NAME: MOMOH ITAMOMOH E.P.

MATRIC: 070403038

DEPARTMENT: ELECTRICAL/ELECTRONIC

COURSE: CEG 202 (LAB)

TITLE: SIMPLY SUPPORTED BEAMS

GROUP: Three

DATE PERFORMED: 20/08/09

DATE SUBMITTED: 3/09/09

Theory:The intention of this report is to develop an understanding of the mathematics and physics involved in determining the amount of weight that is transferred to the supports of a simply supported, single span or overhanging beam, due to any load applied to the beam. Determining these reactions is an important step in an engineer’s analysis of a structural member. The reactions at the supports of a beam indicate the required forces to be resisted by the supporting members and thereafter the foundation. Types of supports and beams are explained as an introduction to the methods used by engineers to perform the analysis. The mechanics of a rigid body, specifically statics, are demonstrated on an example beam in which the equations of equilibrium are presented, and the conditions that must exist for these equilibrium equations to be used are discussed. After this understanding of statics is established, a physical model is presented. Experiments carried out with this model are used to test and support the accuracy of the assumptions, namely Newton’s laws of motion, that are presented by the statical approach to determining the support reactions of the beam. With the experimental data and the theoretical awareness of statics, the mathematical set of equilibrium equations is use to confirm the unknown support reactions for the beam. This proof is illustrated through error analysis and a comparison of the theoretical

reaction values with the actual values recorded during the experiments. Further observations are presented on simplifying complicated structures conclude this report.

DISCUSSIONThe Beam:

A beam is a structural member or an element of a machine that is designed primarily to support forces acting perpendicular to the axis of the member. Generally, the length (L) of abeam is much larger than the other two cross-sectional dimensions, height, and width. Beamscan be straight or curved. A beam with a constant height and width is said to be prismatic. When a beam’s width or height (more common) varies, the member is said to be non-prismatic. Horizontal applications of beams are typically found in bridge and building construction. Vertical beams are also found in various applications. The primary deformation of a beam is in bending. Some beams are loaded, such that only bending occurs. However, beams can be subjected to bending and any combination of axial, shear, and torsional loads. When a slender member is introduced primarily to axial loads, it is considered to be a column. A vertical member found in building construction that is loaded with axial compression, and simultaneously subjected to a horizontal wind or seismic load is commonly

referred to as a beam column. In this project, only straight, prismatic beams are considered.

Types of Beam Supports:The three types of beam supports typically

used by engineers are roller, pinned, and fixed connections. The roller connection (see figure 1a) allows the beam to rotate freely in any direction, and the beam is free to move or translate in the x direction; thus no x component of the force on the beam can be transferred to the support. Therefore, the reaction at the support is a vertical force. The pinned connection resists (see figure 1b) translation of the beam in any direction, but still allows it to rotate in any direction. The resulting reaction is represented by horizontal and vertical components of the force. A fixed connection (see figure 1c) prohibits both rotation and translation of the beam in any direction. The reaction results in horizontal and vertical components of the force, as well as a moment that resists the rotation.

Figure 1.

Types of Loads and Beams

Beams can be catalogued into types based on how they are loaded and how they are supported. Loads that are applied to a small section of the beam are simplified by considering the load to be single force placed at a specific point on the beam. These loads are referred to as concentrated loads (see figure 2a). Distributed loads (w, usually in units of force per lineal length of the beam)( see figure 2b) occur over a measurable distance of a beam. For the sake of determining reactions, a distributed load can be simplified in to an equivalent concentrated load by applying the area of the distributed load at the centroid of the distributed load.

The weight of the beam can be described as uniform load. A moment is a couple as a result of two equal and opposite forces applied at certain section of the beam. A moment induced on any point can be mathematically described as a force multiplied by a perpendicular distance to that point. A beam that is supported by rollers, pins, or a smooth surface at the ends is designated as a simply supported beam (see figure 2a). When the beam extends beyond the supports it is said to be an overhanging beam. The cantilever can occur a either one or both ends of the beam. When the beam is fixed at one end and free at the other (see figure 2c) it also designated a cantilever.

A beam is called a propped beam when it is fixed at one end and simply supported at the other (see figure 2d). A continuous beam (see figure 2e) has more than two simple supports, and a built-in beam (see figure 2f) is fixed at both ends.

The remainder of this report deals only with simple and over-hanging beams loaded with concentrated and uniformly distributed loads.

Statics – Rigid Body Mechanics

Statics is a division of mechanics that deals with bodies that are in equilibrium. When abody is in equilibrium it is subjected to balanced forces, and therefore is at rest or in uniform motion.

Sir Isaac Newton published The Principia in 1687, in which he established the basic lawsthat describe the motion of a particle. Today these laws are stated as follows,

Law 1: In the absence of external forces, a particle originally at rest or moving with a constant velocity will remain at rest or continue

to move with a constant velocity along a straight line.

Law 2: If an external force acts on a particle, the particle will be accelerated in the direction of the force and the magnitude of the acceleration will be directly proportional to the force and inversely proportional to the mass of the particle.

Law 3: For every action or force there is an opposite and equal reaction. The forces of action and reaction between contacting bodies are equal in magnitude, opposite in direction, and collinear. (Riley, 6)

The first law of motion describes a particle in equilibrium. Because the beams we are considering are at rest, the sum of all forces in any one direction must equal zero. If the beams were beams were accelerating in some direction the sum of the forces would equal the mass multiplied by the acceleration.

Beams are described as either statically determinate or statically indeterminate. A beam is considered to be statically determinate when the support reactions can be solved for with only statics equations. The condition that the deflections due to loads are small enough that the geometry of the initially unloaded beam remains essentially unchanged is implied by the expression “statically indeterminate”. Three

equilibrium equations exist for determining the support reactions when forces acting on a beam are in only one plane. That is, if a beam is loaded in the xy-plane, as in figure 2a (page 6), the three equilibrium equations become

if a is in the loading plane. With these three equations, three unknown reaction components can be determined mathematically. When the forces are applied in this plane (xy) and perpendicular to the length of the beam, the ∑F= 0 equation is naturally fulfilled. In order for the beam to bestatically determinate, only two reaction components can exist. The two remaining equilibrium equations become

Simply supported, overhanging, and

cantilever beams are statically determinate. The other types of beams described above are statically indeterminate. Statically indeterminate beams also require load deformation properties to determine support reactions. When a structure is statically indeterminate at least one member or support is said to be redundant, because after removing all redundancies the structure will become statically determinate.

