action on the body, Axis of rolling

Answer: b [Reason:] It is the radius vector. The radius vector is always from the axis of rolling to the point of action of the force on the body. Which means that the radius vector is not on any point on the force vector. Rather it ending at the point on the force vector, where it is being in contact of the body.

15. Rotation is termed as rolling when _______
a) Constant velocity of rolling
b) Variable velocity of rolling
c) Constant speed of rolling
d) Variable speed of rolling

Answer: a [Reason:] The rolling of the cylinder is termed as perfect rolling only when the cylinder is rolling at a constant velocity. Also the rolling should be along the rigid surface. The normal forces are acting perpendicular to the surface of the rolling.

## Set 4

1. ___________ are the structures that are being made to support the loadings applied perpendicular to the axis of that structures.
a) Pillar
b) Box
c) Beam
d) Statues

Answer: c [Reason:] The beams are the structures which are being made so as to support the loadings which are perpendicular to the axis of that structure. This means that the beams can support that weight or that types of weight which are applying the load perpendicular to the axis of the beam.

a) Straight
b) Vertical
c) Rounded
d) Curled

Answer: a [Reason:] Beams are long and straight structures that are being made to support the loadings. They are being made so as to support the loadings which are perpendicular to the axis of that structure. This means that the beams can support that weight or that types of weight which are applying the load perpendicular to the axis of the beam.

3. Beams have a _________ cross sectional area.
a) Non-uniform
b) Rounded
c) Helical
d) Uniform

Answer: d [Reason:] Beams have a uniform cross sectional area. This is because beams are the structures which are being made so as to support the loadings which are perpendicular to the axis of that structure. And to support properly the area must be uniform.

4. Simply supported beams are pinned at both the ends.
a) True
b) False

Answer: b [Reason:] The simply supported beams are supported only at one side by the pin. This means the other side is having the roller. The beams are such designed because they are the structures which are being made so as to support the loadings which are perpendicular to the axis of that structure. Thus not pinned both sides.

5. The other side of the simply supported beam is having pin support, what is the support this side?
a) Roller
b) Pin
c) Hinge
d) Rolling hinge

Answer: a [Reason:] The simply supported beams are supported only at one side the pin. This means the other side is having the roller. The beams are such designed because they are the structures which are being made so as to support the loadings which are perpendicular to the axis of that structure. Thus not pinned both sides.

6. The cantilever beam is having pin supports both sides of it.
a) True
b) False

Answer: b [Reason:] The one end of the cantilever beam is fixed and the other one is having its end as free. This is the other type of the beam which is being designed to support the loadings which are perpendicular to the support. Thus the cantilever is free from one end and fixed at another.

7. The cantilever is having one side of it as fixed while the other one is _________
a) Hinged
b) Roller supported
c) Pinned
d) Free

Answer: d [Reason:] The one end of the cantilever beam is fixed and the other one is having its end as free. This is the other type of the beam which is being designed to support the loadings which are perpendicular to the support. Thus the cantilever is free from one end and fixed at another.

8. The design of the beam requires the knowledge of the variation of the ____________
a) Internal shear force
b) Rolling forces
c) Rotational forces
d) External forces

Answer: a [Reason:] The design of the beam requires the knowledge of the variation of the internal shear force. Because beams are the structures which are being made so as to support the loadings which are perpendicular to the axis of that structure. And to support properly the area must be uniform.

9. The design of the beam requires the knowledge of the variation of the ____________
a) Bending moment
b) Rolling forces
c) Rotational forces
d) External forces

Answer: a [Reason:] The design of the beam requires the knowledge of the variation of the bending moment. Because beams are the structures which are being made so as to support the loadings which are perpendicular to the axis of that structure. And to support properly the variation of the bending moments must be known for each of the section.

10. The bending moment and the shearing stress can be obtained by which method?
a) Bending moment
b) Rolling method
c) Section method
d) Method of section

Answer: d [Reason:] The design of the beam requires the knowledge of the variation of the bending moment. Because beams are the structures which are being made so as to support the loadings which are perpendicular to the axis of that structure. Thus to know these quantities we apply the method of sections so as to determine these values easily.

