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.

SOLUTION OF PRESEMESTER EXAMINATION: ENGINEERING MECHANICS SOLUTION OF PRESEMESTER EXAMINATION: ENGINEERING MECHANICS©subhankar_karmakar

GROUP-A Q.1 Answer the following questions as per the instructions 2x20=20 Choose the correct answer of the following questions: (i) A truss hinged at one end, supported on rollers at the other, is subjected to horizontals load only. Its reaction at the hinged end will be (a) Horizontal;                   (b) Vertical (c) Both horizontal & vertical (d) None of the above. Ans: (c) both horizontal & vertical (ii) The moment of inertia of a circular section of diameter D about its centroidal axis is given by the expression (a) π(D)4/16           (b) π(D)4/32 (c) π(D)4/64           (d) π(D)4/4 Ans: (c) π(D)4/64 Fill in the blanks in the following questions: (iii)The distance of the centroid of an equilateral triangle with each side(a) is …………. from any of the three sides. Ans a /(2√3) (iv)Poisson’s ratio is defined as the ratio between ……. and ………… . Ans: Lateral Strain, Longitudinal Strain (v)If two forces of equal magnitudes P having an angle 2Ө between them, then their resultant force will be equal to ________ . Ans: 2P CosӨ Choose the correct word/s. (vi) Two equal and opposite force acting at different points of a rigid body is termed as (Bending Moment/ Torque/ Couple). Ans: Couple

Choose correct answer for the following parts: (vii) Statement 1: In stress strain graph of a ductile material, yield point starts at the end of the elastic limit. Statement 2: At yielding point, the deformation becomes plastic by nature. (a) Statement 1 is true, Statement 2 is true. (b) Statement 1 is true, Statement 2 is true and they are unrelated with each other (c) Statement 1 is true, statement 2 is false. (d) Statement 1 is false, Statement 2 is false. Ans: (a) Statement 1 is true, Statement 2 is true. (viii) Statement 1: It is easier to pull a body on a rough surface than to push the body on the same surface. Statement 2: Frictional force always depends upon the magnitude of the normal force. (a) Statement 1 is true, Statement 2 is true. (b) Statement 1 is true, Statement 2 is true and they are unrelated (c) Statement 1 is true, statement 2 is false. (d) Statement 1 is false, Statement 2 is false. Ans: (a) Statement 1 is true, Statement 2 is true. (ix) In a cantilever bending moment is maximum at (a) free end            (b) fixed end (c) at the mid span (d) none of these Ans: (b) fixed end (x) The relationship between linear velocity and angular velocity of a cycle (a) exists under all conditions (b) does not exist under all conditions (c) exists only when it moves on horizontal plane. (d) none of these Ans: (a) exists under all conditions

NEXT ARTICLE

SOLUTION OF PRESEMESTER EXAMINATION: ENGINEERING MECHANICS PART-II ©subhankar_karmakar

MULTIPLE CHOICES : ENGG MECHANICS Sub: Engineering Mechanics, Sub Code: EME-202, Semester: 2nd Sem, Course: B.Tech,

MULTIPLE CHOICES:
sub: ENGINEERING MECHANICS

Sub Code: EME - 102 /201
B.Tech First Semester (all branch)
University: UPTU (GBTU), Lucknow, Uttar Pradesh
OBJECTIVE TYPE QUESTIONS

PROBLEM SET - 1

Q1) If the coefficients of friction of an inclined plane be (1/√3), then the angle of repose of the plane will be
a) 90° b) 60° c) 45° d)30°

Q2) Three forces of equal magnitude acts along the side of an equilateral triangle, then the body will be in
a) static equilibrium                                 b) dynamic equilibrium
c) translational motion                             d) rotational motion.