Forces and moments are the internal forces transferred by a transverse cross section (section a, figure 3c) necessary to resist the external forces and remain in equilibrium. Stresses,strains, slopes, and deflections are a result of and a function of the internal forces. The simply supported single span beam in figure 3a is introduced to a uniform load (w) and two concentrated loads (P1) and (P2). Using the equilibrium equations and a free body diagram the support reactions for the beam in figure 3a will be determined. This example will also show how internal forces (shear and moment) can be found at any point along the beam.This same method is applicable to any statically determinate beam.

Finding the support reactions requires a free body diagram that notes all external forces that act on the beam and all possible reactions that can occur. The free body diagram for the entire

beam is illustrated in figure 3b. Applying the equations of equilibrium as defined above, we see that the x component of the support reaction at a is the only x component and must equal zero to satisfy the equilibrium equation.

It is important to note here that when

determining the free body diagram, the direction that the support reaction components are drawn is unimportant. It is however important to set a direction in which forces are positive. This sign convention is denoted in the upper left hand of figures 3b and 3c. We could show the unknown reactions at the support to be down, and when the equations of equilibrium are applied the reaction will result in a negative value. This would tell us that they are actually positive, and acting in the opposite direction. It does enable a less confusing solution if the reactions are initially assumed in the correct direction.The equilibrium equations applied in the y direction are as follows:

(1)

Since P1 and P2 are known values, this equation yields two unknowns, which can not be solved for with only equation. The third equilibrium

equation must now be applied. In order to apply this last equation, a point on the length of the beam should be chosen to sum the moments about. It is rational to choose a point that an unknown passes through, thus eliminating it from the final equation, ultimately simplifying the mathematics thereafter. A point at the left support iscommonly chosen in engineering practice to sum moments. The beam can then be analyzed from left to right. The moment about any point can be visualized as a force at certain distance from the point being considered that will cause the beam to rotate about that point either clockwise or counter-clockwise. Engineers commonly refer to moments as positive when they rotate the beam counter-clockwise. It was stated earlier that a distributed load can be simplifiedby placing the area of the load at the centroid of the load. The area is equal to the load (w) multiplied by the length that it is applied (x). In this example the load is constantly distributed, therefore the centroid occurs at the mid-point of the beam. Summing the moments about the reaction at the left of the beam, the equations of equilibrium result as follows:

Simplifying,

The reaction at b is now solvable when P1,

P2, a, b, w, and the length of the beam (L) are given or known. The reaction can now be solved for, when Ryb (equation (2)) is substituted into equation (1). Doing this yields:

Solving for Rya,

The reaction at a is now solvable when P1, P2, a, b, w, and the length of the beam (L) are given or known.

The internal forces required to resist the external forces at any arbitrary cross section of the beam (section a in figure 3c) can be found statically by breaking the beam into separate pieces at that section. Under the action of internal forces, the separate pieces of the beam are in equilibrium, since the unbroken beam is in equilibrium. By applying the equilibrium

equation ΣFx = 0 to the free body diagram in figure 3c, the internal force, shear (V), can be found for any point along the length of the beam. The resulting internal force is as follows:

By applying the equilibrium equation ΣMza = 0 to the free body diagram in figure 3c, the internal force, moment (M), can be found for any point along the length of the beam. The resulting internal force is as follows:

The Physical ModelIn order to test the assumptions associated with the statical methods described previously, the spring 2002 Math Modeling class (MATH 4583) devised and carried out an experiment with a physical model. The elements involved in the composition of the physical model were two electronic scales, two pencils, two meter sticks, a string, and various weights. The two electronic scales served to measure the amount of force transferred to the supports. A pencil laying flat was placed on each of the two supports to act as a roller support, and a meter stick was placed as a beam or bridge between the two scales. Two meter sticks, one more rigid than the other, were used in separate instances to test if the

properties of the “beam” would effect the distribution offorces to the supports. A string tied in a loop was used to hang the various weights in various combinations from the “beam”. The apparatus is illustrated in figure 4 below.

A series of successive experiments were performed, and data was recorded for the following beam and weight arrangements,

Based on the equilibrium equations, for a simply supported single span beam with a single weight (P) at any point along its length (L) yields values for the reactions for Rya and Ryb as,

Where a and b are the lengths from the left support to the concentrated load and from the right support to the concentrated load, respectively. These statically determined values for experiment 1 appear in the data table below for each reaction. The percent error has also be calculated for each of these compared to their respective reaction. The chart below shows how the weight was distributed to each of the supports in the experiment. Based solely on the equilibrium equation ΣFy = 0, a percent error value is also given for each of these experimental instances. A comparison of the experimental values with the statical solutions is used to support the assumptions made in determining the statical solutions.

The data and charts for the experiments 2 through 5 can be found in the Appendix. For experiment 2, the same equations for the reactions are used for comparison with theexperimental data. Experiment 3 dealt with an overhanging beam. We loaded the cantilever portion of the “beam” in this experiment. Prior to loading the beam, the class realized that the “beam” would lift up from the far support. The beam would essentially act as “teeter-totter” employing afulcrum effect on the support closest to the overhang. In order to compensate for this, we placed a weight on the far support, and “zeroed”

the scale before placing the weight in the overhang.

The result was a negative value for the far support. This confirmed our assumption confirmingthat the reaction would be “negative”.Based on the equilibrium equations, for a simply supported overhanging beam with a single weight (P) at any point on the overhang yields values for the reactions for Rya and Ryb as,

Where L is the length between supports and a is the length from the concentrated load on the overhang to the nearest (b) support.In experiments 4 and 5, we placed two discrete concentrated loads at various points on thebeam. Based on the equilibrium equations, for a simply supported single span beam with twounequal weights (P1 & P2) at any points along its length (L) yields values for the reactions forRya and Ryb as,

Where a and b are the lengths from the left support to the concentrated load (P1) and from

the right support to the concentrated load (P2), respectively.

