solid mechanics crib sheet

7/24/2019 Solid Mechanics Crib Sheet

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Adam D. Foltz

SOLID MECHANICS CRIB SHEET* Memorize


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General State of Stress/Strain

15 Unknowns:

o 6 Stresses

o 6 Strains

o 3 Displacements

15 Governing Equations:

o

3 Equilibrium Equations

o 6 Strain-Displacement Equations

o 6 Strain-Stress Equations

o If all equations are satisfied, the equations of compatibility are satisfied.

3D Strain:

o = o = = o =o

=

o = o 2 = = +o 2 = = + o 2 = = +

Stress-Strain Relationships:

o =1 + + To =1 ( + )+ To

=1

+

+

T

o

2 = = o 2 = = o 2 = = o = (1+)(12) (1 ) + + = 2 + *o

= (1+)(12) (1 ) + ( + ) = 2 + o = (1+)(12) (1 ) + + = 2 +


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Equations of Compatibility:

o 2D:

y +

x =xy*

o 3D:

y +

x =xy

z +

y =

yz

x +z =zx 2

yz =x x + y +z

2zz =

y x y +z

2xy =

z x +y z

Equilibrium Equations:*

o

x +y +

z + Fx = 0o

y +x +

z + Fy = 0o

z +x +y + Fz = 0 Dilation:*

o =x + y + z =12x + y + zo = =

Strain Energy Density:*

o =12 + + + + + Material Property Relationships:

o E (Youngs Modulus/ Elastic Modulus/ Modulus of Elasticity

o (Poissons Ratio)

=

[-1 < v < 0.5 (incompressible)]*

o = 3(12) (Bulk Modulus/ Modulus of Compression)*o

= 2(1+) (Shear Modulus/ Modulus of Rigidity)*o =

(1+)(12) (Lames First Parameter)*


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Plane Stress

Situation:

o Thin plates

Strain:

o

=o =o

= o 2 = = +o 2 = = + o 2 = = +


o =1 o =1 o =1 + = + *o 2

=

=

o = = 0o = = 0o

= (1+)(12) (1 ) + + = 2 + o = (1+)(12) (1 ) + ( + ) = 2 + o = 0


oy +

x =xy (In terms of strain)

o + + =(1 + ) + (In terms of stress)

Equilibrium Equations:o x +y + Fx = 0o

y +x + Fy = 0

o

Fz = 0Plane Strain

Situation:

o Thick plates (long prismatic bodies)

Strain:

o

=

o =o = 0o 2 = = +o

= = 0o = = 0


o

=1 1 o

=11


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o = 0o 2 = = o = = = 0o = = = 0o = (1+)(12) (1 ) + = 2 + (x + y)o = (1+)(12) (1 ) + = 2 + (x + y)o

=

x+

y=

(

x+

y)*


oy +

x =xy (In terms of strain)

o

+ + =11 + (In terms of stress) Equilibrium Equations:

ox +

y + Fx = 0o

y +x + Fy = 0

o Fz = 0Stress Transformations

Uniaxial Stress:

o =2()*o = sin() cos()*

Biaxial Stress:

o =+2 +2 cos(2) + sin(2)o =+2 2 cos(2) sin(2)o = 2 sin(2) + cos(2)

Triaxial Stress:*

o

=

o = [ ]

o

1 = [ ]

o 2 = [ ]

Traction/Normal/Shear:*

o

=

o 1 = o 1 = o =

Plane Equations:*

o

( ) + ( ) + ( ) = 0o Given three points: , , , , ,,

Calculate and = = , , 0 =


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Strain Transformations

Uniaxial Stress:

o =2()o

= sin() cos() Biaxial Stress:

o

=+2 +2 cos(2) + 2 sin(2)o

=+

2 2

cos(2

)

2

sin(2

)

o = sin(2) + cos(2)