Q3) (1/2).E.e² is called
a) total strain energy                                  b) resilience
c) dynamic loading energy                          d) none of this

Q4) The ratio of shear stress and shear strain is known as,
a) modulus of elasticity                                      b) poissons' ratio
c) modulus of rigidity                                         d) bulk modulus

Q5) The bending moment curves generated by UVL is,
a) parabolic                                          b) straight line
c) cubic                                                d) constant

Q6) The centroidal moment of inertia of a quarter circle lamina is
a) 0.11r⁴                                              b) 0.05r⁴
c) 0.4242r⁴                                          d) 0.5868r⁴

Q7) The Point of Contraflexure is a point in the beam where
a) shear force is zero                        
b) bending moment is zero
c) bending moment is zero and it changes sign
d) none of these

Q8) If a body of mass M is moving with an acceleration (a) then the Inertial Force on the body is equals to
a) -Ma          b) Ma        c) |Ma|     d) none of these

Q9) A truss is made of seven linkages and five joints, then the truss is
a) deficient                                              b) redundant
c) perfect                                                d) none of these

Q10) A beam has 3 no.s of supports is known as
a) cantilever beam
b) continuous beam
c) overhanging beam
d) simply supported beam


PROBLEM SET - 2:

Q1) The maximum value of frictional force that comes into play when a body tends to move on a surface called:
a) sliding friction,                                   b) limiting friction,
c) milling friction,                                   d) none of these.

2) The ratio of static friction to dynamic friction is:
a) less than 1,                                      b) equal to 1,
c) greater than 1,                                 d) none of these.

3) The angle of friction is equal to the:
a) ratio of frictional force to the normal reaction.
b) angle of inclined plane when a body tends to slide down.
c) angle of an inclined plane when a body is sliding.
d) none of these.

Q.4) A particle is moving along a circle with constant speed. The acceleration of the particle is:
a. along the circumference                      b. along the tangent
c. along the radius                                 d. zero

Q.5) The area moment of inertia of a quarter circle of radius (r) about the centroidal
axes is
a. (0.055r²)                                        b. (0.11r²)/3
c. (πr²)/4                                           d. π.r²

Q.6) The centroid of a semi circular area of radius r is
a. (r, 3r/4π )                                      b. ( r, 2r/π)
c. (r/π, r)                                           d. (r, r/2π)

Q.7) In the truss analysis, if the no. of linkages is 12 and the no. of joints is 7, then
a. perfect truss                            b. redundant truss
c. deficient truss                          d. indeterminate truss

Q.8) When shear force is zero at a point in a beam, bending moment at a certain point is,
a. zero                                                 b. maximum
c. minimum                                          d. increasing

Q.9) The mass moment of inertia of a solid sphere of mass M and radius R about an diameter of the sphere will be
a. (1/5).M.R²              b. (2/5).M.R⁴           c. (2/5).M.R²       d. (1/5).M.R⁴

Q.10) In a simply supported beam of length L, a concentrated load W acts at the mid span, the maximum bending moment would be
a. W.L/4                                                   b. W.L²
c. W.L/2                                                   d. 0

PROBLEM SET - 3

Q1) A couple can be balanced by
a) a direct Force                                    b) a moment
c) a torque                                            d) an equal and opposite couple.

Q2) Opening a Cold drinks bottle, one has to apply
a) a moment                                            b) a torque
c) parallel forces                                      d) a couple

Q3) A truss is said to be plane truss if
a) all the members lies in one plane
b) any two members lies in one plane
c) all the members are perpendicular to one plane
d) none of these

Q4) The ratio of frictional force normal reaction is called
a) angle of repose
b) angle of friction
c) limiting friction
d) coefficient of friction

Q5) Yielding means
a) elastic deformation
b) plastic deformation
c) fatigue failure
d) shear failure

Q6) The force required to move a body up an inclined plane will be least when the angle of inclination is:
a) equal to friction angle,
b) greater than friction angle,
c) less than friction angle,
d) none of these.

Q7) In a cantilever beam loaded with a point load at the free end, has maximum bending moment,

(i) free end,                                            (ii) fixed end,
(iii) at the mid span,                                (iv) none of the above.

Q8) An idealized truss can only be loaded at
a) only at the mid point of the linkages
b) only at the joints
c) both at the mid points and joints
d) none of the above

Q9) UTM can be used to find
a) Modulus of Elasticity
b) compressive breaking strength
c) ultimate tensile strength
d) all of the above

||| the question paper is made by subhankar karmakar, 2011 |||

Q10) Bulk Modulus is defined as the ratio between
(i) linear stress and strain,                 (ii) shear stress and strain,
(iii) volumetric stress and strain,        (iv) none of the above.