Comparing the statics beam example, and the statical solutions to the different experiments (that were proved to be reasonable, quite near exact in fact), it becomes evident that if two of the same statically determinate beams with different loads are analyzed separately, the resulting reactions for each of the beams could be summed to determine the reactions for a single beam with thatcombination of loads. For example, if a statically determinate “beam 1” has a point load (P) at the mid-span, and a statically determinate “beam 2” of the same length and support type has a point load (P) at its mid-span, the reaction at each support for each beam is 0.5(P). If Both point loads are placed on one statically determinate beam of the same length and support type the point load would be (P + P) or 2P, thus resulting in a reaction at each support of 0.5 +0.5 or P.

Title Of Experiment: Reactions of simply supported beamsAim: (i) To determine the reactions RA and RB for a beam simply supported at its ends. (ii) To determine the values of RA and RB as a given load moves from one end of a simply beam to the other.Apparatus:

1.Two spring balances2. A steel beam of hollow-section 3. Weights4.Scale hangers

Description: It consists of a holding two spring balance suspended vertically upright. A steel beam is hung on the hooks at the bottom of the spring balances. The load is applied to the beam. A spring balance operates under the law of gravity i.e. gravitational force. It measures the force of gravity exerts on it, therefore measuring

the weight in KgF of the object under this force’s field.

Similarly , RA = W (L – x)

LIf the radius of Radius of RA and RB are

plotted against the positions, of the load, the resulting graphs, gives the variation of the values of RA and RB as the load moves from one end of the to the other.

These will all duty be taken into consideration in this experiment.

Procedure:(i) Hang the steel beam on the hooks at

the bottom-end of the spring balances. Make sure beam is perfectly horizontal. Put a load hanger at the mid-point of the beam and read the spring-balances. Put a 2Kgf weight on the load hanger and read the spring balances. Increases the load in steps of 2Kgf up to 16Kgf and read the balances at each incremental loading. Remove all this weights at the end of this part of the experiment.

(ii) Place the load hanger weight under the spring balance A and read the two spring balances. Put an 8Kgf weight on

the load-hanger and read the two balances. Move the load-hanger, with the 8Kgf weight on, to the next mark on the beam and read the weight along the beam until the load-hanger is right under the spring balance B. Read the two balances at each movement.

Results:TABLE OF RESULTSThe values of reactions RA and RB load applied at

the midpointOf the beam, WEIGHT

REACTION A

REACTION B

(KG) (N) (KG) (N) (KG) (N)0 0 0 0 0 02 20 1 10 1 104 40 2 20 2 206 60 3 30 3 308 80 4 40 4 40

10 100 5 50 5 5012 120 6 60 6 60

14 140 7 70 7 7016 160 8 80 8 80

The values of Reactions RA and RB as the load applied was transferred every 100mm from A and B along the beam.

POSITION OF LOAD

REACTION A

REACTION B

(KG) (CM) (KG) CM) (KG) (CM)8 0 0 0 0 08 20 1 10 1 108 40 2 20 2 208 60 3 30 3 308 80 4 40 4 408 100 5 50 5 508 120 6 60 6 608 140 7 70 7 708 160 8 80 8 80

Discussion and Conclusion:From the result, for the first part of the

experiment,a relation between RA and RB is observed. As both reactions are at equidistance from the load applied, they both share the weight of the load. Thus the magnitudes of the reactions are half that of the loading and are equal to each other i.e. RA and RB = ½ (Weight of load).

For the second part of the experiment, the position of the position of the load on the beam varies therefore the two reactions vary as well; as the load is borne as a function of the distance of it, from that reaction.

When the load is at A, RA = Weight of the load while the reaction RB = 0. As the load is at this point the reaction RA is maximium (equal to load). As the load is shifted away from the maximum reaction equal to the load, and RA is null or zero. Here an inversely proportional relation is observed. Comparing the experimental values and those of the theoretical for this part of the experiment, a deviation is seen to occur in

values. Nevertheless, this can be as a result (for the experimental part) of zero error on the metre rule of the spring balance as some approximations were made.

The reactions RA and RB have thus been determined,

(1) When the load is constantly mid-point of the beam and increases in magnitude and

(2) When the magnitude of the load is constant but moves from one end of the beam to the other.The results are already given.

Precautions Taken:Zero error, of the metre rule, in measuring

the length of the beam was avoided. I also made sure that the beam was perfectly horizontal.

Conclusion:It is thus proven that for every action there is

an equal, though opposite reaction.

References:(1) Strength of materials by G.H. Ryder and(2) Strength of materials by R.S. Khurmi.Also,Materials were gotten from www.google.com

EXPERIMENT 2

NAME: MOMOH ITAMOMOH E.P.

MATRIC NO: 070403038 DEPARTMENT: ELECTRICAL/ELECTRONICS ENGINEERING

DATE PERFORMED: 13/08/08

DATE SUBMITTED: 3/09/08

TITLE: DEFLECTION OF SIMPLY SUPPORTED BEAMS

AIM:To determine the deflection of a simply

supported beam.

