Notes 6 "Torsion"

Torsion

• Introduction:

- In the preceding 2 chapters we studied the stress and strain subjected to axial loads .in this chapter we will study the torsion :analyze stress and strain of circular cross section subjected to twisting torques .

•Notes :

- Twist …….rotation………shafts

- Loads…….axial…………..rod

• To deal with torsion there are some conditions:

1.Line  remains line .

2.The circular shape remains in his shape .

3.γ & φ are very small  .

• Notes:

- Φ decreases towards the fixed support

- γ decreases towards the center of the shaft

γ - shear strain

Φ - angle of twist 

Δ - arc

Before we complete  there are some important notes we should take care of them:

• At the center of the shaft the shear = zero

• At the surface of the shaft:

  1.Shear ………………max

  2.γ  ……………………max

• The is a linear relation between γ  and the distance far away the center (p) 

 

J : is the tensional const. Or polar second moment.

G : is the modulus of rigidity.

• Some important relations(τ and t)

α :  is the angel per unit length


• Note : if the Shaft is Hollow :

 

& C2 > C1

•If we have a shaft supports from its 2 edges:

•We free one of supports and solve it as a general ex

• In this type of examples we free one of the supports and we solve it then we use a torque  that opposite to the first torque to achieve the equilibrium .
 

• Example (1):

Solution :
- Free the right support

- Put the opposite torque

- Then complete it by equal these angels

• Example (2):

- Φ2= the angel w.r.t the support at b
- Φ1=the angel w.r.t the shaft A
 
- the is a relative angel between SHAFTS (A&B)…..ώ
- Φ2 R2 = θR2 
- relative angel  ώ=φ2*((R2/R1))
- θ=ώ+φ1
 
N.B:
Lama nin2el angles or torques men tirs l turs tany lazem ni5alibalna men el ratio between R1&R2.

Notes 6 "Slides"

7laha - Stress and Strain

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Related Videos 4

Structure is life - Pt.1

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Related Videos 3

Related Videos 2

" External " Thermal Strain

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" External " Stress & Strain Notes

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" External " Stress & Strain Curves

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" External "Stress on Trusses

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"External" Trusses

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"External" Mechanics and Materials

Part 1

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Part 2

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"External" Equlibrium Problem Solving

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"External" Static equilibrium

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Related Videos

 

7laha - Truss

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7laha - Equilibrium

"7laha - Equilibrium"

Smart structure problems for never stop thinking

Source:  Mechanics of material by

Ferdinand p. Beer and E .Russell Johnston

Solved Example (3-41) :

A continuous cable of total length 4m is wrapped around small pulleys at A,B,C and D if each block has mass =15.6Kg what is the change in the length of the two springs. you should know that the springs are un stretched when d=2m .
K for the springs =500 N/M .

Solution :

Each spring has k=500N/m   each mass =15.6KG
At d=2m the cable has  square form
 

when we put the masses
The change in the form is too slight
So    2 T/√2=15.6 *9.81   so T =108.2N

So ∆=(√2T/500)=0.3m
Note that : this problem can be solved in one step that : mg=K∆

Try this problem :

From the previous problem
At different masses and horizontal springs .Can  the cable take this form?
Consider α=60deg ,θ=30deg ,a and b are known and the weight at pulley D is lighter than the weight at pulley B
If your answer is yes find the mass at pulley D in terms of ∆1 and∆2
If your answer is No say why?

Powerpoint Presentation

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7laha - Force system Resultants

"7laha - Force system Resultants"

Smart structure problems for never stop thinking

Source:  Mechanics of material by

Ferdinand p. Beer and E .Russell Johnston

We all see a rigid body on surface of something .this body has material with special density so at least we study the weight force of the body on the surface of the beam ;……etc.
We start our study for the force of weight  from :
1#choosing axis for our problem at lowest height
2#determining weight force at the beginning and the end of our body/our axis(at X=0   at X has value)
3#determining the force density for area
4#divid the shape of the body into known shapes
5#determinig centers of mass to all shapes
6#determining the resultant of force according to the direction of forces and taking the moment around  any point .
Remember this rule (moment of the resultant =summation of all forces moments around point
Notice :this steps to conclusive problem .some problems does not need all these steps

Solved example on force system resultant :

The granular material exerts the distributed loading 50(1+x)K pa  on the top surface of the beam as shown in the figure .determine the magnitude and location of the equivalent resultant of this load .

solution :

The loading is distributed on the width of the beam so the force distribution on the beam equal   to 50(x+1)*0.2=10(x+1)KN/m .
The trapezoid can be divided into rectangular and triangle .

