Course: Strength of Materials (CAB 2042) Credit Value: Credit Value: 2 2 FACILITATOR: FACILITATOR: DR. TUSHAR KANTI SEN DR. TUSHAR KANTI SEN Chemical Engineering Programme Chemical Engineering Programme Building 04, Level 03-Room-10, Tel. Building 04, Level 03-Room-10, Tel. extn-7578 extn-7578 Course Description Course Description : : Statics of particles, stress Statics of particles, stress and strain, Centroid and Centre of Gravity, Bending and strain, Centroid and Centre of Gravity, Bending moment and shear force, Structural Analysis on Trusses moment and shear force, Structural Analysis on Trusses and machine, Transverse loading, Combined loading, and machine, Transverse loading, Combined loading, Torsion of circular shafts, Mohr’s Circle. Torsion of circular shafts, Mohr’s Circle. COURSE OBJECTIVES COURSE OBJECTIVES : To aquire concept of equilibrium, : To aquire concept of equilibrium, analyse rigid bodies, extend the student’s study of analyse rigid bodies, extend the student’s study of strength of materials beyond statics to the concepts of strength of materials beyond statics to the concepts of stress and strain, mechanical properties of materials, stress and strain, mechanical properties of materials, bending moment and shear force, Continued work on bending moment and shear force, Continued work on problem-solving skills,Enrichment in knowledge during
Transcript
Course: Strength of Materials (CAB 2042)
Credit Value:Credit Value: 22
FACILITATOR:FACILITATOR: DR. TUSHAR KANTI SENDR. TUSHAR KANTI SEN
Chemical Engineering ProgrammeChemical Engineering Programme
Course DescriptionCourse Description:: Statics of particles, stress and strain, Centroid and Statics of particles, stress and strain, Centroid and Centre of Gravity, Bending moment and shear force, Structural Analysis Centre of Gravity, Bending moment and shear force, Structural Analysis on Trusses and machine, Transverse loading, Combined loading, Torsion on Trusses and machine, Transverse loading, Combined loading, Torsion of circular shafts, Mohr’s Circle.of circular shafts, Mohr’s Circle.
COURSE OBJECTIVESCOURSE OBJECTIVES: To aquire concept of equilibrium, analyse rigid : To aquire concept of equilibrium, analyse rigid bodies, extend the student’s study of strength of materials beyond statics bodies, extend the student’s study of strength of materials beyond statics to the concepts of stress and strain, mechanical properties of materials, to the concepts of stress and strain, mechanical properties of materials, bending moment and shear force, Continued work on problem-solving bending moment and shear force, Continued work on problem-solving skills,Enrichment in knowledge during selection of structure and machines.skills,Enrichment in knowledge during selection of structure and machines.
Course: Strength of Materials (CAB 2042)
Course outline, Outcomes and AssessmentCourse outline, Outcomes and Assessment: As per : As per hard copy circulation among students based on OBEhard copy circulation among students based on OBE
TEXTBOOK:TEXTBOOK: (1) (1) Strength of Materials, Strength of Materials, by B. K. Sarkar, by B. K. Sarkar, Tata McGraw-Hill Publishing Company Ltd, New Delhi, Tata McGraw-Hill Publishing Company Ltd, New Delhi, 2003.2003.
(2) (2) Engineering Mechanics STATICSEngineering Mechanics STATICS, by R. C. Hibbeler, , by R. C. Hibbeler, Prentice Hall, Singapore, London, New York, 2004.Prentice Hall, Singapore, London, New York, 2004.
(3) (3) Strength of Materials,Strength of Materials, by R. K. Rajput, S. Chand & by R. K. Rajput, S. Chand & Company Ltd, New Delhi, 2002.Company Ltd, New Delhi, 2002.
(4) (4) Mechanics of MaterialsMechanics of Materials, by F.P. Beer, E. R. , by F.P. Beer, E. R. Johnston and J. T. DeWolf,McGraw-Hill, New York, Johnston and J. T. DeWolf,McGraw-Hill, New York, 2002.2002.
INTRODUCTIONINTRODUCTION
In day-to-day work, an engineer comes across certain materials, i.e. steel girders, angle irons, circular bars, cement etc., which are used in his project work. While selecting a suitable material, for his project, an engineer is always interested to know its strength. The strength of a material may be defined as ability, to resist its failure and behaviour, under the action of external force. A detailed study of forces and their effects, along with some suitable protective measures for the safe working conditions, is known as “Strength of materials”
Basic Quantities:- Length - Time- Mass- ForceAll quantities can be expressed in four systems of units I.e. C.G.S., F.P.S., M.K.S and S.I. Units.