THEORY: BEAMS: STRAIN, STRESS,

DEFLECTIONSThe beam, or flexural member, is frequently encountered in structures and machines, and its elementary stress analysis constitutes one of the more interesting facetsof mechanics of materials. A beam is a member subjected to loads applied transverse to the long dimension, causing the member to bend. For example, a simply-supported beam loaded at its third-points will deform into the exaggerated bent shape shown in Fig. 3.1 Before proceeding with a more detailed discussion of the stress analysis of beams, it is useful to classify some of the various types of beams and loadings encountered in practice. Beams are frequently

classified on the basis of supports or reactions. A beam supported by pins, rollers, or smooth surfaces at the ends is called a simple beam. A simple support will develop a reaction normal to the beam, but will not produce a moment at the reaction. If either, or both ends of a beam projects beyond the supports, it is called a simple beam with overhang. A beam with more than simple supports is a continuous beam. Figures 3.2a, 3.2b, and 3.2c show respectively, a simple beam, a beam with overhang, and a continuous beam. A cantilever beam is one in which one end is built into a wall or other support

so that the built-in end cannot move transversely or rotate. The built-in end is said to be fixed if no rotation occurs and restrained if a limited amount of rotation occurs. The supports shown in Fig. 3.2d, 3.2e and 3.2f represent a cantilever beam, a beam fixed (or restrained) at the left end and simply supported near the other end (which has an overhang) and a beam fixed (or restrained) at both ends, respectively. Cantilever beams and simple beams have two reactions (two forces or one force and a couple) and these reactions can be obtained from a free-body diagram of the beam by applying the equations of equilibrium. Such beams are said to be statically determinate since the reactions can be obtained from the equations of equilibrium.Continuous and other

beams with only transverse loads, with more than two reaction components are called statically indeterminate since there are not enough equations of equilibrium to determine the reactions.

Figure 1

Example of a bent beam (loaded at its third points)

Various types of beams and their deflected shapes: a) simple beam, b) beam with overhang, c) continuous beam, d) a cantilever beam, e) a beam fixed (or restrained) at the left end and simply supported near the other end (which has an overhang), f) beam fixed (or restrained) at both ends.

Examining the deflection shape of Fig. 3.2a, it is possible to observe that longitudinal elements of the beam near the bottom are stretched and those near the top are compressed, thus indicating the simultaneous

existence of both tensile and compressive stresses on transverse planes. These stresses are designated fibre or flexural stresses. A free body diagram of the portion of the beam between the left end and plane a-a is shown in Fig. 3.3. A study of this section diagram reveals that a transverse force Vr and a couple Mr at the cut section and a force, R, (a reaction) at the left support are needed to maintain equilibrium. The force Vr is the resultant of the shearing stresses at the section (on plane a-a) and is called the resisting shear and the moment, Mr, is the resultant of the normal stresses at the section and is called the resisting moment.

Figure 3.3 Section of simply supported beam.The magnitudes and senses of Vr and Mr may be obtained form the equations of equilibrium ∑Fy = 0 and ∑MO = 0 where O is any axis perpendicular to plane xy (thereaction R must be evaluated first from the free body of the entire beam). For the present

the shearing stresses will be ignored while the normal stresses are studied. The magnitude of the normal stresses can be computed if Mr is known and also if the law ofvariation of normal stresses on the plane a-a is known. Figure 3.4 shows an initially straight beam deformed into a bent beam. A segment of the bent beam in Fig. 3.3 is shown in Fig. 3.5 with the distortion highly exaggerated. The following assumptions are now made

(i) Plane sections before bending, remain plane after bending as shown in Fig. 3.4 (Note that for this to be strictly true, it is necessary that the beam be bent only with couples (i.e., no shear on transverse planes), that the beam must be proportioned such that it will not buckle and that the applied loads are such that no twisting occurs.

Figure 3.4 Initially straight beam and the deformed bent beam.

Figure 3.5 Distorted section of bent beamii) All longitudinal elements have the same length such the beam is initially straight and has a constant cross section.iii) A neutral surface is a curved surface formed by elements some distance, c, from the outer fibre of the beam on which no change in length occurs. The intersection of the neutral surface with the any cross section is the neutral axis of the section.

StrainAlthough strain is not usually required for engineering evaluations (for example,failure theories), it is used in the development of bending relations. Referring to Fig. 3.5,the following relation is observed:

where dy is the deformation at distance y from the neutral axis and dc is the deformation at the outer fibre which is distance c from the neutral axis. From Eq. 3.1, the relation forthe deformation at distance y from the neutral axis is shown to be proportional to the deformation at the outer fibre:

Since all elements have the same initial

length, Dx , the strain at any element can be determined by dividing the deformation by the length of the element such that:

Figure 3.6 Undeformed and deformed elements

Note that e is the in the strain in the x direction at distance y from the neutral axis and that e =e x . Note that Eq. 3.3 is valid for elastic and inelastic action so long as the beam does not twist or buckle and the transverse shear stresses are relatively small.

An alternative method of developing Eq. 3.3 involves the definition of normal strain. An incremental element of a beam is shown both undeformed and deformed in Fig. 3.6.Note once again that any line segment Dx located on the neutral surface does not changes its length whereas any line segment Ds located at the arbitrary distance y fromthe neutral surface will elongate or contract and become Ds' after deformation. Then by definition, the normal strain along Ds is determined as:

Strain can be represented in terms of

distance y from the neutral axis and radius of curvature r of the longitudinal axis of the element. Before deformation ds = dx but after deformation Dx has radius of curvature r with center of curvature at point O'. Since Dq defines the angle between the cross sectional sides of the incremental element, ds = dx = rDq . Similarly, the deformed length of Ds becomes Ds'= (r - y) Dq .

Substituting these relations into Eq. 3.4 gives:

Eq. 3.5 can be arithmetically simplified as e = -y / r . Since the maximum strain occurs at the outer fibre which is distance c from the neutral surface, emax = -c / r = ec , the ratio of strain at y to maximum strain is

which when simplified and rearranged gives the same result as Eq. 3.3:

Note that an important result of the strain

equations for e = -y / r and emax = -c / r = ec indicate that the longitudinal normal strain of any element within the beam depends on itslocation y on the cross section and the radius of curvature of the beam's longitudinal axis at that point. In addition, a contraction (-e ) will occur in fibres located "above" the neutral axis (+y) whereas elongation (+e ) will occur in fibres located "below" the neutral axis (-y).

Stress:The determination of stress distributions of beams in necessary for determining the level of performance for the component. In particular, stress-based failure theories

require determination of the maximum combined stresses in which the complete stress state must be either measured or calculated.

