AC 2011-1028: INTRODUCTION OF A GLOBAL PERSPECTIVE USINGA TEAM PROJECT IN A STRENGTH OF MATERIALS COURSEDean Q. Lewis, Penn State Erie, The Behrend College Dean Lewis has been a lecturer in mechanical engineering at Penn State Erie, The Behrend College for five years teaching courses in design, mechanics, and mechanical engineering. His research interests include attachment design for plastic parts and engineering education.Mary Lynn Brannon, Pennsylvania State University, University Park Mary Lynn Brannon, Instructional Support Specialist at the Leonhard Center for the Enhancement of Engineering Education at the Pennsylvania State University, has a Master of Arts Degree in Education and Human Development
elementsimproves the quality of approximation to the real behavior. However in this simple formthe model is sufficient to demonstrate a couple of typical tasks relevant to the operationwith Multibody problems. The following sections explain the derivation of the equations ofmotion. Because of its compact theoretical formulation section 2.1 starts with the Lagrangeequation of second kind, section 2.2 demonstrates the more relevant procedure with theNewton-Euler equations.2.1 Lagrange equation of second kindFor the derivation of equations of motion this section uses the Lagrange equation of secondkind with the vector of minimal coordinates q = (ϕ1 , ϕ2 , ϕ3 )T , the kinetic energy T , thepotential energy V and the vector of generalized forces u
presenting the total externalwork and total strain energy equations beginning first with a single load P applied to a planartruss with one load sequence. Then loads P and Q are applied using two load sequences in whichthe load Q is applied at the location and in the direction of the desired displacement. From thisbasis of understanding, an additional load S is included in both load sequences to discuss itsinfluence on the displacement expression. This leads to a general understanding of the influencethat any number of additional loads would have on the displacement expression, and that theeffect of the load Q remains unchanged as these loads are applied. It then becomes evident thatBarry T. Rossonthe desired displacement due to all the applied loads
the unit vector nˆ3 .Bodies A and B are slender uniform rods with mass mA and mB , and length LA and LBrespectively. A torsional spring with the spring constant K A acts between body A and the ground.q and q are generalized coordinates. The basis vectors aˆ , bˆ , and nˆ (i = 1,2,3) are fixed on 1 2 i i ibody A, body B and the ground respectively. A force FQ is applied to point Q in the direction bˆ1 . Figure 1: A Simple Case Study – the Rigid Body Double PendulumThough the double pendulum case is simple, it contains basic features that are necessary to Page
] matrixis typically denoted as the [Q] stiffness matrix for loading that coincides with the fiber direction ] _(i.e., 0° angle) and is denoted as Q for loading at a non-zero angle with respect to the fiberdirection.The first example modeled by the students in this course is for an isotropic material in loaded inplane stress at a 0° angle where the material properties are given as E11 := 207·GPa, E22 :=207·GPa, ν12 := 0.33 and G12 :? E11 ? 77.82 GPa . Using these material property 2 * ∗1 − π 12 +values, the students determine the values of the [Q] matrix elements: 232.297 76.658 0
.Bar Structure Stochastic Static Analysis8Before doing any computer solutions, let us discuss a simple one element baranalysis without making any reference to any computer programs or results.Figure 4 shows a cantilever bar of length L, cross sectional area A, the materialmodulus of elasticity E, and the bar is subject random axial load Q. In addition,let us assume that one of the parameters of A, L E and Q is random at a time. Page 15.922.8 7 EA Q L qFigure 4: Cantilever Bar
