Name: Momoh Itamomoh E.P

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.

DISCUSSION

The 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 laws

that 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 be

statically 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 is

commonly 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 simplified

by 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 Model

In 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 of

forces 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.