The coupling is used to connect the two shafts together. Assuming that the shear stress in the bolts is uniform, determine the n
umber of bolts necessary to make the maximum shear stress in the shaft equal to the shear stress in the bolts. Each bolt has a diameter dd. Express your answer in terms of some or all of the variables rrr, RRR, ddd, and TTT. nn
1 answer:
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Answer:
Explanation:
Given:-
- The diameter of the cylinder, d = 50 mm.
- The compressive load, F = 80 KN.
Solution:-
- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.
- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).
εx = - [ σx - ν( σy + σz ) ] / E
εy = - [ σy - ν( σx + σz ) ] / E
εz = - [ σz - ν( σy + σx ) ] / E
- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:
εx = [ ν*σz ] / E
εy = [ ν*σz ] / E
εz = - [ σz ] / E .... Eq 1
- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.
- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:
εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx
εx' = - [ σx' - ν( σy' + σz ) ] / E = [ ν*σz ] / 2E
εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy
εx' = - [ σy' - ν( σx' + σz ) ] / E = [ ν*σz ] / 2E
- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.
- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).
Note: σx' = σy', The cylinder is radially enclosed around the entire surface.
Therefore,
- [ σx' - ν( σx'+ σz ) ] = [ ν*σz ] / 2
σx' ( 1 - v ) = [ ν*σz ] / 2
σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]
- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):
εz' = - [ σz - ν( σy' + σx' ) ] / E
εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E
εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E ... Eq2
- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:
... Answer
Answer:
minimum length of a surface crack is 15.043 mm
Explanation:
given data
strain fracture toughness K = 78 MPa
tensile stress = 345 MPa
Y = 1.04
to find out
minimum length of a surface crack
solution
we find here length of critical interior flaw from formula that is
α = ....................1
put here value we get
α =
α = 15.043 mm
so minimum length of a surface crack is 15.043 mm