ofthree students labeled P, Q, and R whom I believe to be representative, in a macro sense, of thedifferent ways in which students played the Novice level of the Spumone Drop challenge. Theletters P, Q, and R were chosen for convenience. Any relation that may exist between theseletters and the students’ actual names is coincidental.Student P. The way that that Student P played the novice and practice levels of the Dropchallenge is depicted in Figure 8. The horizontal lines represent timelines on two separate days.Vertical lines above the horizontal axis represent instances in which the student played the game.The practice level is essentially the same as the novice level, except there are no flowers to killthe spuCraft. The purpose of the
derivative of the strain energy with respect to a load applied at that point. If an externalload is not present at the point of interest, then a dummy load can be applied there for thepurpose of deflection determination. After the partial derivative is calculated with respect to thedummy load, that dummy is set to zero in the moment equation. In equation form, we write: Page 24.946.4 U M Q 0 M L Q dx Eq.2 Q 0 EI QThe variable Q is used to
b0 ? 1 for b 0 (9) b0 ? 0 for b>0 (10)Referring to the beam ab in Fig. 1, we may, for illustrative purposes, employ the rudiments ofsingularity functions and observe the defined sign conventions for beams to write the loadingfunction q, the shear force V, and the bending moment M for of this beam as follows:6-8 q ? Va > x @ /1 - M a > x @ /2 / P > x / xP @ /1 - K > x / xK @ /2 w1 / w0 > x / xw @ 0 / > x / xw @1 (11) L / xw
particularly satisfactory response toconcept question 1. The amount of time each student spent on the first question is tabulated inTable 1. Students’ approaches are outlined below. Table 1. Amount of time that students spent on concept question 1, all three parts. Student P Q R S T U V Time (min:sec) 1:20 5:20 8:30 1:33 2:20 9:00 8:50Student S and Student T had similar approaches to the concept question. Neither of them drew afree body diagram (FBD), even though they always drew FBDs on problem-solving questionsthey encountered on their midterm and final exams. For parts A and B, they simply observed thattension from the string creates
engineering majors would take MEEN 221 as their main staticsengineering course. Starting in the Fall 2015 term, the Mechanical Engineering Departmentdeveloped a new Statics course exclusively for MEEN students, MEEN 225. This new course wasdesigned to better prepare students for subsequent MEEN curriculum. Topics covered are verysimilar, however MEEN 225 uses group projects in addition to homework and major exams forassessment. The students must also attend a 3-hour recitation every week in MEEN 225.Since the Fall 2015 semester when the divergence of the MEEN 221 and MEEN 225 coursesbegan, the department has seen a much higher percentages of students earn a grade of a D, an F,or Q-drop in MEEN 221 creating higher DFQ rates. A Q-drop prevents a
S, D Q, S, D S, Q, D5.2 Find items under bending5.3 Bending members with commoncross-sections5.4 Feel craft sticks bending5.5 Stress Opticon: Bending stressdistribution S, Q, D5.6 Quantify flexure in a craft stick5.7 Stress Opticon: simple support5.8 Photoelastic beam bending S, D S, Q, DStress Transformation7.1 Directional Strength (Craft Stick) S7.2 Directional Orientation inStructures7.3 Photoelasticity: Beam with holes S, Q, D7.4 Matching loads and failure planes7.5 Brittle and Ductile Failure
C Rp P P P (a) x (b) 3R p (c) R e2 (t ) q p e1 O
(θ2) f2 = xB*xB + yB*yB s = a*sin(θ2) for θ2 = 0 to 360 f = sqrt(f2) f2 = r*r + s*s Q = cos(θ2) γ = atan2(yB, -xB) δ = acos((A - f2)/B) A = K3 - K1 - (K2 - 1)*Q β = acos((f2+C)/(2*f*c)) g = b – c*cos(δ) B = -2*sin(θ2); θ4 = π – (γ + β); h = c * sin(δ) C = K3 + K1 - (K2 + 1)*Q xC = c*cos(θ4); θ3 = atan2(h*r - g*s,g*r + h*s) D = K5 - K1 + (K4 + 1)*Q yC = c*sin(θ4); θ4 = θ3 + δ