• Equilibrium• At rest or move with constant
velocity
Introduction to StaticsIntroduction to Statics
Idealizations:Idealizations:
Particle - has a mass but with negligible size Rigid Body - does not deform under load Concentrated Force - load acting on a point of body
Newton’s Three Laws of Motion:• First Law - at rest; constant velocity; balanced
force• Second Law - unbalanced force F; acceleration
aF = ma
• Third Law - mutual forces of action and reaction: equal, opposite, and collinear
Exponential form Prefix SI symbol 109 giga G 106 mega M 103 kilo k 10-3 milli m 10-6 micro 10-9 nano n
Prefixes:Prefixes:
Dimensional Homogeneity:
All terms in an equation must be expressed in the same units
Where A,B,C are length of the sides, and a, b, c are the corresponding angles opposite the sides.
Some General Trigonometric IdentitiesSome General Trigonometric Identities
Examples Problems on TrigonometryExamples Problems on Trigonometry
Example 1. A 40 ft long ladder leaning against a wall makes an angle of 60° with the ground. Determine the vertical height to which the ladder will reach.4.6
Ans: 34.64 ‘
Examples Problems on TrigonometryExamples Problems on Trigonometry
Example 2: In the roof truss shown in the Fig, the bottom chord members AD and DC have lengths of 18 ft. and 36 ft respectively. The height BD is 14 ft. Determine the lengths of the top chords AB and BC and find the angles at A and C
Ans: AB = 22.8 ft; BC = 38.6 ft; A = 37.90, C = 21.250.
Basic Vector ReviewBasic Vector Review
1. Definitions: Scalar: Any quantity possessing magnitude (size) only, such as mass, volume, temperature Vector: Any quantity possessing both magnitude and direction, such as force, velocity, momentum
2.Vector Addition:Vector addition may be done several ways including, Graphical Method, Trigonometric Method, and Component Method. We will be reviewing only the Component Method, as that is the method which will be used in the course. Other methods are detailed in your textbook.3.Vector Addition -Component Method:(2-dimensional)
The component method will follow the procedure shown below
Choose an origin, sketch a coordinate system, Choose an origin, sketch a coordinate system, and draw the vectors to be added (or and draw the vectors to be added (or summed).summed).
Break (resolve) each vector into it's "x" and Break (resolve) each vector into it's "x" and "y" components, using the following "y" components, using the following relationships:relationships: AAxx = A cosine Ø, and, A = A cosine Ø, and, Ayy = A sine Ø, where A is = A sine Ø, where A is the vector, and Ø is the vector's angle.the vector, and Ø is the vector's angle.
Sum all the x-components and all the y-Sum all the x-components and all the y-components obtaining a net resultant Rcomponents obtaining a net resultant Rxx, and , and RRyy vectors. vectors.
RRxx = A = Axx + B + Bxx + C + Cxx + . . ., &, R + . . ., &, Ryy = A = Ayy + B + Byy + C + Cyy + . . . + . . . 4. Recombine Rx and Ry to obtain the 4. Recombine Rx and Ry to obtain the final resultant vector (magnitude and final resultant vector (magnitude and direction) usingdirection) using 2 2 RyR R R and Tangentx y Rx
Example-Vector AdditionExample-Vector Addition
Three ropes are tied to a small metal ring. At the end of each rope three rope three students are pulling, each trying to move the ring in their direction. If we look down from above the students, the forces and directions they are applying the forces are as follows; (See diagram to the right).
Find the net (resultant) force (magnitude and direction) on the ring due to the three applied forces.
SolutionSolution
Choose origin, sketch coordinate system and vectors (done above)
Resolve vectors into x & y components (See Diagram)
Solution (Contd)Solution (Contd)Sum x & y components to find resultant Rx and Ry forces. Rx = 24.0 lbs - 35.4 lbs - 40.0 lbs = -51.4 lbsRy =18.1 lbs + 35.4 lbs - 69.3 lbs = -15.8 lbs 'Recombine' (add) Rx and Ry to determine final resultant vector.
Thus the resultant force onthe ring is 53.8 poundsacting at an angle of 197.1degrees.