Normal Stress: Having derived the proportionality relation for strain, ex , in the xdirection, the variation of stress, s x , in the x-direction can be found by substituting s for e in Eqs. 3.3 or 3.7. In the elastic range and for most materials uniaxial tensile andcompressive stress-strain curves are identical. If there are differences in tension and compression stress-strain response, then stress must be computed from the straindistribution rather than by substitution of s for e in Eqs. 3.3 or Note that for a beam in pure bending since no load is applied in the z-direction, sz is zero throughout the beam. However, because of loads applied in the y-direction to obtain the bending moment, sy is not zero, but it is small enough compared to s x to neglect. In addition, s x while varying linearly in the y direction is uniformly distributed inthe z-direction. Therefore, a beam under only a bending load will be in a uniaxial, albeit a non uniform, stress state.

Figure 3.7 Stress (force) distribution in a bent beamNote that for static equilibrium, the resisting moment, Mr, must equal the appliedmoment, M, such that ∑MO = 0 where (see Fig. 3.7):

and since y is measured from the neutral surface, it is first necessary to locate this surface by means of the equilibrium equation σFx = 0 which gives ∫ σdA = 0 A . For the case of elastic action the relation between s x and y can be obtained from generalized Hooke'slaw σx.

and the observation that

The resulting stress-strain relation is for the uniaxial stress state such that σx =Eεx which when substituted into Eq. 3.3 or 3.7 gives

Substituting Eq. 3.9 into Eq. 3.8 gives:

Note that the integral is the second moment of the cross sectional area, also known as the moment of inertia, I, such that

Figure 3.8 Action of shear stresses in unbonded and bonded boards

Substituting Eq. 3.11 into Eq. 3.10 and rearranging results in the elastic flexure stress equation:

where s x is the normal bending stress at a distance y from the neutral surface and acting on

a transverse plane and M is the resisting moment of the section. At any section of thebeam, the fibre stress will be maximum at the surface farthest from the neutral axis such that.

where Z=I/c is called the section modulus of the beam. Although the section modulus can be readily calculated for a given section, values of the modulus are often included intables to simplify calculations.

Shear Stress: Although normal bending stresses appear to be of greatest concern for beams in bending, shear stresses do exist in beams when loads (i.e., transverseloads) other than pure bending moments are applied. These shear stresses are ofparticular concern when the longitudinal shear strength of materials is low compared to the longitudinal tensile or compressive strength (an example of this is in wooden beams with the grain running along the length of the beam). The effect of shear stresses can be visualized if one considers a beam being made up of flat boards stacked on top of one another without being fastened together and then loaded in a direction normal to the surface of the boards. The resulting deformation will appear somewhat like a deck of cards when it is bent (see Fig. 3.8a). The lack of such relative sliding and deformation inan actual

solid beam suggests the presence of resisting shear stresses on longitudinal planes as if the boards in the example were bonded together as in Fib. 3.8b. The resultingdeformation will distort the beam such that some of the assumptions made to develop the bending strain and stress relations (for example, plane sections remaining plane) are not valid as shown in Fig. 3.9

Typeequationhere .

Figure 3.9 Distortion in a bend beam due to shear

The development of a general shear stress relation for beams is again based on static equilibrium such that ∫F = 0. Referring to the free body diagram shown in Fig.3.10, the differential force, dF1 is the normal force acting on a differential area dA and is equal to σ dA . The

resultant of these differential forces is F1 (not shown). Thus, F1 = ∫ σ dA integrated over the shaded area of the cross section, where s is the fibre stress at a distance y from the neutral surface and is given by the expression

Figure 3.11 Shear and normal stress distributions in a uniform cross section beam

Finally, the maximum shear stress for certain uniform cross section geometries can be calculated and tabulated as shown in Fig. 3.12. Note that a first order approximationfor maximum shear stress might be made by dividing the shear force by the cross sectional area of the beam to give an average shear stress such that tav ≈ V/A. However,if the maximum shear stress is interpreted as the critical shear stress, than an error of 50% would result for a beam with a rectangular cross section where tmax ≈ 3V/2A which is 1.5 times tav ≈ V/A.

Figure 3.12 Maximum shear stresses for some common uniform cross sections DeflectionsOften limits must be placed on the amount of deflection a beam or shaft may undergo when it is subjected to a load. For example beams in many machines must deflect just the right amount for gears or other parts to make proper contact. Deflections of beams depend on the stiffness of the material and the dimensions of the beams as well asthe more obvious applied loads and supports. In order of decreasing usage four commonmethods of calculating beam deflections are:

1) double integration method, 2)superposition method, 3) energy (e.g., unit load) method, and 4) area-moment method.The double integration method will be discussed in some detail here.

Deflections Due to Moments: When a straight beam is loaded and the action is elastic, the longitudinal centroidal axis of the beam becomes a curve defined as "elasticcurve." In regions of constant bending moment, the elastic curve is an arc of a circle of radius, r, as shown in Fig. 3.13 in which the portion AB of a beam is bent only withbending moments. Therefore, the plane sections A and B remain plane and the deformation (elongation and compression) of the fibres is proportional to the distancefrom the neutral surface, which is unchanged in length. From Fig. 3.13:

from which

and finally

moment, M, the stiffness of the material, E, and the moment of inertia of the cross section, which relates the radius of curvature of the neutral surface of the beam to the bending

Figure 3.13 Bent element from which relation for elastic curve is obtained

Figure 3.15 Sign conventions used for deflection

Eqs. 3.28 and 3.29 show that except for the factor EI, the area under the moment diagram between any two points along the beam gives the change in slope between thesame two points. Likewise, the area under the slope diagram between two points along a beam gives the change in deflection between these points. These relations have beenused to construct the series of diagrams shown in Fig. 3.16 for a simply supported beam with a concentrated load at the center of the span. The geometry of the beam was used tolocate the points of zero slope and deflection, required as the starting points for the construction.

