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

PARALLEL AXIS THEOREM AND IT'S USES IN MOI

Moment Of Inertia of an Area.

download pdf file of solved numerical

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

CENTROID OF AN AREA

 

CENTROID OF AN AREA

Engineering Mechanics EME-102

Geometrical Center of an area (A) is often termed as Centroid or Center of an Area.

Suppose we have an area A in a certain X-Y coordinate system, we divide the area into n parts and named them as A₁, A₂, A₃, .... A_n,. Let the coordinates of those tiny elemental areas are as (X₁,Y₁), (X₂,Y₂), (X₃,Y₃) ..... (X_n,Y_n).

As area can be represented by a vector, hence, Area A can be treated as the resultant of the tiny elemental vectors A₁, A₂, A₃, .... A_n... Let the direction of the resultant vector passes through the point G(X_g,Y_g) on the plane of the area. The point G(X_g,Y_g) is called the CENTROID of the area A. (The direction of any area is along the perpendicular to the area drawn at the centroid of the area).

Like other vectors, an area has a moment about an axis and be represented by the product of the radial distance between the area and the axis and the area itself. So if an elementary area A₁ has a coordinate (X₁,Y₁) it means the area is at a distance X₁ from the Y axis and Y₁ from the X axis. Therefore the moment produced by A1 about Y axis is X₁A₁ and about X axis is Y₁A₁.

 Therefore the summation of all the moments produced by each and every elemental area about Y axis will be ∑A_iX_i and about X axis will be ∑A_iY_i.

Again, the resultant area A passes through the point G(X_g,Y_g). Therefore the moment produced by the area A about Y axis will be AX_g and about X axis will be AY_g.

Like other vectors, it will obey the Moment Theorem which states the total moment produced by individual vectors will be exactly equal to the moment produced by the resultant vector about a certain axis.

Therefore,
AX_g = A₁X₁ + A₂X₂ + A₃X₃ + ...... + A_nX_n

and
AY_g = A₁Y₁ + A₂Y₂ + A₃Y₃ + ...... + A_nY_n

           

                 For an area, a centroid G(X_g,Y_g) can be defined using calculus by the equations,

X_g = (1/A)∫x.dA   ------ (i)

Where dA = elemental area and A= total area.

Y_g = (1/A)∫y.dA   ------ (ii)

HOW TO DERIVE THE VALUES OF X_g and Y_g FOR BASIC GEOMETRIC FIGURE:

STEPS TO FIND X_g


i) Draw the figure in a Coordinate System.

ii) Draw a thin strip of area of thickness (dx) parallel to Y axis and at a distance (x) from Y axis.

iii) Find the height of the strip. Either the height will be constant or the height is a function of (x), that can be calculated from the equation of the figure.

iv) Calculate the elemental area of the strip, and named as dA. Hence, dA = hdx

v) integrate the expression ∫xdA, but dA = hdx. Therefore, we shall integrate  ∫hxdx over the total area.

vi) X_g = (∫ xdA)/A = (∫ hxdx)/A ; where A = total area = ∫dA = ∫hdx


STEPS TO FIND Yg

i) Draw the figure in a Coordinate System.

ii) Draw a thin strip of area of thickness (dy) parallel to X axis and at a distance (y) from X axis.

iii) Find the length (b) of the strip. Either the length will be constant or the length is a function of (y), that can be calculated from the equation of the figure.

iv) Calculate the elemental area of the strip, and named as dA. Hence, dA = bdy

v) integrate the expression ∫ydA, but dA = bdy. Therefore, we shall integrate ∫bydy over the total area.

vi) Y_g = (∫ ydA)/A = (∫bydy)/A





CENTROID OF A COMPOSITE AREA:




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

2D FORCE ANALYSIS : HOW TO FIND REACTIONS IN A CASE OF CONCURRENT FORCE SYSTEM ACTING ON A BODY IS IN EQUILIBRIUM

DEFINITION : CONCURRENT FORCE SYSTEM

If the lines of actions of all the forces in a force system pass through a common point, then the force system is called as Concurrent Force System. The equilibrium conditions for a concurrent force system is

ΣF_x = 0 and   ΣF_y = 0

 

The steps to find out reactions when a coplanar concurrent force system acting on a body in equilibrium condition :

 

 

STEP 1 :

 

(i) Draw the diagram and identify all the contact points the body makes with other bodies including ground.

(ii) Draw a tangent at each contact point with the object. These tangents are called Contact Surfaces.

(iii) Draw a perpendicular to the contact surface at each and every contact points. These perpendiculars will be the directions of reactions at each and every contact point.

(iv) Find the angles made by the reactions with respect to horizontal with the help of Geometry.

 

 

STEP 2 :

 

(i) Draw the Free Body Diagram (FBD) that consists of the external forces acting on the object. (applied forces, forces of gravity and reactions all are external forces)

(ii) Assign reactions by symbols like R₁, R₂ ....... and resolve all the external forces along X-axis and Y-axis.

(iii) Now add all the horizontal component forces as ΣF_x and put ΣF_x = 0 ---- eqn (1)

and add all the vertical component forces as ΣF_y and put ΣF_y = 0 --------eqn (2)

(iv) Solving these two equations we shall get values of  R₁, R_2.

 

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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