EGR 315 Final Paper
EGR 315
Final Paper
Joe Hyde
There are 5 student learning outcomes covered in this semester. These leaning outcomes where the concentreation of time was spent in the semester and were deemed important by the powers that be. By grasping the concepets involved in these leaning outcomes they are what should be taken from this course and are to be applied in the workplace in the future. Not all of the topics can be defined in 10 pages. So the topics that follow are ones that I found importance in and were used very frequently to solve the problems in this semester. The topics to come are Shear Stresses for Beams in bending, Castigliano’s Theorem, Distortion Energy Theorem for Ductile Materials, Mechanics of Power screws, and Fatigue loading …show more content…
thus yielding
Equation 3
This stress from equation 3 is known as the transverse shear stress, and is always accompanied with bending stress. Defining the variables in this equation, b is the width of section at y=y1, and I is the second moment of area of the entire section about the neutral axis.
Being that cross shears are equal, and area A’ being finite, the shear stress can be calculated with equation 3.
The shear stress distributing in a beam depends on how Q/b varies as a function of y1. For a beam with a rectangular cross sectional area, subjected to a shear force V and a bending moment M. as a result of the bending moment a normal stress is developed on a cross section, which is compression above the neutral axis and it is tension below the neutral axis. To investigate the shear stress at a distance y1 above the neutral axis. Then dA=bdy, so equation 2 becomes
Equation 4
Subbing this value in for Q into equation 3 gives Equation 5
Equation 5 is known as the general equation for shear stress in a rectangular beam. The second moment of area for a rectangular section from appendix A-18, …show more content…
Topic:
Castigliano’s Theorem
One of the simplest ways to approach deflection analysis with the energy method is to use Castigliano’s theorem. Castigliano’s theorem states “when forces act on elastic systems subject to small displacements, the displacement corresponding to any force, in the direction of the force, is equal to the partial derivative of the total strain energy with respect to that force”.
The terms force and displacement are broadly interpreted to also apply equally to moments and angular displacements. The mathematical formula for this theorem is:
Equation 1
Where, δi is the displacement of the point of application of the force Fi in the same direction of this force. For rotational displacement the theorem can be written as:
Equation 2
Were, θi is the rotational displacement, in radians, of the beam where the moment Mi exists in the direction of that moment.
Using Castiglianos theorem to get axial and torsional deflections yields:
Equation(s) 3,4
Transverse shear can be considered zero if, the l/d ration is less than
Final Paper
Joe Hyde
There are 5 student learning outcomes covered in this semester. These leaning outcomes where the concentreation of time was spent in the semester and were deemed important by the powers that be. By grasping the concepets involved in these leaning outcomes they are what should be taken from this course and are to be applied in the workplace in the future. Not all of the topics can be defined in 10 pages. So the topics that follow are ones that I found importance in and were used very frequently to solve the problems in this semester. The topics to come are Shear Stresses for Beams in bending, Castigliano’s Theorem, Distortion Energy Theorem for Ductile Materials, Mechanics of Power screws, and Fatigue loading …show more content…
thus yielding
Equation 3
This stress from equation 3 is known as the transverse shear stress, and is always accompanied with bending stress. Defining the variables in this equation, b is the width of section at y=y1, and I is the second moment of area of the entire section about the neutral axis.
Being that cross shears are equal, and area A’ being finite, the shear stress can be calculated with equation 3.
The shear stress distributing in a beam depends on how Q/b varies as a function of y1. For a beam with a rectangular cross sectional area, subjected to a shear force V and a bending moment M. as a result of the bending moment a normal stress is developed on a cross section, which is compression above the neutral axis and it is tension below the neutral axis. To investigate the shear stress at a distance y1 above the neutral axis. Then dA=bdy, so equation 2 becomes
Equation 4
Subbing this value in for Q into equation 3 gives Equation 5
Equation 5 is known as the general equation for shear stress in a rectangular beam. The second moment of area for a rectangular section from appendix A-18, …show more content…
Topic:
Castigliano’s Theorem
One of the simplest ways to approach deflection analysis with the energy method is to use Castigliano’s theorem. Castigliano’s theorem states “when forces act on elastic systems subject to small displacements, the displacement corresponding to any force, in the direction of the force, is equal to the partial derivative of the total strain energy with respect to that force”.
The terms force and displacement are broadly interpreted to also apply equally to moments and angular displacements. The mathematical formula for this theorem is:
Equation 1
Where, δi is the displacement of the point of application of the force Fi in the same direction of this force. For rotational displacement the theorem can be written as:
Equation 2
Were, θi is the rotational displacement, in radians, of the beam where the moment Mi exists in the direction of that moment.
Using Castiglianos theorem to get axial and torsional deflections yields:
Equation(s) 3,4
Transverse shear can be considered zero if, the l/d ration is less than