Triaxial Stress:

o =

Principal Stresses

Uniaxial Stress:

o

1 =o 2 =3 = 0o

= 0

Biaxial Stress:

o 1 =+2 +22 + 2o 2 =+2 22 + 2o 3 = 0o =2 2 + 2o tan2 =2o = + 45*

Triaxial Stress:

o

[1,2,3] =([])o [1,2,3] =([])o =12 |1 3|

Stress Invariants:

o 3 12 + 2 3 = 0o

1 =([])o 2 = + + 2 2 2o

3 =([])Principal Strains

Uniaxial Strain:o 1 =o 2 =3 = 0o = 0

Biaxial Strain:

o 1 =+2 +22 + 22o 2 =+2 22 + 22o 3 = 0


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o

= 2 + 2o tan2 = o = + 45

Triaxial Strain:

o [1, 2, 3] =([])o [1,2,3] =([])o

= |

1 3|

Strain Invariants:

o 3 12 +2 3 = 0o1 =([])o2 = + + 2 2 2o3 =([])

Mohrs Circle (Stress)

avg =+2 =22 + 2

Biaxial Stress: Triaxial Stress:

Stress Transformation*o = + (2)o =(2)o

(*If the seam lies in the upper half-plane, rotate the x-axis counter-clockwise until parallel to the seam,

to find ; otherwise rotate the x-axis clockwise until parallel;is always positive.)Mohrs Circle (Strain)

avg =+2 =22 + 22 Biaxial Strain: Triaxial Strain:


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Strain Transformation:

o = + (2)o =(2)

Elementary Formulas for Stress/Strain

Axial Loading:*

o

=

o = = = ()()() (Ex. () =for column deflecting from own weight)o =1(%1) + 2(%2) (Rule of Mixtures for Binary Composites)o Saint-Venants Principle

o

=o

= Torsion:*

o = o = = = ()()() o = (relating torques in shafts connected by gears)o Stress Distribution

o

=o

= Shear and Moment Diagrams:*

o

Types of Beams:


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o Sign Convention:

o Steps:

Get

(

)from a F.B.D. (A

(

)must be found for each section with new loads)

Find the area under each section ofto get , sketch the slope.o *Make sure the shear and moment diagrams match the end boundary conditions.o *Replace a distributed loaded with an equivalent point load when performing calculations.

o *The shear diagram is discontinuous; the moment diagram is continuous unless there is a moment load.

o *For a beam on the ground, assume a uniform reaction of =(solve for w).o *The shear diagram doesnt change going through a moment, neither does the slope of the moment.

o *Add a negative moment for the moment diagram, and subtract a positive moment.

o *For a square distributed load, () =2 (from the left end)o *For a triangular distributed load, () =6 (from the left end)

Bending:*

o = o

1 =

o Stress Distribution (the neutral axis lies on the centroid of the cross-section)

Shear:*

o

=

o =o Diagram

o Stress Distribution:

o Maximum Shear Stresses:

=32 (Rectangular Cross-Section) =43 (Circular Cross-Section) =2

(Concentric Circular Cross-Section)


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Thin-Walled Pressure Vessels:*

o Cylindrical:

=1 = (hoop stress) =2 =2 (longitudinal stress)

=2 (out-of-plane, maximum) =4 (in-plane) Diagram:

o Spherical:

=1 =2 (hoop stress) =2 =2 (longitudinal stress) =4 (out-of-plane, maximum) = 0 (in-plane)

o Strain:

= *. = =

Combined Loading:*

Geometrical Properties

First Moment of Area:*

o Purpose: used to find the centroid

o Q = x Ao

Q = y A


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o y =Q = o x =Q = o Common formulas:

*

*


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Second Moment of Area/ Rectangular Moment of Inertia/ Area Moment of Inertia:

o Purpose: used to find the resistance to bending

o

I = x2Ao

I = y2Ao

= +2*o

Common Formulas:

Polar Moment of Inertia:

o Purpose: used to find the resistance to torsion

o = 2 =(2 + 2) = I + I

*

*


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Radius of Gyration (of an Area):*

o = o

= o =

Advanced Formulas for Stress/Strain

Airys Stress Function:*

o Definition:

Scalar potential function that can be used to find stress for two-dimensional problems.