PROBLEM SET - 4:

Q.1 Answer the following questions as per the instructions 2x20=20

Choose the correct answer of the following Questions:

(i) The magnitudes of the force of friction between two bodies, one lying above the another depends upon the roughness of the
(a) Upper body                               (b) Lower body
(c) Both the body                           (d) The body having more roughness

(ii) The moment of inertia of a circular section of diameter D about its centroidal axis is given by the expression
(a) π(D)⁴/16
(b) π(D)⁴/32
(c) π(D)⁴/64
(d) π(D)⁴/4

Fill in the blanks in the following questions:

(iii) The distance of the centroid of an equilateral triangle with each side (a) is …………. From any of the three sides.

(iv) Poisson’s ratio is defined as the ratio between ……………. and…………………………

(v) If two forces of equal magnitudes P having an angle 2Ө between them, then their resultant force will be equal to ________

Match the following columns for the following two parts:

(vi) Match the column I to an entry from the column II:

COLUMN I----                            ---COLUMN II

(i) BMD of an UDL                      (a) stored strain energy per unit volume
(ii)Resilience is                            (b) brittle materials
(iii) Bulk Modulus                        (c) parabolic in nature
(iv) YieldPoint                             (d) volumetric stress and strain
                                                  (e) Ductile materials
                                                  (f) Shear stress

(vii) Match the Following columns:

COLUMN I  ------------------       COLUMN II

(i) Square of side b                                              (p) π b⁴ / 64

(ii) Equilateral Triangle of side b                           (q) b⁴ / 12

(iii) Circle of diameter b                                       (r) b⁴/ 36

(iv) Isosceles right angle triangle of base b            (s) b⁴/(32√3)

Column II gives the value of Moment of Inertia I_xx about a centroidal axis.

Choose correct answer for the following parts:

(viii) Statement 1:
In stress strain graph of a ductile material, yield point starts at the end of the elastic limit.
Statement 2:
At yielding point, the deformation becomes plastic by nature.

(i) Statement 1 is true, Statement 2 is true.
(ii) Statement 1 is true, Statement 2 is true and they are unrelated with each other
(iii) Statement 1 is true, statement 2 is false.
(iv) Statement 1 is false, Statement 2 is false.

(ix) Statement 1:
It is easier to pull a body on a rough surface than to push the body on the same surface.
Statement 2:
Frictional force always depends upon the magnitude of the normal force.
(i)Statement 1 is true, Statement 2 is true.
(ii)Statement 1 is true, Statement 2 is true and they are unrelated with each other
(iii) Statement 1 is true, statement 2 is false.
(iv) Statement 1 is false, Statement 2 is false.

Choose the correct word/s.
(x) Two equal and opposite force acting at different points of a rigid body
is termed as (Bending Moment/ Torque/ Couple).

PROBLEM SET - 5

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 ratio of the stress generated by a dynamic loading to the stress developed by the gradually applying the same load is
a) 1           b) 2               c) 3                    d) none of the above

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

PROBLEM SET - 6:

Q.1) Moment of Inertia of a body means
a) The resistance of the body against any movement or the tendency of movement of the body
b) The resistance of the body against any rotation of the body about a certain point
c) The force multiplied by the distance of the force from the body
d) None of these

Q.2) A truss is a
a) load bearing structure
b) framed mechanical structure
c) structure made of linkages and joints
d) all of the above

Q.3) The type of joints that can resist a moment is called as
a. roller joint                                            b. pin joint
c. hinged joint                                          d. fixed joint

Q.4) The slope of the normal stress-strain graph is equal to the
a. modulus of rigidity
b. bulk modulus
c. modulus of elasticity
d. stiffness constant.