11. We can make the graph of shear force acting on the beam. But not for the bending moment.
a) The first part of the statement is false and other part is true
b) The first part of the statement is false and other part is false too
c) The first part of the statement is true and other part is false
d) The first part of the statement is true and other part is true too

Answer: a [Reason:] Both of them can be plotted on the graph as the main things are the distance variation of the force, which can be made by simple calculation. The main motto of the calculations are to design the graph and know where the maximum force and the moment are. Thus the graph of both the quantities are possible.

12. We apply the equations of ________ to determine various forces acting on the beams.
a) Equilibrium
b) Rotation moment
c) Linear moment
d) Translation

Answer: a [Reason:] The force developed by a support doesn’t allow the translation of its attached member. This is the basic condition for the equilibrium of the forces in any dimension. And many other are applied at the points where the forces are to be determined. Thus equilibrium equations are being applied at the points where the main forces are to be determined.

13. Which of the following is true for a beam having various forces acting over it?
a) If we are considering the clockwise direction to be positive then the rotation along the counter clockwise direction is negative
b) If we are considering the clockwise direction to be negative then the rotation along the counter clockwise direction is also negative as no such assumptions can be taken
c) If we are considering the clockwise direction to be positive then the rotation along the counter clockwise direction is also positive as no such assumptions can be taken
d) If we are considering the clockwise direction to be negative then the rotation along the counter clockwise direction is neither positive nor negative as we only pre assumed the clockwise direction

Answer: a [Reason:] We can take such assumptions. We are free to do so. We just need to specify if clockwise is positive then the counter clockwise is to be negative. Also such assumption are very useful in 2D calculation of the moments caused by various forces on the body.

14. Determine the shear force generated.

a) 1.35KN
b) 25KN
c) 22KN
d) 23KN

Answer: a [Reason:] The loading acting upward is considered to be positive. This is done as to decrease the time used for the calculations. If the method of sections is being applied at the complex sections for the determination of the unknown forces, much time is required for it. Thus the main motto is to make the calculations easy and make the less use of time.

15. Determine the moment generated at R.

a) 18.75KNm
b) 8.75KNm
c) 1.75KNm
d) 175KNm

Answer: a [Reason:] The loading acting upward is considered to be positive. This is done as to decrease the time used for the calculations. If the method of sections is being applied at the complex sections for the determination of the unknown forces, much time is required for it. Thus the main motto is to make the calculations easy and make the less use of time.

## Set 5

1. The determination of the internal loading in the beams for shear stress diagrams is usually done so as to:
a) Break the beam
b) Know the length
c) Know the diameter
d) Design the beam

Answer: d [Reason:] The determination of the internal forces in the beam is done so as to design the beams as in the application purpose the beams will be subjected to many loads. This will help us to make the beam properly. And also this will ensure that the beams will not break after the loading is done on them. For this another method is the distribution load method.

2. The change in moment in shear stress diagrams is equal to _________
a) Rotational moment
b) Bending moment
c) Total weight
d) Area under shear diagram

Answer: d [Reason:] After the application of the force equation of equilibrium to the segment of the beam, we have the above result. This is done on the very small part of the beam. That is the minimal section of the beam is to be considered and then the application of the equilibrium equations are done so as to calculate the final result.

3. Determine the shear force of the beam shown.

a) 450N
b) 50N
c) 40N
d) 45N

Answer: a [Reason:] The loading acting upward is considered to be positive. This is done as to decrease the time used for the calculations. If the method of sections is being applied at the complex sections for the determination of the unknown forces, much time is required for it. Thus the main motto is to make the calculations easy and make the less use of time.

4. The couple moment and the force in shear stress diagrams is divided to get the distance of the axis from the point of action of the force.
a) True
b) False

Answer: a [Reason:] The couple is the cross product of the force and the perpendicular distance between the forces. And if the division of the both is done than the resulting quantity is the distance only. But care must be taken that the product is the cross product and hence sine angle must be divided too to get the distance.