learning, all examples will be first solved by the traditional method ofintegration (MoI) with the use of singularity functions then solved again by the method ofmodel formulas (MoMF). As usual, the loading function, shear force, bending moment, slope,and deflection of the beam are denoted by the symbols q, V, M, y , and y, respectively.Example 1. A cantilever beam AB with constant flexural rigidity EI and length L is acted on bya concentrated force of magnitude P at C, and two concentrated moments of magnitudes PL and2PL at A and D, respectively, as shown in Fig. 2. Determine the slope A and deflection yA atend A. Fig. 2. Cantilever beam carrying a force and two momentsSolution by MoI. Using the symbols defined earlier
and the object subjected to the general rigid bodymotion and the deformation is presented in fig.2. Consider a point P in the non-deformedoriginal body translated to point p through rigid-body translation ⃗ . Due to the additional rigidbody rotation and deformation, the target point Q displaced to point q. Subsequently, the smallelement vector changed to .The position vectors of point Q and q are given by ⃗ (5)and ⃗ ⃗ ( ⃗) . (6)By comparing the position vector ⃗ and , the small element
included in this studycompleted both of the exams. Page 26.849.3 Table 2: Ratings of student performance for the study semesters. Statics Mean (SD) Dynamics Mean (SD) ID Q/HW Exam R Survey Q/HW Exam R Survey H1 67(22) 79(12) 0.200 88(5) 64(28) 74(13) 0.181 85(5) H2 81(12) 79(14) 0.003 88(6) 75(25) 75(12) 0.254 88(5) Q1 75(15) 75(17) *0.708 88(4) 73(18) 63(21) *0.478 88(5) Q2 76(11) 81(11) *0.598 90(4
Optical Flow8,9 and Block Matching algorithms10,11 are relatively easy toimplement. However, both algorithms are derived based on the assumption that target objectdoes not rotate and deform through motion. Clearly, this assumption is not applicable toproblems encountered in the mechanics of material. With the assist of finite strain theory andnonlinear optimization theory, the concept of DIC can be described as follows.The deformation of an elastic body takes place between non-deformed and deformedconfigurations after motion and deformation is schematically presented in Fig. 1. Consider apoint P in the non-deformed body translates to point p through translation 𝑈⃗ . Due to theadditional rigid body rotation and deformation, the target point Q in
automobile as: 1 1 T = m x& 2 + J θ& 2 . (1) 2 2The potential energy is described in Equation 2 as: 1 1 k 1 ( y + x − l1θ ) + k 2 ( x + l 2θ ) . 2 2 U = (2) 2 2Rayleigh’s dissipation function describing viscous dissipation in the dampers is: 1 1 Q= 2 ( c1 y& + x& − l1θ& ) 2
i( x ) = i4 i ( x ) i = 1,2,... (11) dx 4If (9) and (10) are substituted into (1) using relation (11) the result is a set of uncoupled ordinarydifferential equations in the normal coordinates or EI i4 qi + cq + qi = A0 f i g( t ) 1 = 1,2,... (12)Division through by and noting the definitions of (4), (6) and (7) the result is a set of ordinarydifferential equations in the modal coordinates or A0 f i qi + 2 i i q i + i2 qi = g( t ) i = 1,2,... (13)For the forcing functions considered, this equation can be easily solved
-Dimensional Equilibrium of a Particle Page 11.788.6 Figure 5: 3-Dimensional Equilibrium of a ParticleBasic vector operations moduleThis panel provides a comprehensive tool for the basic vector operations such as cross product,scalar product, unit vectors, and dot product. Figure 6 illustrates the structure of this module. Page 11.788.7 Figure 6: Vector product moduleThe cross product, also known as vector product, is an operation on coplanar vectors P and Q isdefined as 10: f f f
. Note that Page 11.878.4Eq. (4) is extremely useful and important in solving problems by the virtual work method!III. Relevant Fundamental ConceptsIn teaching and learning the virtual work method, it is well to recall the following relevant fun-damental concepts:̇ Work of a forceIf a force F acting on a body is constant and the displacement vector of the body from positionA1 to position A2 during the action is q, then the work U1› 2 of the force F on the body is2-6, 8,9 U1› 2 ? F © q ? FqE (5)where F is the magnitude of F and qE is the scalar component of q parallel to