FORCEFORCE Resultant Force: If a number of forces P, Q, R-------etc., are acting
simultaneously on a particle, then a single force, which will produce the same effect as that of all the given forces, is known as a resultant force. The forces P, Q, R—etc. are called component forces.
Composition of Forces: It means the process of finding out the resultant force of the given component forces. A resulting force may be found out analytically, graphically or by the following laws:
(a) Parallelogram Law of Forces: “If two forces acting simultaneously on a particle be represented, in magnitude and direction, by the two adjacent sides of a parallelogram, their resultant may be represented, in magnitude and direction, by the diagonal of the parallelogram passing through the point of their interaction”.
(b) Triangle Law of Forces: “If two forces acting simultaneously on a particle be represented, in magnitude and direction, by the two sides of a triangle taken in order, their resultant may be represented, in magnitude and direction, by the third side of the triangle taken in opposite order”.
FORCEFORCE
Resultant Force: If a number of forces P, Q, R-------etc., are acting simultaneously on a particle, then a single force, which will produce the same effect as that of all the given forces, is known as a resultant force. The forces P, Q, R—etc. are called component forces.
Composition of Forces: It means the process of finding out the resultant force of the given component forces. A resulting force may be found out analytically, graphically or by the following laws:
(a) Parallelogram Law of Forces: “If two forces acting simultaneously on a particle be represented, in magnitude and direction, by the two adjacent sides of a parallelogram, their resultant may be represented, in magnitude and direction, by the diagonal of the parallelogram passing through the point of their interaction”.
FORCEFORCE
(a) Triangle Law of Forces: “If two forces acting simultaneously on a particle be represented, in magnitude and direction, by the two sides of a triangle taken in order, their resultant may be represented, in magnitude and direction, by the third side of the triangle taken in opposite order”.
(c) Polygon Law of Forces: “If a number of forces acting simultaneously on a particle be represented, in magnitude and direction, by the sides of a polygon taken in order, their resultant may be represented, in magnitude and direction, by the closing side of the polygon taken in opposite order”.
(d) Moment of a Force: It is the turning effect, produced by the force, on a body on which it acts. It is mathematically equal to the product of the force and the perpendicular distance between the line of action of the force and the point about which the moment is required.
How Forces are RepresentedHow Forces are RepresentedThere are two ways in which forces can
be represented in written form: Scalar Notation (Figure 1.2.1) Vector Notation (Figure 1.2.2) The method used depends on the type of
problem being solved and the easiest approach to finding a solution.
Scalar notation is useful when describing a force as a set of orthogonal force components. For example: Fx = 15N, Fy = 20N, Fz = 10N
How Forces are RepresentedHow Forces are Represented
Vector notation is useful when vector mathematics are to be applied to a problem, such as addition or multiplication. Vector notation is somewhat simple in form:
F = 15i + 20j + 10k N. The N term represents the unit of
force, Newtons in this instance.
Multiple forces can be applied at a point. These forces are known as concurrent forces and can be added together to form a resultant force. If the component forces are orthogonal, then the magnitude of the resultant force can be determined by taking the Square Root of the Sum of the Squares (SRSS). The SRSS method is an extension of the Pythagorean Theorem to three dimensions. Figure 1.3.1 illustrates the calculation of a vector magnitude using the SRSS method.
Addition of Addition of ForcesForces
What is a Distributed Force What is a Distributed Force A distributed force can be thought of as a force
spread out over a specific area. Suppose we had a 3'x3'x3' cube of concrete (very heavy!). We might say that the cube exerts a force of: (3ft)^3 * 145lbs/ft^3 = 3,220 lbs downward
We could also consider the distributed force beneath the cube:
3,220 lbs/(3ft x 3ft) = 356 lbs/ft^2Figure 3.1.1 illustrates the two ways we might
consider the concrete weight.Can you see the relationship between the
concentrated force and the distributed force? The concentrated force represents a single force vector applied at the centroid of the object. The centroid here means the center of gravity.
Distributed forces come in many shapes and sizes. Actually, in nature there are more distributed forces than there are concentrated forces. Even the load beneath your shoes as you walk is distributed over the sole and heel. Most realistic problems will involve distributed forces.
