The Journal of Rowan Engineering

The Journal of Rowan Engineering/

Mechanical

Fall 2001

ME-F01-06

Visual beams II

Joseph Plitz, ME / Michael Resciniti, ME
Aditya Chaubal, ECE / Frank Brown, CEE

Abstract

The goal of the Visual Beams II project was to design, build, and test an instrumented simply supported beam with a computer interface. By the end of the semester, two structurally sound supports were designed and constructed and a computer interface was designed and programmed to meet and surpass the original goal. The result was a visualization of the reactions of an infinite number of combinations of beam shapes, point load and support positions, and point load forces shown on a user-friendly computer program.

introduction

Visual Beams was an idea started in the spring semester of 2001 at Rowan University with the introduction of a cantilever beam designed with load and torque cells for data acquisition. The successful design made it possible for professors, students, and other interested users to see the different reaction forces and torques with an applied force or torque. Though the mechanical design of the cantilever beam was sound, the data acquisition had room for improvement. The Visual Beams II project kept the same principles of having an interactive visualization tool to aid students and others interested in beam bending, but took the visualization aspect farther.

The data acquisition of the Visual Beams I project plotted many different numbers on a single graph. These numbers represented the reaction forces of whatever weight or force the user put on the beam. Again, these are only numbers and not a visual tool. The beam used, however, was made of a flexible enough plastic so the user could actually see and get a feel for deflection, but no connection between the actual bending and theoretical bending was made.

The purpose of Visual Beams II was to have the user actually see the connection between what happens in reality and what happens on paper when a force is on a beam. As the user applies weight and sees some deflection in the beam, a computerized version of the beam would show that deflection magnified to any extent. In addition, the moment and shear stress diagrams for that load could be seen by clicking to a different window.

It is hard for some students to believe that equations have been made that can model a beam with a force on it. Similarly, some students have trouble visualizing what kind of deflection or stresses a force can create. With the aid the Visual Beams, these problems and many more will be solved.

In order to accomplish the goal of Visual Beams II, many tasks needed to be completed.

MECHANICAL DESIGN

Knowing the beam is simply supported, the actual mechanical design had to be as ideal as possible. In a simply supported beam one support must be a pivot and the other support must be a roller. The beam needed to be able to pivot and slide without friction. Below is a schematic of the engineering model.

A specialty hinge was designed to accomplish the task of a pivot and a support for the load cell. The hinge is a simple pin joint, but is designed to support both the slider, and the load cell. It was also designed to resist any forces applied normal to the side view.

Also, the beam needs to slide with no friction; therefore, a linear slider made of self-lubricating frelon-j was used to give this effect. The advantage of using this slider was its very low friction and it could be used to either lock the bar in place (pivot support) or allow the bar to slide freely (roller support). A disadvantage of using this slider is it can only adjust to adapt to a small range of bar widths. Also the length of the slider is 3.125 inches. This is a problem when the beam is bending. The actual deflection is not going to be exactly the same as the calculations, because the beam cannot bend inside of the rigid box.

Figure 1: Frictionless Slider

When the DAQ system reads the load cell, the reaction forces must show only the vertical components. Many people assume that the forces would always be vertical, but if the beam is under heavy deflection, the reaction forces will also have a horizontal component. Therefore, steps were taken to ensure the load cells would read only the vertical load. To accomplish this, four clearance holes were drilled into the support block and the bottom hinge. Then four steel dowel pins were inserted into the holes. This allowed the hinge and slider assembly to slide vertically on steel dowel pins with minimal friction. To reduce the chance of the pins and the bottom hinge from binding up, the holes were drilled very deep and also reamed to size. This allows a longer pin to fit deeply inside causing a smaller angle to form between the bottom hinge and pins. The smaller the angle, the less force is on the top of the pin, therefore the less chance the pin will bind up.

Figure 2: Short pin in hinge

Figure 3: Long pin in hinge

Also, this design must be user friendly and must be able to handle a wide variety of applications. A track system with locking thumbscrews was installed to allow the user to move either support to the desired position with ease. Below is a solid works model of the final design.