Q.5) The shear stress in a circular shaft under torsion varies
a. linearly                                                  b. parabolically
c. hyperbolically                                        d. uniformly

Q.6) If for a material poissons ratio is 0.2 and modulus of elasticity is 200 GPa, then the value of modulus of rigidity would be
a) 100.33 GPa                                          b) 93.33 GPa
c) 83.33 GPa                                            d) 100 GPa

Q.7) If the section modulus of a beam decreases, then bending stress will
a. decrease                                   b. increase
c. remain same                             d. bending stress is independent of section modulus

Q.8) Which one of the following statements is correct?
a. Energy and work are scalars
b. Force and work are vectors
c. Energy, momentum and velocity are vectors
d. Force, momentum and velocity are scalars

Q.9) Stress may be defined as:
a. load per unit area
b. external force per unit area
c. internal resistance per unit area
d. same as pressure

Q.10) Hooke's law is valid up to the
a. yield point                                           b. elastic limit
c. proportional limit                                 d. ultimate point.

PROBLEM SET - 7:

Q.1) The first law of motion provides the definition of
a. momentum                                         b. force
c. energy                                               d. acceleration

Q.2) A man in a lift will weigh more when the lift is
a. accelerated upwards
b. accelerated downwards
c. descends freely
d. lift going up is slowing down

Q.3) The motion of a bicycle wheel is
a. linear                                                   b. rotary
c. translatory                                           d. rotary as well as translatory

Q.4) A particle is moving along a circle with constant speed. The acceleration of the particle is :
a. along the circumference
b. along the tangent
c. along the radius
d. zero

Q.5) The mass moment of inertia of a thin disc of mass (m) and radius (r) about the centroidal axes is
a. (m.r²)/2                                                  b. (m.r²)/3
c. (m.r²)/4                                                  d. m.r²

Q.6) The centroid of a semi circular arc of radius r is
a. (3r)/π                                                     b. 2r/π
c. r/π                                                        d. r/(2π)

Q.7) In the method of sections for trusses, the section must be passed so as to cut not more than
a. two members
b. three members
c. four members
d. five members

Q.8) When bending moment at a certain point is maximum, shear force is
a. zero b. maximum
c. minimum d. increasing

Q.9) The mass moment of inertia of a solid sphere of mass M and radius R about an diameter of the sphere will be
a. (1/5).M.R²
b. (2/5).M.R⁴
c. (2/5).M.R²
d. (1/5).M.R⁴

Q.10) In a cantilever beam of length L, a concentrated load W acts at the free end, the bending moment at the free end would be
a. W.L                              b. W.L²
c. (W.L)/2                         d. 0

Multiple Choice type Question:

TOPIC : FRICTION

1) The maximum value of frictional force that comes into play when a body tends to move on a surface called :
a) sliding friction,
b) limiting friction,
c) milling friction,
d) none of these.

2) The ratio of static friction to dynamic friction is :
a) less than 1,
b) equal to 1,
c) greater than 1,
d) none of these.

3) The angle of friction is equal to the :
a) ratio of frictional force to the normal reaction.
b) angle of inclined plane when a body tends to slide down.
c) angle of an inclined plane when a body is sliding.
d) none of these.

4) The coefficient of friction depends upon :
a) area of contact,
b) shape of the body,
c) nature of contact surfaces,
d) none of these.

5) Kinetic friction is :
a) limiting friction,
b) friction when a body is moving,
c) friction when a body is stationary,
d) none of these.

©subhankar_karmakar

PROBLEM SET - 8

Q1) The enclosed area in the shear force diagram for a beam is the magnitude of
a) shear force
b) bending moment
c) applied load
d) bending stress.

Q2) If a particle is moving in a circular path at constant velocity on a plane, then the direction of the angular velocity will be
a) along the tangent to the circular path
b) along the radial direction
c) normal to the plane
d) none of this.

There will be 2 statements, which may be right or may be wrong, read those sentences and then choose the right choice

Q3) statement 1: Two forces act on a particle are collinear forces.
statement 2: The ratio of longitudinal strain and lateral strain is the value of poisson's ratio.

a) both of them are true
b) statement 1 is true, but statement 2 is false
c) statemen 2 is true while stateme 1 is false
d) both of them are false.

MOMENTUM : AN IMPORTANT CONCEPT

NEWTON'S LAW OF MECHANICS:

Although we know there are three laws of motion proposed by Issac Newton, but it can be shown that the 2nd law of motion is the fundamental laws of motion, and the other two laws are nothing but special cases of second law.

The second law states that the rate of change of momentum is equal to force, which is another physical quantity and it is a vector.