Forces in PlaneForces in PlaneCase 1:• Given two forces determine
MAGNITUDE and DIRECTION of resultant force FR
• Tool #1: Parallelogram:
a
b
FR
FR
a
b
Tool #2: Law of Cosines (To find the magnitude or direction):
(for a, b, and are known)
Tool #3: Law of Sines (To find the direction or magnitude):
(where is unknown)
cos222 abbaFR
sinsinRFa
Case 2: Given one force resolve the force into its COMPONENTS Tool: Vector Triangle
Fx = F cos Fy = F sin· See examples 2.1, 2.2, 2.3, 2.4
Fy
Fx
x
y F
F
Fx
Fy
Summary on Force VectorsSummary on Force Vectors
Goals:To show how to add forces and resolve them into componentsTo express force and position in Cartesian vector formTo explain how to determine a vector’s magnitude and directionScalars and Vectors:A scalar is a real number e.g., mass, time, volume and length A vector has both magnitude and direction e.g. force, velocity and accelerationVector OperationsMultipication of a vector by a scalar: The product of a vector A and a scalar a is a vector aA with magnitude mod aA = mod a mod A.
Summary on Force VectorsSummary on Force Vectors
VECTOR ADDITION: Two vectors A and B can be
added to form a resultant vector R = A + B by using the parallelogram law. If the two vectors are collinear (both vectors have the same line of action), the resultant is formed by an algebraic or scalar addition.
A
B(a) B
AR=A+B
(b) Parallelogram law
A B
R = A +B(c) Triangle construction
R = B + A
BA
(d) Triangle construction
A BR
Addition of collinear vector
Summary on Force VectorsSummary on Force Vectors
RESOLUTION OF A VECTOR: A Vector may be resolved
into components having known lines of action by using the parallelogram law.
R
a
b(a) Extend parallel lines from the head of R to form components
RA
Bb
a
Components
Resultant
(b)
Summary on Force VectorsSummary on Force Vectors
VECTOR ADDITION OF FORCES
Forces are added together or resolved into components using the rules of vector algebra
Two common problems in statics involve either finding the resultant force given its components or resolving a known force into components
Often the magnitude of a resultant force can be determined from law of cosines, while its direction is determined from the law of sines.
b
a
cA B
C
2 2
:
sin :
2
SinelawA B C
Sina Sinb SincCo elaw
C SQRT A B ABCosc
Concept of EquilibriumConcept of Equilibrium
Any kind of object or component must have a net balance of zero forces and moments applied to it in order to remain at rest. This type of balance is called static equilibrium, where nothing is moving. Figure 4.1.1 illustrates a weight, pully and block situatuion where all of the forces, including friction beneath the block, sum to zero.
Upon inspection of the forces in Figure 4.1.1, you will see that they sum to zero. Hence, there is equilibrium and no movement.
Concept of EquilibriumConcept of Equilibrium
Determining the forces on a body is an important part of mechanics. It helps us answer questions such as: Will an object move? How much force or stress is in a part? Will something break? Do you think the stationary block in Figure 4.1.1 would move if the hanging mass was 50kg instead of 25kg? What if the cable were to break?
There are two vital items which must be understood in order to evaluate forces. These two items are boundary conditions and the free body diagram.
Equilibrium of a Rigid BodyEquilibrium of a Rigid Body
Two equations of equilibrium:
0
0
oM
F
The Free Body Diagram (FBD) is a graphical tool for evaluating forces and equilibrium. A FBD is a drawing of an analysis problem showing the components, relevant dimensions, and every applied force and moment present. The rules of equilibrium apply to the FBD. All of the forces in the drawing will sum to zero in the X, Y, and Z directions. Also, any set of moments at any location on the FBD will also sum to zero. It is necessary to have the dimensions drawn in order to figure out moments on the FBD.
Free-Body Diagram:
Equilibrium of a Rigid BodyEquilibrium of a Rigid Body
Two equations of equilibrium:
0
0
oM
F
• A sketch of the outlined shape of the body, free from its surroundings
• Show all the known/unknown forces and couple moments acting on the body due to (1)Applied loading – only external forces(2)Reactions at the supports/at the points of
contact with other bodies (3)Weight of the body – at the center of gravity
Free-Body Diagram:
Equilibrium in Two-Dimensions:Equilibrium in Two-Dimensions:
1. If a support prevents translation of body in a given direction, then a force is developed in that direction
2. If a support prevents rotation, then a couple moment is exerted on the body
The topic of statics deal with objects or structures which are in equilibrium, that is structures that are at rest or in uniform, (non-accelerated) motion. We will be normally looking at structures which are at rest. For these structures we will be interested in determining the forces (loads and support reactions) acting on the structure and forces acting within members of the structure (internal forces). To determine forces on and in structures we will proceed carefully, using a well defined methodology. This is important as most problems in statics and strength of materials are not the kind of problem in which we can easily see the answer, but rather we must relay on our problem solving techniques.