Figure 3.16 Illustration of various elastic relations for a beam in three-point loadingIt is important to remember that the calculation of deflections from elastic curve relations is based on the following assumptions:1) The square of the slope of the beam is assumed to be negligible compared to unity2) The beam deflection due to shear stresses is negligible (i.e., plane sections remain plane)3) The value of E and I remain constant for any interval along the beam.The double integration method can be used to solve Eq. 3.24 for the deflection y as a function of distance along the beam, x. The constants of integration are evaluated by applying the applicable boundary conditions. Boundary conditions are defined by a known set of values of x and y or x and dy/dx at a specific point in the beam. One boundary condition can be used to determine one and only one constant of integration. A roller or pin at any point in a beam (see Figs.3.17a and 3.17b) represents a simple support which cannot deflect (y=0) but can rotate (dy/dx¹0). At a fixed end (see Figs. 3.17c and 3.17d) the beam can neither deflect orrotate (y=0 and dy/dx=0).

Matching conditions are defined as the equality of slope or deflection, as determined at the junction of two intervals from the elastic curve equations for bothintervals.

An example of the use of integration methods is as follows for a simply supported beam in three-point loading. The loading condition, free body, shear and momentdiagrams are shown in Fig. 3.19.

Figure 3.19 Loading condition, free body, shear and moment diagrams

There are two boundary conditions: at x=0, y1=0 and at x=L, y2=0

There are two matching conditions: at x=a, y'1=y'2 and at x=a, y1=y2

RESULTS:12*12*1000mm Steel barLOAD DEFLECTION

Kg N Dial Gauge

Reading

(mm)

0 0 00.0 0.002 20 96.0 0.964 40 200 2.006 60 298 2.988 80 397 3.9710 100 497 4.9712 120 601 6.0114 140 696 6,9616 160 798 7.98

12*12*1200mm Steel barLOAD DEFLECTION

Kg N Dial Gauge

Reading

(mm)

0 0 00.0 0.002 20 167.0 1.674 40 340 3.406 60 512 5.128 80 687 6.8710 100 858 8.5812 120 1029 10.2

914 140 1199 11.9

916 160 1375 13.7

5

25*6*1200mm Steel bar

LOAD DEFLECTIONKg N Dial

Gauge Reading

(mm)

0 0 00.0 0.002 20 36.0 0.364 40 80.0 0.806 80 130.0 1.308 100 175.0 1.7510 120 224 2.2412 140 275 2.7514 160 314 3.1416 160 372 3.7218 180 425 4.2520 200 477 4.77 22

220 528 5.28

24 240 579 5.79

REPORT:The experiments carried out to determine the

deflection of simply supported beams were done using:

1) A 25mm by 6mm flat bar of 1200mm span;

2) A 12mm by 12mm square bar of 1000mm span; and

3) A 12mm by 12mm square bar of 1200mm span.

Thus, the results and the graphs obtained in the former pages.

DISCUSSION AND CONCLUSION:From the theory, it is already stated that the deflection depends on the applied load, the length of the beam (span), the cross- sectional area, the moment of inertia and the results obtained from the experiments carried out, it will be thus proven.

Cross- Sectional Area:Comparing tables 3 and 2. Table 3 shows the deflection of a beam with 25mm bby 6mm flat bar with 1200mm span while table 2 is of a beam 12mm by 12mm square with 1200mm span. Both have the same span but different cross- sectional areas.The beam of 25mm by 6mm given an area of 150mm2 has a lesser deflection at a 2kg load than that of the 12mm by 12mm, area 144mm2 beam. This is shown still as the load is increased.

Span or Length of the Beam:Table 1 depicts the deflection of a 1000mm span while table 2 depicts a 12mm by 12mm beam of 1200mm span. Both have the same areas but different spans. Comparing the two tables, the beam of 1200mm span has a greater deflection than that of 1000mm span, at a load of 2kg and consecutive

loads. This proves that deflection depends on the span.

Load:From all 3 tables, it is seen that as the load increases, the deflection increases and so on.

Moment of Inertia:From all calculations, using the slope obtained from the graph of each beam, it is seen that the deflection does depend on the moment of inertia.

Precautions Taken:1) I continually made sure that the tip of the

dial gauge was in contact with the beam first touching at its centre.

2) I made sure the instrumental parts were not touched not to create false deflection.

3) I avoided error due to parallax when measuring the deflection on the beam.

Accuracy:

As a result of errors due to parallax and on the metre rule, dial gauge metres which are unavoidable, the results obtained might not be as accurate, but the error incurred is definitely minimum.

Use of Results:The results obtained can be used to determine the deflection in a material. I.e. how much load the material can carry, how long it can last under loading without any sign of failure or deformation as if it fails, the load supported will be destroyed.It is applied in bridge construction and buildings.

Difficulties Encountered:Placing the load on the hangers slowly so as to read the deflection was hard.With a heavy load, the rate of deflection (the spinning of the pointer) was fast so it was difficult to read the deflections.

References:1) Strength of Materials by R.S. Khurmi2) Strength of Materials by G.H. Ryder3) Strength of Material by Beer & Johnson4) www. Wikipedia.com/deflection of beam

NAME: FAPOHUNDA OLUSHOLA

MATRIC: 070403038

DEPARTMENT: ELECTRICAL/ELECTRONIC

COURSE: CEG 202 (LAB)

TITLE: TENSILE STRESS ON STEEL

GROUP: Three

DATE PERFORMED: 23/08/08

DATE SUBMITTED: 8/09/08

AIM:

(I) TO DETERMINE THE YOUNG’S MODULUS OF ELASTICITY OF STEEL RODS FROM TENSILE TEST ON THESE RODS.

(II) TO DETERMINE THE TENSILE STRENGTH (I.E. THE ULTIMATE TENSILE STRESS) OF A STEEL ROD.

APPARATUS:

(1) AVERY UNIVERSAL TESTING MACHINE(2) STEEL OR BRASS BEAM

(3) VERNIER CALIPERS(4) FIRM – JOINT CLIPPERS(5) ALUMINIUM RODS

THEORY:

On applying load to a material, deformation occurs. If the material recovers its original position immediately after removing the load, the deformation is said to be elastic. Purely elastic deformation in associated with the stretching of primary bonds in materials.