SO WHAT DOES MOMENTUM MEAN?

MOMENTUM

Tuesday, 16. November, 02:41

Objects in motion are said to have a momentum. This momentum is a vector. It has a size and a direction. The size of the momentum is equal to the mass of the object multiplied by the size of the object's velocity. The direction of the momentum is the same as the direction of the object's velocity.

Momentum is a conserved quantity in physics. This means that if you have several objects in a system, perhaps interacting with each other, but not being influenced by forces from outside of the system, then the total momentum of the system does not change over time. However, the separate momenta of each object within the system may change. One object might change momentum, say losing some momentum, as another object changes momentum in an opposite manner, picking up the momentum that was lost by the first.

IMPORTANCE OF MOMENTUM

Momentum is a corner stone concept in Physics. It is a conserved quantity. That is, within a closed system of interacting objects, the total momentum of that system does not change value. This allows one to calculate and predict the outcomes when objects bounce into one another. Or, by knowing the outcome of a collision, one can reason what was the initial state of the system.

MOMENTUM IS MASS TIMES VELOCITY

When an object is moving, it has a non-zero momentum. If an object is standing still, then its momentum is zero. To calculate the momentum of a moving object multiply the mass of the object times its velocity. The symbol for momentum is a small p.

MOMENTUM IS A VECTOR QUANTITY

Momentum is a vector. That means, of course, that momentum is a quantity that has a magnitude, or size, and a direction. The above problem is a one dimensional problem. That is, the object is moving along a straight line. In situations like this the momentum is usually stated to be positive, i.e., to the right, or negative, i.e., to the left.

MOMENTUM IS NOT VELOCITY

.

Sometimes the concept of momentum is confused with the concept of velocity. Do not do this. Momentum is related to velocity. In fact, they both have the same direction. That is, if an object has a velocity that is aimed toward the right, then its momentum will also be directed to the right. However, momentum is made up of both mass and velocity. One must take the mass and multiply it by the velocity to get the momentum.

MOMENTUM IS DIRECTLY PROPORTIONAL TO VELOCITY

If the mass is kept constant, then the momentum of an object is directly proportional to its velocity. In the example at the left, the mass is kept constant at a value of 2.0 kg. The velocity changes from 0 m/s to 10 m/s while the momentum changes from 0 kg-m/s to 20 kg-m/s. This creates a straight line graph when momentum is plotted as a function of velocity. (The symbol for momentum as a function of velocity would be p(v).) The straight line graph demonstrates the direct proportion between momentum and velocity.


That is, if one were to double the velocity of an object, then the momentum of that object would also double. And, if one were to change the velocity of an object by a factor of 1/4, then the momentum of that object would also change by a factor of 1/4.

MOMENTUM IS DIRECTLY PROPORTIONAL TO MASS

If the velocity is kept constant, then the momentum of an object is directly proportional to its mass. In the example at the left, the velocity is kept constant at a value of 3.0 m/s. The mass changes from 0 kg to 10 kg while the momentum changes from 0 kg-m/s to 30 kg-m/s. This creates a straight line graph when momentum is plotted as a function of mass. (The symbol for momentum as a function of mass would be p(m).) The straight line graph demonstrates the direct proportion between momentum and mass.

That is, if one were to triple the mass of an object, then the momentum of that object would also triple. And, if one were to change the mass of an object by a factor of 1/2, then the momentum of that object would also change by a factor of 1/2.

TWO DIMENSIONAL FORCE SYSTEM

Q: WHAT DO YOU UNDERSTAND BY THE TERM "FORCE"? WHAT IS THE EFFECT OF FORCE ON A PARTICLE AND A RIGID BODY? EXPLAIN WITH SUITABLE EXAMPLES.

Answer:

FORMAL DEFINITION:

A FORCE is that which can cause an object with mass to ACCELERATE. Force has both MAGNITUDE and DIRECTION, making it a vector quantity. According to Newton's second law, an object with constant mass will accelerate in proportion to the net force acting upon it and in INVERSE PROPORTION TO ITS MASS (M). An equivalent formulation is that the net force on an object is equal to the RATE OF CHANGE OF MOMENTUM it experiences. Forces acting on three-dimensional objects may also cause them to rotate or deform, or result in a change in pressure. The tendency of a force to cause angular acceleration about an axis is called TORQUE. Deformation and pressure are the result of stress forces within an object.