. Gere , Barry J. 7(12) Goodno 7. Mechanics of Materials (2000) Anthony Bedford, 5(12) Kenneth M. Liechti 8. Introduction to Mechanics of Materials (1989) William F. Riley, 7(11) Loren W. Zachary 9. Mechanics of Solids (1995) Gerald Wempner 2(9) 10. Mechanics of Materials (Fourth Revised Edition) James M. Gere, 6(10) Stephen P. Timoshenko 11. Mechanics of Materials (Sixth Edition) William F. Riley, Leroy D. 2(10) Sturges, Don H. Morris 12. Mechanics of Materials (Second Revised Edition) Roy R. Craig, Jr. 8(12) 13. Mechanics of Materials (1985) David Q. Fletcher 2(14) 14. Mechanics of
itemsbut are dependent on each other. If students find that they include a dimension or force in theequations of equilibrium that is not in the FBD then they must go back and add it in.Emphasizing the idea that the equations of equilibrium are driven by the free body diagram hashelped students develop a clearer understanding of how and why FBDs are important to theproblem solving process. In practice, instructors write ABCD ⇒E on the board frequently whendrawing free body diagrams as a constant reminder.Let’s take a look at an example from Hibbeler’s text at how this maps to what’s already done.(Pearson has granted permission to use these images.)To avoid confusion with the discussion of the technique, the point locations are renamed to be P,Q
evidencefrom an offering of the course instead of perceptions of faculty members about what they mightwant. This process also provided some insight into the alignment of skills engineering faculty Page 22.153.5felt were necessary to be successful in the course and those that are actually utilized in thecourse. From this analysis, a list of skills in mathematics and physics mechanics was constructed(see Figure 2).Figure 2. Portion of Q-matrix Used to Determine Skills in Homework and Exam Problems Homework Problems 3-1 3-5 3-6 3-47
8. Computing Projects Solutions 9. Posted Recitation 9. Posted Recitation Solutions 10. Exam Self-Assessment 10. Exam Self-Assessment 10. Exam Self-Assessment 11. Q&C 11. Q&C 11. Q&C 12. UGTAs 12. UGTAs 13. RE and MA Solutions 13.12
upward displacement. Page 12.240.3 A positive slope is a counterclockwise angular displacement.III. Derivation of Model FormulasAny beam element of differential width dx at any position x may be perceived to have a left faceand a right face. Using singularity functions,8-10 we may write, for the beam ab in Fig. 1, theloading function q, shear force V, and bending moment M acting on the left face of the beamelement at any position x for this beam as follows: q ? Va > x @/1 - M a > x @/2 / P > x / xP @/1- K > x / xK @/2/ w0 > x / xw @0 w / w0
upward linear displacement [e.g., ya and yb in Fig. 1(b)].■ Methodology and pedagogy of the method of model formulasThe four model formulas in Eqs. (1) through (4) were derived in great detail in the paper thatpropounded the MoMF.12 For convenience of readers, let us take a brief overview of how thesemodel formulas are obtained. Basically, it starts out with the loading function q,9 written in termsof singularity functions for the beam ab in Fig. 1; as follows: q = Va < x >− 1 + M a < x >− 2 − P < x − x P >− 1 + K < x − x K >− 2 − w0 < x − x w >0 w − w0 w − w0 − 1 < x − x w >1 + w1< x − u w > 0 + 1 < x − u w >1
balls after one match and to assess how the coefficient ofrestitution varied with initial drop heights. Page 15.1331.12 Energy Dissipated - .., 30 Q,) cu 25 a. I/) I/) 20 - "C c Q,) (.) ' 15 10 • Height1 • Height2 Q,) 5 a.. 0
area (Q) can be obtained using the following equation: b(h 2 / 4 / y 2 )Q? (25) 2Determination of the Principal Stresses:A square differential element of beam subjected to the normal stresses sx and sy and a shearingstress txy is shown in Figure 4. To develop the expression for the principal normal and shearingstresses acting on the element of the beam, the equation for the normal stress s and shearingstress t on an inclined plane with an angle of inclination of s are obtained first. uy A v xy ux