5. The slope of the shear diagram is equal to__________
a) Rotational moment
b) Bending moment
c) Total weight

Answer: d [Reason:] After the application of the force equation of equilibrium to the segment of the beam, we have the above result. This is done on the very small part of the beam. That is the minimal section of the beam is to be considered and then the application of the equilibrium equations are done so as to calculate the final result.

6. The equation of change in moment in shear stress diagrams equals area under shear diagram is not applied when?
a) Rotational moment
b) Bending moment
c) Total weight
d) A concentrated force acts

Answer: d [Reason:] As this is done with the help of distributed load method thus when a concentrated load is applied this equation is not valid. This is done on the very small part of the beam. That is the minimal section of the beam is to be considered and then the application of the equilibrium equations are done so as to calculate the final result.

7. Change in the shear in shear stress diagrams is equal to ____________
a) Rotational moment
b) Bending moment
c) Total weight

Answer: d [Reason:] After the application of the force equation of equilibrium to the segment of the beam, we have the above result. This is done on the very small part of the beam. That is the minimal section of the beam is to be considered and then the application of the equilibrium equations are done so as to calculate the final result.

8. Determine the normal force of the beam shown.

a) 0N
b) 50N
c) 40N
d) 45N

Answer: a [Reason:] The loading acting upward is considered to be positive. This is done as to decrease the time used for the calculations. If the method of sections is being applied at the complex sections for the determination of the unknown forces, much time is required for it. Thus the main motto is to make the calculations easy and make the less use of time.

9. The equation of change in moment in shear stress diagrams equals area under shear diagram is not applied when?
a) Rotational moment
b) Bending moment
c) Total weight
d) Couple moment acts

Answer: d [Reason:] As this is done with the help of distributed load method thus when a couple moment this equation act is not valid. This is done on the very small part of the beam. That is the minimal section of the beam is to be considered and then the application of the equilibrium equations are done so as to calculate the final result.

10. The couple in the beam is simplified easily by the help of right hand rule in shear stress diagrams. But the forces simplification is not possible as there is no such system of rules so that the forces can be simplified.
a) The first part of the statement is false and other part is true
b) The first part of the statement is false and other part is false too
c) The first part of the statement is true and other part is false
d) The first part of the statement is true and other part is true too

Answer: c [Reason:] Both of them are vector quantities. And both of them can be easily simplified. If taken in the vector form then the task is even easier. Thus it is not necessary for the force or the couple to be vector only, even if the magnitude is taken, the simplification is done in the 2D. Forces are simplified on the basis of the algebra.

11. Slope of the moment diagram is equal to _________
a) Rotational moment
b) Bending moment
c) Total weight
d) Shear

Answer: d [Reason:] After the application of the force equation of equilibrium to the segment of the beam, we have the above result. This is done on the very small part of the beam. That is the minimal section of the beam is to be considered and then the application of the equilibrium equations are done so as to calculate the final result.

12. The internal loading in shear stress diagrams can be found by ___________
b) Method of area
c) Method of line
d) Method of volume

Answer: a [Reason:] The determination of the internal forces is done by various methods among which the distribution of load method is the one. Also we have method of joints, which is done so as to determine the internal forces which are being developed in the trusses or the frames. This is done as to design them. In this method the loading acting upward is considered to be positive.

13. In the making of the shear force diagram or the bending moment diagrams, one method is used, distributed load. In this the distributed load will be considered positive if?

Answer: a [Reason:] The loading acting upward is considered to be positive. This is done as to decrease the time used for the calculations. If the method of sections is being applied at the complex sections for the determination of the unknown forces, much time is required for it. Thus the main motto is to make the calculations easy and make the less use of time.

14. The shear diagram will jump downward if _________
a) Rotational moment
b) Bending moment
c) Total weight
d) Force will act downwards