Satisfies equilibrium in the absence of body forces.

o = o = o =o

2

+

=

4

= 0

Castigliones Theorem:*o General:

=2 + 2 + 2 + 2 ( ) =1 +1 +1 +1 =1 +1 +1 +1 *Add dummy load to point of interest, set load to zero at the end.

*Add reaction force to point of interest, set deflection/slope to zero at the end.

*Add (x-a) for a load shifted a to the right.

*

=6

5 (

),

=10

9 (

)

o

Trusses:

= 2=1 = = =1 N is the axial force in each member, found from the method of joints or the method of sections.

Thin-Walled Cross Sections:*

o Open:

= 13 3 (b is the length of a segment) =

=

o

Closed:

=2 = = 2 = 12

o Closed/Multiply Connected:

= 2 = 0 (calculate for each intersection) = 2 (calculate for each enclosed area)


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Impact Loading/ Dynamic Loading:*

o = 1 +1 +2 (impact factor given height(h))o = (impact factor given velocity(v))o =o =o

=

Failure Theories

Maximum Normal Stress Fracture Criterion/ Maximum Principal Stress Fracture Criterion:*

o Fails if: (|1|, |1|, |1|) >o Useful for: predicting fracture in brittle materials under tension-dominated loading

Maximum Shear Stress Yield Criterion/ Tresca Criterion:*

o

Fails if: (|1 2|, |2 3|, |3 1|) >o Useful for: predicting yield in ductile materials

Maximum Distortion Energy Yield Criterion/ Von Mises Criterion:*

o Fails if: 12( )2 + 2 + ( )2 + 62 + 2 + 2 >

o Fails if: 12 [(1 2)2 + (2 3)2 + (3 1)2] >o Useful for: predicting yield in ductile materials

=*Fatigue and Fracture

Cyclic Loading:

o

=+2 =2 (1 + ) (mean stress)o = (range stress)o

=

2=

2=

2(1

) (alternating stress)

o = =11+ (stress ratio)o = =11+ (amplitude ratio)

Stress Concentration Factor:

o = Stress Intensity Factor:

o = (= geometry factor) Typical Stress versus Life (S-N) Curve:

Crack Deformation Types:

1.

Mode 1: Opening (a)

2.

Mode 2: Sliding (b)

3.

Mode 3: Tearing (c)


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Columns

General Equations:*

o = o

= ( is the slenderness ratio, r is the smallest radius of gyration)

o

Effective lengths:

Deflection of Beams

Method of Integration:*

o

=()o =() =o = =()o

=() Boundary Conditions:*


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Common Beam Deflections

*

*

* *

**

* *


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Statically Indeterminate/Geometric Constraint/Deflection Techniques

Multi-Load/Multi-Segment Axial/ Torque Load:*

1.

Find reaction forces/torques from a F.B.D.

2.

Start from one side, and find the axial load/torque for each segment from a F.B.D.

3.

For each segment, calculate the deflection, and sum deflections at the end.

Linear Deflection (Two Links):*

1.

Find reaction forces from a F.B.D.

2.

Find the deflection in each of the elastic links from the reaction forces.

3.

The deflection at any other point on the rigid frame falls on the deflection line.

Linear Deflection (> Two Links/ Statically Indeterminate):*

1.

Find a relationship between the reaction forces from a F.B.D. Use both a force and moment relationship

(two equations).

2.

The deflection at any other point on the rigid frame falls on the deflection line:() = + . Find anequation for the deflection of each link. (Same concept for a rotation)

3.

Find a relationship between the reaction force and deflection of each individual link, and solve the

system of equations.

Axial Clamp Deflection:*

1.

Find the deflection as a function of the axial load in each link (use symmetry).

2.

Find the constrain equation (total deflection = deflection of link in same direction deflection of link in

opposite direction, should sum together).

3.