Some major terms used in tensile test of steel include:

Ductility: extent to which a material can sustain plastic deformation without rupture. Elongation and reduction in area are common indices of ductility in a tensile test.

Elastic limit: greatest stress that can be added to a material without causing permanent deformation. For metals and other materials that have a significant linear portion on their stress/strain curve, elastic limit is approximately equal to proportional limit. For materials that do not exhibit a significant proportional limit, elastic limit is an arbitrary approximation.

Elasticity: ability of a material to return to its original shape when load causing deformation is removed.

Elongation: measure of ductility of a material determined in a tension test. It is the increase in gauge length (measures after fracture) divided by original gauge length. Higher elongation indicates higher ductility.

Flow stress: stress required to cause plastic deformation.

Fracture stress: true stress generated in a material at fracture.

Modulus of elasticity: rate of change of strain as a function of stress, i.e. the slope of the straight line portion of a stress strain curve. Tangent modulus of elasticity is the slope of the stress strain curve at any point. Secant modulus of elasticity is stress divided by strain at any given value of stress or strain.

It is also called stress-strain ratio. Tangent and secant modulus of elasticity are equal up to proportional limit. Depending on the type of loading represented by the stress-strain curve, modulus of elasticity may be reported as: compressive modulus of elasticity; flexural modulus of elasticity; shear modulus of elasticity; tensile modulus of elasticity; torsional modulus of elasticity.

Modulus of elasticity may be determined by dynamic testing, where it can be derived from complex modulus. Modulus alone generally refers to tensile modulus of elasticity. Shear modulus is almost equal to torsional modulus and both are called modulus of rigidity. Moduli of elasticity in tension and compression are approximately equal and are known as Young’s Modulus.

Necking: localized reduction of cross-sectional area of a specimen under tensile load. It is disregarded in calculating engineering stress but is taken into account in determining true stress.

Offset yield strength: arbitrary approximation of elastic limit. It is the stress that corresponds to the point of intersection of a stress-strain curve and a line parallel to the linear portion of the curve. Offset refers to the distance between the origin of the stress –strain curve and the point of intersection of the parallel line and the stress axis. Offset is expressed in terms of strain.

Plastic deformation: deformation that remains after the load causing it is removed. It is the permanent part of the deformation beyond the elastic limit of a material. It is also called plastic strain and plastic flow.

Plasticity: tendency of a material to remain deformed, after reduction of the deforming stress, to value equal to or less than its yield strength.

Proportional limit: highest stress at which stress is directly proportional to strain. It is the highest stress at which curve in a stress-strain diagram is a straight line. It is equal to elastic limit for many materials.

True strain: instantaneous percentage of change in length of specimen in mechanical test. It is equal to the natural logarithm of the ration of length at any instant to original length.

True stress: applied load divided by actual area of the cross section through which load operated.

It takes into account the change in cross section that occurs with changing load.

 

Stress in the force per unit area or more expansively, a system of action and reaction acting over the cross-sectional area of a member, denoted as

σ = F/A where F = force, A = Area.

Strain is concerned with geometrical operations. If a force is applied to a material, not only do we induce in the fibres a state of stress, but in some respect, we alter the size or shape of the material Thus, strain is defined as the change in length per unit original length of the material. It is denoted as

ε = ΔL/L where ΔL = deformation/elongation, L = length.

As observed by Hooke, the relationship between stress and strain in a direct proportional relation

with a condition that it must be within an elastic limit

σ α ε

σ = Eε

where E is the modulus of elasticity (Young’s Modulus) which is related to the potential energy well of the inter atomic bond.

Hooke’s law relates these parameters as stated, σ = Eε. It is implicit here that only axial stresses and strains are of interest. Note, it is assumed σ = 0, where ε = 0 so that σ =Eε represents the first part of the load-displacement curve, a straight line that passes through the origin with E as the slope. If permanent deformation occurs, it is called a plastic. The onset of plastics deformation corresponds to a stress level necessary to initiate the motion of dislocation ( a type of defect ) in crystalline materials. The stress necessary to produce permanent deformation is the yield strength of the material. Some materials exhibit a sharp yield point, while

others, a slow change in slope at the end of the elastic range. Conventionally, the yield strength is defined as the stress necessary to produce a plastic strain of 0.2% (elongation). In some materials, part of all of the elongation remaining after the load has been removed may be gradually recovered with time, thus, the deformation is inelastic.

When small load or little stress is applied to a material and it produces a large deformation (strain), the material is said to be ductile. If the material is reluctant to strain, and even at high stress yields only a small strain, it is said to be brittle or rigid. Thus in ductile materials, the strain to fracture is relatively large compared with brittle materials.

Plastic deformation of ductile material can require progressively higher stress because dislocations multiply in the process and their motion becomes more difficult due to the increased degree of interaction among them.

This process is called work-hardening. Sometimes it is possible to observe bands propagating along the specimen during work-hardening. These are luders bands indicating the multiplication and motion of dislocation. They will not be visible unless the specimen is highly polished. Uniform elongation of the gauge length occurs when the hardening rate is faster than the decrease in cross-sectional area. Thus

dθ/θ ≥ dA/A

If the hardening is too low, a runaway situation called necking develops. This corresponds to the load reaching a maximum, at which point, the tensile deformation is homogenous and the strain is no longer uniform. The corresponding stress is called the Ultimate Tensile Strength or UTS.

The elongation of failure, which is the permanent engineering strain after fracture, is an expression of material ductility. It does not include elastic strain but does include uniform strain and the localized necking strain. The elongation to failure

is usually stated as percent strain over a given gauge length. A second indication that is related to the ductility of a material is the percent reduction of area (A) which is defined as

RA = Original Cross Sectional Area – Minimum Final Area

Original Cross Sectional Area

= Ao - Amin = decrease in area x 100 Ao original area

This deformation process is terminated by fracture. In a brittle material, this occurs by the propagation of cracks initiated at the microscopic flows in the materials. Cracks propagated by cleavage, which involves breaking of atomic bonds along specific crystallographic plane with the work of fracture spent primarily on creating a new surface (i.e. surface energy). On the other hand, however, ductile materials tend to fail by nucleation of micro-voids at second phase

particles and the subsequent growth and coalescence of these micro-voids. Since plastic deformation consumers significant amounts of energy in the little form of creation and motions of dislocation, ductile tearing is usually associated with a higher work of fracture.