EXPLANATION OF MECHANICAL FORCE AND IT'S EFFECT ON A PARTICLE:

CHANGE IN POSITION:

To know force well, first we have to understand what do we mean by Change. What does it mean when we say the position of the body has been changed? Whenever we find the state of object becomes different than that of the same object before some time say Δt, then we say that there exists a change in the state of the object. Suppose the change occurs in the position of the body. But to find the initial position of a body, we need a co-ordinate system.

THE CAUSE OF CHANGE:

It has been seen that to induced a change or to make a change in the position of an object we must have to change the energy possess by the body. To transfer energy into the object we shall have to apply FORCE on the body. Therefore Force is the agency that makes a change in position of a body.

THE CONCLUSION: GALILEO'S LAW OF INERTIA OR NEWTON'S FIRST LAW OF MOTION.

So, if there is no force on an object the position of the object won't change with respect to time. It means if a body at rest would remain at rest and a body at uniform motion would remain in a steady motion. This law is known as Galileo's Law of Inertia or Newton's first law of motion.

• 2 DIMENSIONAL FORCE

In physics, force is a vector quantity that is used to describe the interaction between two objects. In a two-dimensional system, forces can act in two different directions, which are typically labeled as the x-axis and the y-axis.

When dealing with two-dimensional force, it is essential to use vector addition to determine the net force acting on an object. The net force is the vector sum of all the forces acting on the object. The direction of the net force is determined by the angle of the resultant force vector.

To calculate the net force in two dimensions, we must first break down each force into its x and y components. The x-component of a force is the amount of force acting in the x-direction, and the y-component is the amount of force acting in the y-direction. Once we have the x and y components for each force, we can add them together to find the net force.

The magnitude of the net force can be calculated using the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. In this case, the hypotenuse is the magnitude of the net force, and the other two sides are the x and y components of the net force.

In summary, when dealing with two-dimensional force, it is essential to use vector addition to determine the net force acting on an object. To calculate the net force, we must first break down each force into its x and y components and then add them together. The magnitude and direction of the net force can be determined using trigonometry.

• ORTHOGONAL RESOLUTION OF A FORCE

Orthogonal resolution of a force is a technique used in physics to break down a force vector into its components along two orthogonal axes, typically the x and y axes. This technique is useful in analyzing the motion of an object under the influence of a force and can help determine the net force acting on an object.

To perform orthogonal resolution of a force, we first need to identify the angle that the force vector makes with respect to one of the axes, usually the x-axis. We can then use trigonometry to determine the components of the force vector along the x and y axes.

If the angle between the force vector and the x-axis is θ, the x-component of the force can be found using the equation F_x = F cos(θ), where F is the magnitude of the force. Similarly, the y-component of the force can be found using the equation F_y = F sin(θ).

Once we have the x and y components of the force, we can use vector addition to determine the net force acting on an object. The net force is the vector sum of all the forces acting on the object and can be found by adding the x and y components of each force separately.

Orthogonal resolution of a force is a powerful technique that is used in many areas of physics, including mechanics, electromagnetism, and fluid dynamics. By breaking down a force vector into its components, we can better understand the forces acting on an object and predict its motion under different conditions.

WHAT IS A FORCE SYSTEM? CAN WE CLASSIFY FORCE SYSTEMS?

ANSWER:
                         A force system may be defined as a system where more than one force act on the body. It means that whenever multiple forces act on a body, we term the forces as a force system. We can further classify force system into different sub-categories depending upon the nature of forces and the point of application of the forces. Almost any system of known forces can be resolved into a single force called a resultant force or simply a Resultant. The resultant is a representative force which has the same effect on the body as the group of forces it replaces. (A couple is an exception to this) It, as one single force, can represent any number of forces and is very useful when resolving multiple groups of forces. It is important to note that for any given system of forces, there is only one resultant.