For static equilibrium problems, we will be able to apply the Conditions of Equilibrium to help us solve for the force in and on the structures. There are two general equilibrium conditions: Translational Equilibrium, and Rotational Equilibrium.
The Translational Equilibrium condition states that for an object or a structure to be in translational equilibrium (which means that the structure as a whole will not experience linear acceleration) the vector sum of all the external forces acting on the structure must be zero. Mathematically this may be expressed as: Fall = 0 or,Fx = 0, Fy =0, Fz =0 in 3-dimensions: , ,
That is, forces in the x-direction must sum to zero, for translational equilibrium in the x-direction, and, the forces in the y-direction must sum to zero, for translational equilibrium in the y-direction, and, the forces in the z-direction must sum to zero, for translational equilibrium in the z-direction.
To see the application of the first condition of equilibrium and also the application of a standard problem solving technique, let's look carefully at introductory examples. Select: Example 1- Concurrent Forces.
structure, we have a weight supported by two cables, which run over pulleys (which we will assume are very low friction) and are attached to 100 lb. weights as shown in the diagram. The two cords each make an angle of 50o with the vertical. Determine the weight of the body. (The effect of the pulleys is just to change the direction of the force, it may be considered to not effect the value of the tensions in the ropes.)
Concurrent, Coplanar ForcesConcurrent, Coplanar Forces If we examine the first diagram
for a moment we observe this problem may be classified as a problem involving Concurrent, Coplanar Forces. That is, the vectors representing the two support forces in Cable 1 and Cable 2, and the vector representing the load force will all intersect at one point, just above the body. When the force vectors all intersect at one point, the forces are said to be Concurrent. Additionally, we note that this is a two-dimensional problem, that forces lie in the x-y plane only. When the problem involves forces in two dimensions only, the forces are said to be Coplanar
To "Solve" this problem, that is to determine the weight To "Solve" this problem, that is to determine the weight supported by forces (tensions) in cable 1 and cable 2, we will supported by forces (tensions) in cable 1 and cable 2, we will now follow a very specific procedure or technique, as follows:now follow a very specific procedure or technique, as follows: (Example-1 contd) Diagram 2 (Example-1 contd) Diagram 2
1. Draw a Free Body Diagram (FBD) of the structure or a portion of the structure. This Free Body Diagram should include a coordinate system and vectors representing all the external forces (which include support forces and load forces) acting on the structure. These forces should be labeled either with actual known values or symbols representing unknown forces. The second diagram 2 is the Free Body Diagram of point just above the weight where with all forces come together
2. Resolve (break) forces not in x or y direction into their x and y components. Notice for Cable 1, and Cable 2, the vectors representing the tensions in the cables were acting at angles with respect to the x-axis, that is, they are not simply in the x or y direction. Thus the forces Cable 1, Cable 2, we must be replaced with their horizontal and vertical components. In the third diagram, the components of Cable 1 and Cable 2 are shown.
Example 1 (contd)Example 1 (contd)
3. Apply the Equilibrium Conditions and solve for unknowns. In this step we will now apply the actual equilibrium equations. Since the problem is in two dimensions only (coplanar) we have the following two equilibrium conditions: The sum of the forces in the x direction, and the sum of the forces in the y direction must be zero. We now place our forces into these equations, remembering to put the correct sign with the force, that is if the force acts in the positive direction it is positive and if the force acts in the negative direction, it is negative in the equation. Fx = 0 or, -100 cos 40o + 100 cos 40o = 0 (Just as we would expect, the x-forces balance each other.) Fy = 0 or, 100 sin 40o + 100 sin 40o - weight of body = 0In this instance, it is very easy to solve for the weight of the body from the y-equation; and find:Weight of body = 128.56 lb.
Learning Outcomes from this chapterLearning Outcomes from this chapter
Fundamentals on Forces & their Classification How Forces are representedAddition of ForcesExplain the concept of equilibrium and analyze the equilibrium condition of rigid bodies.Free Body diagramProblem solving skills
