MAE 241 –Statics Fall 2006 Jacky C. PruczLecture 1
MEC 302 * ADU
Course Learning outcomes
Upon successful completion of this course, a student will be able to:
Demonstrate an understanding of the fundamental principles of mechanics of deformable bodies. (a3)
Conduct experiments properly and safely, as well as analyze and interpret results. (b2)
Explain how bars and beams deflect under externally applied forces and calculate deflections. (a3)
Explain how structures internally resist externally applied loads and calculate internal forces or stresses. (a3)
Determine maximum stresses and its directions to determine possible failure planes. (c3)
Derive the relations between deflections and internal forces. (e1)
Determine required sizes of bars and beams (design) given the properties of the materials and the loading conditions. (c1)
Determine critical buckling load of columns. (e1)
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MEC 302 * ADU
Catalog Description: Stress and strain; Material behavior; Hooke's law; Axial loading; Safety factors; Shear force and bending moment diagrams; Bending stresses and deflections; Shear stresses in beams; Torsion of circular members; Combined stresses; Mohr's circle; Buckling of columns; Engineering applications.
Prerequisites: CIV 201 Statics.
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MEC 302 * ADU
Absences: On those days when you will be absent, find a friend or an acquaintance to take notes or visit the blackboard. Absences checked in the class is final. No any excuse will change it.
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Results
Quiz : You will solve one or two questions every 2 weeks in the class as a 10-15 min quiz.
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MEC 302 * ADU
Final Project: There is a final project, it will consist of a design of a product by applying knowledge gained from the course. It will be due on Sunday Nov. 22 2014 followed by a presentation on Thursday.
The total course grade is comprised of homework assignments, quizzes, partial exams, final exam, and a project as follows:
Homework 5%
Course Grading
Cheating: You are allowed to cooperate on homework by sharing ideas and methods. Copying will not be tolerated. Submitted work copied from others will be reported to the OAI and will get zero marks. Zero tolerance for violating AI. NOTE: all papers will be run through TURNITIN.
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Most Course Materials (Course Notes, Handouts, Homework, Final Project, and Communications) are on the Blackboard.
Updated Power Point Lectures will posted every week or two weeks.
Office Hours:
Mon., Wed. @ 10-11 AM & 2 to 4 PM
By appointment (No walk-in please)
No office hours on Thursdays
At my office D-2F236
1.1-1.7
2.1-2.10
2.11-2.19
5.1-5.4
6.1-6.7,
7.1-7.3
7.4-7.9
James M. Gere, (2004), Mechanics of Materials, seventh edition, Brooks/Cole, Thomson Learning.
Concept of Stress
The main objective of the study of the mechanics of materials is to provide the future engineer with the means of analyzing and designing various machines and load bearing structures.
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The structure is designed to support a 30 kN load
Perform a static analysis to determine the internal force in each structural member and the reaction forces at the supports
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Structure Free-Body Diagram
Structure is detached from supports and the loads and reaction forces are indicated
Ay and Cy can not be determined from these equations
Conditions for static equilibrium:
Component Free-Body Diagram
In addition to the complete structure, each component must satisfy the conditions for static equilibrium
Consider a free-body diagram for the boom:
substitute into the structure equilibrium equation
Results:
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Method of Joints
The boom and rod are 2-force members, i.e., the members are subjected to only two forces which are applied at member ends
For equilibrium, the forces must be parallel to to an axis between the force application points, equal in magnitude, and in opposite directions
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From a statics analysis
FAB = 40 kN (compression)
FBC = 50 kN (tension)
dBC = 20 mm
From the material properties for steel, the allowable stress is
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Design of new structures requires selection of appropriate materials and component dimensions to meet performance requirements
An aluminum rod 26 mm or more in diameter is adequate
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Axial Loading: Normal Stress
The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the member axis.
The detailed distribution of stress is statically indeterminate, i.e., can not be found from statics alone.
The normal stress at a particular point may not be equal to the average stress but the resultant of the stress distribution must satisfy
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Centric & Eccentric Loading
The stress distributions in eccentrically loaded members cannot be uniform or symmetric.
A uniform distribution of stress in a section infers that the line of action for the resultant of the internal forces passes through the centroid of the section.
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Shearing Stress
Forces P and P’ are applied transversely to the member AB.
The resultant of the internal shear force distribution is defined as the shear of the section and is equal to the load P.
Corresponding internal forces act in the plane of section C and are called shearing forces.
Shear stress distribution varies from zero at the member surfaces to maximum values that may be much larger than the average value.
The shear stress distribution cannot be assumed to be uniform.
The corresponding average shear stress is,
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Bearing Stress in Connections
Bolts, rivets, and pins create stresses on the points of contact or bearing surfaces of the members they connect.
The resultant of the force distribution on the surface is equal and opposite to the force exerted on the pin.
Corresponding average force intensity is called the bearing stress,
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Would like to determine the stresses in the members and connections of the structure shown.