Relate the axial forces in each link using a F.B.D. and solve the system of equations.

4.

*A pitch of 1 in. deflects 1 in. per turn of a screw/nut.


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Multi-Segment Axial/ Torque Load with Geometric Constraint:*

1.

Find an equation relating the reaction forces in each of the segments to the applied load using a F.B.D.

2.

Find equations for the deflection of each segment.

3.

Create a geometric constraint that requires the deflections to be equal, and solve the system of

equations.

Wedged Rod:*

1.

Find an equation relating the two unknown reaction forces.

2.

Pick a point in the center of the rod, and find an equation for the deflection of the point coming from

the left, and find an equation for the deflection of the point coming from the right. (Use same

coordinate system for both; for a temperature change, each rod extends away from the rigid wall).

3.

Due to the geometric constraint, the deflections should be equal, so set the equations equal to each

other, and then solve the system of equations. (If there is a gap in-between, either subtract the gap

from the deflection coming from the right, or add the gap to the deflection coming from the left).

Indeterminate Beam (Boundary Condition):*

1.

Find an equation for the moment.

2.

Integrate the equation twice, and treat the redundant reaction as an unknown.

3.

Use three boundary conditions to find the three unknowns (

,

1,

2).

Indeterminate Beam (Method of Superposition):*

1.

Find an expression for the deflection in the link.

2.

Use the method of superposition to find the deflection from the reaction force and the distributed load

in separate cases.3.

Set the expression for the deflection of the link equal to the summation of the two deflections of the

beam at the connection to the link from the reaction force of the link and the distributed load.


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Elementary Statics

Newtons Laws:*

o 1st: Every object in a state of uniform motion tends to remain in that state of motion unless an external

force is applied to it. This we recognize as essentially Galileo's concept of inertia, and this is often

termed simply the "Law of Inertia".

o 2nd: = (for static cases, = 0)o

3rd: For every action, there is an equal and opposite reaction.

Equilibrium of a Particles:*

o Fx = 0o

Fy = 0o

Fz = 0 Direction Cosines:*

o =+ + o 2 + 2 + 2 = 1o

= o = o =

Force Vector:*o = + + o

=2 + 2+2o () =

Moment:*

o =o =

Equilibrium of Rigid Bodies:*

o Fx = 0o Fy = 0o

Fz = 0o Mx = 0o My = 0o Mz = 0

Definitions of Structures:

o Trusses-designed to support loads and are usually stationary, fully constrained structures. They consist

exclusively of straight members connected at joints located at the end of each members. Members of a

truss, therefore are two-force members, or members acted upon by two equal and opposite forces

directed along the member.

o Frames- designed to support loads and are also, fully constrained structures. Unlike trusses, frames

contain at least one multi-force member, or a member acted upon by three or more forces which, in

general, are not directed along the member.

o

Machines- designed to transmit and modify forces and are structures containing moving parts.

Machines, like frames, always contain at least one multi-force member.

Trusses:*

o Tension (a) / Compression (b)


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o = 2() 3 (for a determinate truss)o Method of Joints:

Used to determine the force in every member.

Up to two unknown members.

o Method of Sections:

Used to determine the force in specific members.

Up to three unknown members.


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Dry Friction/ Coulomb Friction:*

o Between Two Surfaces:

Static Friction (Impending):

= Kinetic Friction (Sliding):

= (~.8)

Cases:

Solution Method:

1.

Solve for F and N.

2.

If , equilibrium is maintained, if , motion occurs and =o

Belt/Rope Friction: = (2 >1,2, > 2, =2 + )

o Impending Tipping vs. Slipping:


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Assume slipping occurs:

1.

Set =2.

Solve for x, P, and N

3.

Check that 0 2

Assume tipping occurs:

1.

Set =22.

Solve for P, N, and F

3.

Check that

Hydrostatic Forces on a Plane Surface:*

o

=o =(specific weight)o =(for a square surface)o = + o = + (for a square surface)


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Reactions*


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