Hi Every one,
It is immense pleasure to meet all you in this technical forum.
I have a basic query in strength of materials
This is about of short beam
Beam dimensions are
L=7.375 mm
Dia =15.05 mm
Tip Force=27514 N
For the above beam, following are the geometrical properties
area=177.894 mm^2
IZZ=2518 mm^4
For this kind of beam, how should we approach so as to determine the maximum stress in the component.
If i use convensional bending equation, it gives the bending stress of 606 MPa.
If i use shear stress formula, i am getting the shear stress of 155 Mpa. As this beam is shorter one, i considered vertical shear instead of longitudinal one.
My query is, which stress would cause failure in this structure. As a matter of fact, component failure appears to be shear kind. But, as per the above calculations, bending stress appears to be high.
All i want to know is, how to calculate maximum failure stress in the short beam?
Can any one throw some light regarding this.
Or else the link which contains answer for the above query would be appreciated.
Thanks
Krishnan
It is immense pleasure to meet all you in this technical forum.
I have a basic query in strength of materials
This is about of short beam
Beam dimensions are
L=7.375 mm
Dia =15.05 mm
Tip Force=27514 N
For the above beam, following are the geometrical properties
area=177.894 mm^2
IZZ=2518 mm^4
For this kind of beam, how should we approach so as to determine the maximum stress in the component.
If i use convensional bending equation, it gives the bending stress of 606 MPa.
If i use shear stress formula, i am getting the shear stress of 155 Mpa. As this beam is shorter one, i considered vertical shear instead of longitudinal one.
My query is, which stress would cause failure in this structure. As a matter of fact, component failure appears to be shear kind. But, as per the above calculations, bending stress appears to be high.
All i want to know is, how to calculate maximum failure stress in the short beam?
Can any one throw some light regarding this.
Or else the link which contains answer for the above query would be appreciated.
Thanks
Krishnan
Forums
Short Cylindrical Shaft Problem
Assuming that this is a cantilevered cylindrical shaft, the conventional formula for the shear stress distribution across the section is
Tau = VQ/(Ib)
where in this case,
V = 27514 N
Q = Integrate[Integrate[y,{x,-(r^2-y^2)^(1/2),(r^2-y^2)^(1/2)}],{y,y,r}]
= (2/3)*(r^2-y^2)^(3/2)
I = Integrate[Integrate[y^2,{x,-(r^2-y^2)^(1/2),(r^2-y^2)^(1/2)}],{y,-r,r}]
= (Pi/4)*r^4
b = 2*((r^2-y^2)^(1/2))
2r = 15.05 mm
which results in a parabolic shear stress distribution
Tau = (4/3)*(V/A)*(1-(y/r)^2)
that assumes a maximum of 206.22 MPa halfway through the section (at y=0).
This problem sits between two extremes: a beam-like variation of the stress distribution (ref. Euler-Bernoulli, Timoshenko theory) and a uniform stress distribution (ref. Mohr's Circle). A computer-aided mechanical analysis would verify this.
The criteria for failure depends on the material and is typically in terms of the principal stresses (ref. Tresca & von Mises Envelopes). The above shear stress (assuming no axial stresses at y=0) is the same as the absolute value of the principal stresses at that point (which are equal in magnitude and opposite in sign/direction).