Different types of force system:

• (i) COPLANAR FORCES:

If two or more forces rest on a plane, then they are called coplanar forces. There are many ways in which forces can be manipulated. It is often easier to work with a large, complicated system of forces by reducing it an ever decreasing number of smaller problems. This is called the "resolution" of forces or force systems. This is one way to simplify what may otherwise seem to be an impossible system of forces acting on a body. Certain systems of forces are easier to resolve than others. Coplanar force systems have all the forces acting in in one plane. They may be concurrent, parallel, non-concurrent or non-parallel. All of these systems can be resolved by using graphic statics or algebra.

• (ii) CONCURRENT FORCES:

A concurrent coplanar force system is a system of two or more forces whose lines of action ALL intersect at a common point. However, all of the individual vectors might not actually be in contact with the common point. These are the most simple force systems to resolve with any one of many graphical or algebraic options. If the line of actions of two or more forces passes through a certain point simultaneously then they are called concurrent forces. concurrent forces may or may not be coplanar.

• (iii) LIKE FORCES:

A parallel coplanar force system consists of two or more forces whose lines of action are ALL parallel. This is commonly the situation when simple beams are analyzed under gravity loads. These can be solved graphically, but are combined most easily using algebraic methods. If the lines of action of two or more forces are parallel to each other, they are called parallel forces and if their directions are same, then they are called LIKE FORCES.

• (iv) UNLIKE FORCES:

If the parallel forces are such that their directions are opposite to each other, then they are termed as "UNLIKE FORCE".

• (v) NON COPLANAR FORCES:

The last illustration is of a "non-concurrent and non-parallel system". This consists of a number of vectors that do not meet at a single point and none of them are parallel. These systems are essentially a jumble of forces and take considerable care to resolve.

on 20th November, 2010: ©subhankar

SECOND SESSIONAL TEST (odd SEMESTER 2009-10) B.Tech…first Semester Sub Name: Engineering Mechanics

SECOND SESSIONAL TEST (odd SEMESTER 2009-10)

B.Tech…first Semester

Sub Name: Engineering Mechanics                                         Max. Marks: 30

Sub Code: EME-201                                                          Max. Time: 2: 00 Hr

Group A

Q.1 Choose the correct answer of the following questions 1x6=6

(i) If two forces of equal magnitudes have a resultant force of the same magnitude then the angle between them is

(a) 0⁰          (b) 90⁰         (c) 120⁰          (d) 135⁰

Ans: ( c)

Explanation: P² = P² + P² +2P.P.cos θ

ð      P²= P² (1+2cos θ)

ð      cos θ = -½

ð      θ = 120°

(ii) If a ladder is kept at rest on a vertical wall making an angle θ with horizontal. If co-efficient of the friction in all the surfaces be µ, then the tangent of the angle θ will be equal to

(a) (1-µ²)/2µ                                  (b) (1-µ)² /2µ

(c) (1-µ)/2µ                                   (d) none of the above

Explanation: (a) (1-µ²)/2µ

(iv) Varignon’s theorem is related with _____________ .

Answer: moment

(v) If two forces of equal magnitudes P having an angle (90°- θ) between them, then their resultant force will be equal to ________.

Answer: √2P (1+sin θ)

(vi) A fixed joint produces

(a) 1             (b) 2             (c) 3               (d) 4         reactions

Answer: (c ) 3

(vii) The equilibrium conditions of concurrent force system is_________.

Answer: ∑Fx =0; ∑Fy=0.

Group B                                                                                                                  8x3=24

Attempt any three questions

Q.2. State and explain Varignon’s theorem of moment. Three forces of magnitudes 3 KN, 4 KN and 2KN act along the three side of an equilateral triangle ∆ABC in order. Find the position, direction and magnitude of the resultant force.         4+4

Answer: resultant force: √3 KN

Q.3  (a) Two cylindrical rollers are kept at equilibrium inside a jar or channel as shown in the figure. The channel width is 1000 mm, where as the rollers have diameters 600 mm and 800 mm respectively. The weights are 2 kN and 5 kN respectively. Find all the reactions at contacts.

                                                                                            4+4

(b) What is pure bending? If a stone is thrown with a velocity 400 m/s then find the maximum height that the stone can reach.

Q:4)  a) Classify different types of joints in beam with proper explanations.

(b) Find the reactions at the support for the beam as shown in the figure.                                           4+4