Figure 4: Final Design

To load the beam, a design was created that allowed the user to add standard gym weights to a point load without interfering with the supports system. This design also allows the user to remove the load or move it to any place on the bar with ease. Below is a solid works model of the weight hanger.

Figure 5: Hanger

The support blocks were chose to be relatively small to prevent additional torque to the base and also to prevent high school students from breaking or knocking the visual beam over during usage. After building and testing the project, it was then realized that if the blocks had been taller, heavy weights could be more easily applied. As it stands, only standard 10 lb gym weights can be placed on the bar. It might have been more convenient to use heavier weights for loading, but the design still allow for maximum loading conditions.

Figure 6: Complete Design

The load cells were chose for both positive and negative loading. This was necessary because of the semi-cantilever case. The range of the load cells was chosen because the beam material could not with stand too heavy of a load and also they were cost effectiveness. These load cells were also very simple to adapt to the assembly.

Beam material selection

Various materials and shapes had to be considered for the beam. A solid square bar, a hollow square tube, and an I-beam (cross sections are shown below, Figure 7) were selected to illustrate how the arrangement of mass in the beam can affect its strength and therefore performance. In addition, these shapes are some of the most commonly used in structures and can be easily be associated with. A solid beam with certain overall dimensions is very strong but uses much material. The same amount of material can be better arranged in an I-beam or hollow tube shape in order to reduce the stresses and therefore increase the overall strength. The main constraint of the shape of beams was that the overall dimension had to be no larger than 1.5 inches square in order to accommodate the slider that would mount them.

Figure 7: Beams

Using the various shapes, the stresses for different materials were calculated. Bending and shear stresses were calculated as a measure for beam strength. These stresses were based on the worst case loading scenarios. The loading cases were created from the arrangement of the supports and the point load. The worst case-loading scenario included the arrangement of the two supports about an inch apart with the rest of the beam overhanging one support. The support toward the end of the beam acted as a pin, while the other support near the overhang performed as a roller. The load was applied at the end of the overhang which produced the maximum moment and the maximum shear force for a beam loaded with a given point load. These maximums were experienced in the overhanging section of the beam and therefore caused the maximum bending and shear stresses. Calculations included the use of various materials, such as aluminum, PVC, and acetal-copoly. Plastic materials were preferred due to their deflection characteristics and lightweight. It was decided to use acetal-C and PVC material early in the project and as a result calculations focused on the use of these materials. Microsoft Excel spreadsheets were used to make calculations concerning the use of acetal-c in different loading conditions and also used for some general strength properties. A comparison between charts show how stresses are kept below the maximum values for a given material. It is important to note that research shows that PVC and acetal-c have approximately the same strength.

Other considerations included the availability of shaped or raw material stock. Certain materials only come in certain shapes and machining them to other shapes was often difficult and labor intensive. For this reason, a square hollow tube and two square solid bars of acetal-c were ordered. One of the acetal bars would be used in its existing shape, while the other bar would be machined into an I-beam. The choice of these materials and shapes completed the beam material selection portion of the project.

bEAM Deflection mODELING

Although many geometries can be found for the simply supported beam with a single point load, each can be derived from two more general geometries as seen below.

Geometry 1

Geometry 2

Where ‘Ra’ and ‘Rb’ are the reaction forces of the two supports; ‘P’ is the point load; and ‘a’ and ‘d’ and ‘b’ and ‘c’ are the distances from the ends to the outside forces and the distance between the outside forces and the inside force respectively.

Deflection can be solved for both geometries with the following steps.

1) Newton’s 2nd Law: entire bar

The point load can be written in terms of the reaction forces by summing the forces and summing the moments at a certain point.

SF = -Ra + Rb - P = 0

P = Rb - Ra

Geometry 1

SMA = -P (b) + Rb (b+c) = 0

P = Rb (b+c)/(b)

Geometry 2

SMA = Rb (b) - P (b+c) = 0

P = Rb (b)/(b+c)

Since the reaction forces are known from the load cells, the magnitude and placement of the point load can be found with these equations.