The loading is distributed on the
area of the shape
F1=9*10=90KN(length #width)
F2=0.5*9*90=405KN
Parallel forces  so the resultant =495KN=Fr   

Note that : X`2=3m X`1=4.5 m so he last thing to know  X`r
MA=Fr*X`r=405*3 +90*4.5
So  the location of resultant force =3.27m
 

Try this problem :

if the distribution of the ground reaction on the pipe (the diameter=1.6m  )per meter of length can be approximated as (0.5(1+cosα)KN/m  )shown figure .determine the magnitude of the resultant force due to this loading .

Powerpoint presentation :

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Thermal Loading

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Stress and strain - Axial loading

Notes 4

 "Stress and strain - Axial loading"

1-If it has a uniform cross section
2-loaded at its ends
3-Rod is homogenous ( E constant )

Stress and strain - Axial loading

It is  not always possible to determine the forces in the members of structure by applying only the principles of statics , because it is based on the assumption of undeformable rigid structure .
 

By considering engineering structures as deformable and analyzing the deformation in their various members .
 

It will be possible for us to compute forces that are statically indeterminate . Indeterminate within the framework of statics .
 

You will consider the deformations of a structural member  such as a rod , bar or plate under axial loading  .

Normal strain is the deformation of the  member per unit length or stress per modulus of elasticity  .

Powerpoint Presentation

" Solved Problems Included "

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The generalized Hook’s law

Notes 3

 "Equlibrium"

Introduction :

Equilibrium is the foundation of any science .Each science sees it with different point of view and present it in away which suits the nature of the proposed problem. In the case of studying static structure, equilibrium is essential to model any structural problem, and it means that the structure will be able to sustain the applied external load without movement ,in other words, the external generalized forces equals the support reactions and the structure internal forces equal the external loads.

STRESS AND STRAIN :

Average stress =force /area     (vector )
There are two types of stress :

1-Axial stress ( axis line is parallel to surface area ) .
2- shear stress (axis line is normal to surface area –Tangential stresses )( friction betweeen two surfaces causses shear stress) .

Shear Bolt :

 << Poisson Ratio >>

* where ν is poisson ratio ..

HOOK'S LAW :

The generalized hook’s law : 
a

Applying  a hydrostatic  pressure :
a

a
SO ,the dilatation (reduction in the  volume / original volume ) will be :aa

a
For  the problem to be physically acceptable the volume must decrease , so :
a

a

Volumetric Strain :

Volume =Length x weight  x  height
 

      V=LWH

a

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" Solved Problems Included "

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Trusses

Notes 2

 "TRUSSES"

About Trusses :

1-Axial loads only
2-Loads applies at end points only
3- Elements are joined by pins only

1-Determine the force in each member of the truss ?

2-Determine the force in each member of the truss and indicate whether the the members are in tension or compression ?

3-Determine the force in each member of the truss and indicate whether the the members are in tension or compression ?

At joint A :

At joint C :

At joint D :

4-Determine the force in each member of the truss and indicate whether the the members are in tension or compression ?

At joint B :
a

At joint C :
a

At joint A :
a

* The Method of Sections :

6-Determine the force in each member of the truss and indicate whether the the members are in tension or compression ?

 Method of joints : We will begin analyzing the Equilibrium of joint D , and the proceed to analyze joints C & D .

Note : The equilibrium analysis of joint A can be used to determine the components of support reaction at A .

Powerpoint Presentation

" Solved Problems Included "

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