Stress Analysis & Design Example
Must consider maximum normal stresses in AB and BC, and the shearing stress and bearing stress at each pinned connection
From a statics analysis:
FAB = 40 kN (compression)
FBC = 50 kN (tension)
MechMovies website http://web.mst.edu/~mecmovie/index.html
Rod & Boom Normal Stresses
The rod is in tension with an axial force of 50 kN.
The boom is in compression with an axial force of 40 kN and average normal stress of –26.7 MPa.
The minimum area sections at the boom ends are unstressed since the boom is in compression.
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The cross-sectional area for pins at A, B, and C,
The force on the pin at C is equal to the force exerted by the rod BC,
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Pin Shearing Stresses
Divide the pin at B into sections to determine the section with the largest shear force,
Evaluate the corresponding average shearing stress,
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Pin Bearing Stresses
To determine the bearing stress at A in the boom AB, we have t = 30 mm and d = 25 mm,
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Methods of Problem Solution
You should approach a problem in mechanics of materials as you would approach an actual engineering situation. By drawing on your own experience and intuition, you will find it easier to understand and formulate the problem.
Once the problem has been clearly stated. Your solution must be based on the fundamental principles of statics and on the principles you will learn in this course.
Every step you take must be justified on these basis, leaving no room for your “intuition.”
After an answer has been obtained, it should be checked. Here again, you may call upon your common sense and personal experience.
If not completely satisfied with the result obtained, you should carefully check your formulation of the problem, the validity of the methods used in its solution, and the accuracy of your computations.
Lecture 1
Methods of Problem Solution
The statement of the problem should be clear and precise. It should contain the given data and indicate what information is required.
A simplified drawing showing all essential quantities involved should be included.
The solution of most of the problems you will encounter and will necessitate that you first determine the reactions at supports and internal forces and couples.
And the drawing of one or several free-body diagrams, as was done in Sec. 1.2, from which you will write equilibrium equations.
These equations can be solved for the unknown forces, from which the required stresses and deformations will be computed.
After the answer has been obtained, it should be carefully checked.
Lecture 1
Methods of Problem Solution
Mistakes in reasoning can often be detected by carrying the units through your computations and checking the units obtained for the answer.
For example, in the design of the rod discussed in Sec. 1.4, we found, after carrying the units through our computations, that the required diameter of the rod was expressed in millimeters, which is the correct unit for a dimension.
if another unit had been found, we would have known that some mistake had been made.
Errors in computation will usually be found by substituting the numerical values obtained into an equation which has not yet been used and verifying that the equation is satisfied.
The importance of correct computations in engineering cannot be overemphasized.
Lecture 1
Stress in Two Force Members
Will show that either axial or transverse forces may produce both normal and shear stresses with respect to a plane other than one cut perpendicular to the member axis.
Axial forces on a two force member result in only normal stresses on a plane cut perpendicular to the member axis.
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Pass a section through the member forming an angle q with the normal plane.
Stress on an Oblique Plane
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Maximum Stresses
The maximum normal stress occurs when the reference plane is perpendicular to the member axis,
The maximum shear stress occurs for a plane at + 45o with respect to the axis,
Normal and shearing stresses on an oblique plane
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Stress Under General Loadings
A member subjected to a general combination of loads is cut into two segments by a plane passing through Q
For equilibrium, an equal and opposite internal force and stress distribution must be exerted on the other segment of the member.
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Stress components are defined for the planes cut parallel to the x, y and z axes. For equilibrium, equal and opposite stresses are exerted on the hidden planes.
It follows that only 6 components of stress are required to define the complete state of stress
State of Stress
The combination of forces generated by the stresses must satisfy the conditions for equilibrium:
Consider the moments about the z axis:
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DESIGN CONSIDERATIONS
In the preceding sections you learned to determine the stresses in rods, bolts, and pins under simple loading conditions.
In later chapters you will learn to determine stresses in more complex situations.
In engineering applications, however, the determination of stresses is seldom an end in itself.
Rather, the knowledge of stresses is used by engineers to assist in their most important task, namely, the design of structures and machines that will safely and economically perform a specified function.
Lecture 1
a. Determination of the Ultimate Strength of a Material
An important element to be considered by a designer is how the material that has been selected will behave under a load.
For a given material, this is determined by performing specific tests on prepared samples of the material.
For example, a test specimen of steel may be prepared and placed in a laboratory testing machine to be subjected to a known centric axial tensile force, as described in Sec. 2.3.
As the magnitude of the force is increased, various changes in the specimen are measured, for example, changes in its length and its diameter.
Lecture 1
a. Determination of the Ultimate Strength of a Material
Several test procedures are available to determine the ultimate shearing stress, or ultimate strength in shear, of a material.
The one most commonly used involves the twisting of a circular tube (Sec.3.5).
A more direct, if less accurate, procedure consists in clamping a rectangular or round bar in a shear tool (Fig. 1.39) and applying an increasing load P until the ultimate load PU for single shear is obtained.
Lecture 1
a. Determination of the Ultimate Strength of a Material
If the free end of the specimen rests on both of the hardened dies (Fig. 1.40), the ultimate load for double shear is obtained.
Lecture 1
uncertainty in material properties .
importance of member to integrity of whole structure.
risk to life and property.
influence on machine function.