2) Newton’s 2nd Law: sections of bar

By taking four “slices” of the bar and taking the sum of the forces and the sum of the moments at each slice, the shear forces and the moments can be found. Below are the slices for Geometry 1. Geometry 2 is solved in a similar fashion.

Section 1

SF = -V1 = 0

V1 = 0

SM = M1 = 0

M1 = 0

Section 2

SF = Ra - V2 = 0

V2 = Ra

SM = -Ra (x-a) + M2 = 0

M2 = Ra (x-a)

Section 3

SF = Ra - P - V3 = 0

V3 = Ra - P

SM = -Ra (x-a) + P (x-a-b) + M3 = 0

M3 = Ra (x-a) - P (x-a-b)

Section 4

SF = Ra - P + Rb - V4 = 0

V4 = Ra - P + Rb

SM = -Ra (x-a) + P (x-a-b) - Rb (x-a-b-c) + M4 = 0

M4 = Ra (x-a) - P (x-a-b) - Rb (x-a-b-c)

Where ‘x’ is the distance from the left end, V is the shear force, and M is the moment. Note that each equation only works for the section it is in.

3) Solve for deflection

Since M = E I d2y/dx2 where ‘E’ is the modulus of elasticity and ‘I’ is the moment of inertia, the deflection ‘y’ of each section can be attained by integrating.

M1 = 0 = E I d2y1/dx2

E I dy1/dx = C1

E I y1 = C1 x + C2

M2 = Ra (x - a) = E I d2y2/dx2

E I dy2/dx = Ra (x2/2 - ax) + C3

E I y2 = Ra (x3/6 - ax2/2) + C3 x + C4

M3 = Ra (x - a) - P (x - a - b) = E I d2y3/dx2

E I dy3/dx = Ra (x2/2 - ax) - P (x2/2 - ax - bx) + C5

E I y3 = Ra (x3/6 - ax2/2) - P (x3/6 - ax2/2 - bx2/2) + C5 x + C6

M4 = E I d2y4/dx2 = Ra (x - a) - P (x - a - b) - Rb (x - a - b - c)

E I dy4/dx = Ra (x2/2 - ax) - P (x2/2 - ax - bx)

- Rb (x2/2 - ax - bx - cx) + C7

E I y4 = Ra (x3/6 - ax2/2) - P (x3/6 - ax2/2 - bx2/2)

- Rb (x3/6 - ax2/2 - bx2/2 - cx2/2) + C7 x + C8

Plug boundary conditions into these equations to solve for the constants, ‘Cn’.

i. @ x = a : y1 = 0

ii. @ x = a : dy1/dx = dy2/dx

iii. @ x = a : y2 = 0

iv. @ x = a + b : y2 = y3

v. @ x = a + b : dy2/dx = dy3/dx

vi. @ x = a + b + c : y3 = 0

vii. @ x = a + b + c : y4 = 0

viii. @ x = a + b + c : dy3/dx = dy4/dx

Once the constants are known, the deflection equations are as follows for Geometry 1.

y1 = (a-x)(b3Ra+3b2cRa+3bc2Ra-c3Rb)/(6(b+c)E I)

y2 = -(a-x)(-b3Ra-3b2 cRa+a2(b+c)Ra+c3Rb-2a(b+c)Rax

+cRax2+bRa(-3c2+x2))/(6(b+c)E I)

y3 =(a+b+c-x)(a2(b+c)Rb+b3(Ra+Rb)-b2(3Ra+2Rb)x

+cRbx(c+x)+bx(-c(3Ra+Rb)+Rbx)

+a(b+c)(b(3Ra+2Rb)-Rb(c+2x)))/(6(b+c)E I)

y4 = -(a+b+c-x)(2b3Ra+6b2cRa+3bc2(Ra-Rb)-2c3Rb)

/(6(b+c)E I)

These equations are used to govern the data acquisition.

DATA ACQUISITION