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**Unformatted text preview: **PERIDO, JOHN PAULO M.
MSAE – FPM
2018-68796 AENG 260 -EXPERIMENTAL STRESS ANALYSIS
PROBLEM SET IV
DR. DELFIN C. SUMINISTRADO 3.31. A steel (E = 207 GPa and v = 0.30) ring is shrunk onto another steel ring, as shown in the
figure below. Determine the maximum interference possible without yielding one of the rings if
the yield strength of both rings is 900 MPa and the dimensions of the rings are as follows: θ , , , Case 1
100
125
150 Case 2
100
115
200 Case 3
100
150
175 Case 4
0
100
150 Interference between two different materials = [ + + ( + )
+
(
− )] − − Where: - interference, mm - pressure generated with the designed connection or yield strength, MPa - Modulus of elasticity of the outer materials, GPa - Modulus of elasticity of the inner materials, GPa - outer radius of the larger ring, mm - inner radius of the smaller ring, mm - inner radius of the larger ring or outer radius of the smaller ring, mm - Poisson’s ratio of the inner ring - Poisson’s ratio of the outer ring Interference of same the material − =
[ ( − )( − )
Where: - interference, mm - pressure generated with the designed connection or yield strength, MPa - Modulus of elasticity of the inner materials, GPa - outer radius of the larger ring, mm - inner radius of the smaller ring, mm - inner radius of the larger ring or outer radius of the smaller ring, mm CASE 1.
( )( )
( ) − ( )
=
[ (() − ( ) )(( ) − ( ) ) = . CASE 2.
( )( )
( ) − ( )
=
[ (() − ( ) )(( ) − ( ) ) = . CASE 3. = ( )( )
( ) − ( )
[ (( ) − ( ) )(( ) − ( ) ) = . CASE 4.
( )( )
( ) − ( )
=
[ (( ) − ( ) )(( ) − ( ) ) = . Reference
Press Fit Pressure Calculator – Optimize Your Interference/Transition Fit Design (2012, October
0000021). Retrieved from - calculator00000optimize.html. 3.16. Determine the stresses in the uniformly loaded cantilever beam shown in the figure below
by using Airy’s stress function approach. Assume a unit thickness. Boundary conditions
At x = 0; = 0 = 0
At x = L;
ℎ ∫ = −ℎ
ℎ ∫ = 0
−ℎ
ℎ ∫ =
−ℎ At y = h; = − = 0 1 2 2 At y = -h; = 0 = 0
Stress function, superimposing polynomials = 2 + 3 + 4 + 5. Also, since the cross section is symmetric about the y-axis and = − at y = h and = 0 at y = -h, should be
an odd function of y. Thus the stress function should contain odd functions of y, hence, should also contain odd functions of y since the net axial force is zero. Second, since is
constant as a function of x on the top and bottom faces, the stress function should not contain
powers of x greater than 2 . = 2 + + 2 + 3 + 2 + 2 + 3 + 4 + 3 + 2 2 + 3 + 4
+ 5 + 4 + 3 2 + 2 3 + 4 + 5
Removing all terms with a power of x greater than two and y with even power. Rearranging all
terms will be. = + 2 + 2 + 3 + 3 + 2 3 + 5
Since ∇4 = 0,
∇4 = 24 + 120 = 0
Thus = −5
Equations of Stresses 2 =
= 6 + 6 − 30 2 + 20 3 2 = 2
= 2 + 2 − 10 3 2 2
=
= −( + 2 + 3 2 − 30 2 ) For at = 0, + 3 2 = 0
this can be true for all values of y when,
==0
For at = ±ℎ and equating to zero,
0 = −(2 − 30ℎ2 ) = 15ℎ2
Substituting to equation of , = 2 + 30ℎ2 − 10 3 For = − and = 0 at = ℎ and = −ℎ,
− = 2 + 30ℎ3 − 10ℎ3 = 2 + 20ℎ3 And
0 = 2 − 30ℎ3 + 10ℎ3 = 2 − 20ℎ3 Solving simultaneously,
=− =− 4 40ℎ3 Substituting to the equation G, = 15 (− 3
2
)
ℎ
=
−
40ℎ3
8ℎ Substituting E and G to = 6 + 6(0) − 30(− = 6 +
ℎ ) 2 + 20(−
) 3
3
40ℎ
40ℎ3 3 2 −
3
3
4ℎ
2ℎ3 1 For ∫−ℎ = 2 2 at = ,
ℎ ∫ (6 +
−ℎ 3 2 1 − 3 ) = 2
3
3
4ℎ
2ℎ
2 ℎ ∫ (6 2 +
−ℎ (2 3 + 3 2 2 1 2
4
) − = 4ℎ3
2ℎ3
2 2 3 1 2
5 − ) = 4ℎ3
10ℎ3
2 {2[ℎ3 − (−ℎ3 )] + 2 3 1 [ℎ − (−ℎ3 )] −
[ℎ5 − (−ℎ5 )]} = 2
3
3
4ℎ
10ℎ
2 {2[ℎ3 − (−ℎ3 )] + 2 3 1 [ℎ − (−ℎ3 )] −
[ℎ5 − (−ℎ5 )]} = 2
3
3
4ℎ
10ℎ
2 2 ℎ2 1 2
4ℎ + −
= 2
5
2
3 4ℎ3 = = ℎ2
5 20ℎ Substituting the constants back to the equations for stresses, 2
) + 6(0) − 30 (−
) +
20
(−
) 3
20ℎ
40ℎ3
40ℎ3 = 6 ( = 3
3 2 + −
3
3
10ℎ
4ℎ
2ℎ3 = ( + ) = 2 (− − 3 ) − 10 (−
) + 2 (−
) 3
4
8ℎ
40ℎ3 = − = − = −(0 + 2 (− 3 3
−
+
2 4ℎ 4ℎ3 ( + − ) 3 ) + 3(0) 2 − 30 (−
) 2 )
8ℎ
40ℎ3 = = 3 3 2
−
4ℎ 4ℎ3 ( − ) ...

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- Fall '19