The area under the engineering stress strain curve is a measure of the energy needed to fracture the specimen. It has units of work per unit volume of the gauge length and is sometimes a measure of materials’ toughness

W / (AOLO) = ∫ε σdε

Engineering stress is the force per unit original cross sectional area of the specimen σ = F/Ao while engineering strain is the elongation per unit original length of the specimen ΔL/Lo. The true stress and strain are determined from the instantaneous dimensions during the test. Consequently, the engineering stress strain curve does not give a true indication of the deformation characteristics of a metal because it is based

entirely in the original dimensions of the specimen and these dimensions change continuously during the test.

Also, a ductile metal, which is pulled in tension becomes unstable and necks down during the course of the test. Because the cross-sectional area of the specimen is decreasing rapidly at the stage in the test, the load required for continuous deformation falls off. The mean stress based on original area likewise decreases, and this produces the fall off in the stress-strain curve beyond the point of maximum load. Actually the metal continues to strain-harden, all the way up to the fracture, so that the stress required to produce further deformation should also increase. If the true-stress, based on the actual cross sectional area of the specimen is used, it is found that the stress-strain curve increases continuously up to a point (fracture) as shown in figure 2. If the strain measurement is also based on instantaneous measurements, the curve, which is obtained, is known as a true stress, true-

strain curve. This is also known as a flow-curve since it represents the basic plastic-flow characteristics of the material. Any point on the flow curve can be considered the yield stress for a metal strained in tension by the amount shown on the curve. Therefore, if the load is removed at this point and then reapplied, the material will behave elastically throughout the entire range of reloading.

There is no significant difference in the engineering and true strains when all measurement are of small strains (typically when deformation is still elastic). For the instantaneous true strain increment dε, we have,

dε = dL/L by integrating

∫

For strains of about 1%, the error in using the engineering strain, versus the order 10-4.

The yield stress, ultimate tensile stress and young’s modulus of a material can all be

determined from the stress the engineering stress – strain curve for that material. The curve shown in the figure below is titally of metallic behaviour. At small strain values (the elastic region) the relationship between stress and strain is almost linear. Within this region, the slope of the stress-strain curve is defined as the elastic modulus. Since many metals lack a sharp yield point, i.e. a sudden observable transition between the elastic region and the plastic region, the yield point often defined as the stress that gives rise to 2% permanent plastic strain. By this convention, a line is drawn parallel to the elastic region of the material, starting at a strain level of 0.2% (or 0.002mm). The point at which this line intersects the curve is called yield point or the yield stress. The ultimate tensile strength (stress), in contrast is found by determining the maximum stress reached by the material.

PROCEDURE:

The rod was cut to size and centre punched (very lightly) at 100mm centre distance in the middle. It was fixed into the jaws of the universal testing machine in such that there was access to the two centre punched marks i.e. there was a span of 200mm. The firm joint callipers was then used to once again read the actual distance (still 200mm) and the vernier calliper, to read the initial diameter of the rod.

The Avery universal testing machine was then switched on and 0.1tons of tensile force was applied to the rod. As the load was increased for 0.1ton, the distance between the two marks was read. Also, the lower yield point, upper yield point, ultimate stresses were noted. The final temperature measured and the diameter at the break point measured with the two parts carefully fixed together.

RESULTS:

Length (mm)Initial length of the rod 100.0Final length of the rod 114.0

Elongation 14.0Initial diameter of the rod 12.0Final diameter of the rod 9.0

Reduction in diameter 3.0Lower yield point 5.54 tonsUltimate stress 6.62 tons

Upper yield point ( Breaking point)

4.26 tons

REPORT:The equipment to determine the Young’s modulus of elasticity of steel was done from the tensile test in the rod. The tensile strength (the ultimate stress) was also

determined.

As direct reading was not taken of extension against load, a crude graph is drawn using the lower yield point and the origin.

Precautions

Zero error on the metre scale was avoided when using the vernier callipers.

Error due to parallax (zero error) of the metre rule was avoided.

Error due to parallax was avoided when loading the material in the testing machine.

APPLICATIONTensile stress test is used to measure the ability of different materials to carry or withstand load pressures. Tensile test is used on beams, slabs, and other materials as found in bridges and other materials subjected to heavy loads and pressure.

In bridges, for instance, tensile test is used to estimate the maximum load the bridge can carry at an instance or at a particular time such that whenever the maximum strength is exceeded, the bridges collapses.Also, tensile stress is used as beams and columns to estimate the maximum load that can be supported by them. In this case, if anything was wrong. It could result in collapse of the particular portion of the building, or the entire building.In

view of this, re-enforcement is needed to avert any havoc which could result from bending of the material.

DISCUSSION AND CONCLUSION:It has been studied already that whenever an external force acts on a body, the body will deform. Obeying Hooke’s law, if the force acts upon the material within elastic limit, the material will regain its original form (deformation completely disappears). However, beyond this limit, it has been found that deformation doesn’t completely disappear. The behavior of a material to applied

stress can be of four kinds namely

(1)Perfectly Elastic

(2)Brittle

(3)Inelastic

(4)Ductile

The above terms were defined above in the theory. From the results obtained and graphs drawn, The lower yield point is 4.26tons meaning that up to this point, the material behaves elastic i.e. regains its original form (and or shape).

Above this point up to the ultimate stress point which is 6.62tons, the material still exhibits some material elastic properties although it won’t achieve its original shape or form on removal of load.

Beyond this point the material becomes inelastic i.e. shows an appreciate strain even without increase in load.

References:1) Strength of Materials by R.S. Khurmi2) Strength of Materials by G.H. Ryder3) Strength of Material by Beer & Johnson4) www. Wikipedia.com/deflection of beam