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Bourbon daddy
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I was wondering if there were any mechanical engineers that can answer a few questions I have regarding an assignment that I have been set. We have to choose a suitable beam to support a monorail. we are looking for a moment of deflection of around 10mm. Using the universal beams table bs 4 1993 I have found a beam that satisfies the requirements when using the formulas:
R=E.I/M
We're R is radius
E is Elastic modulus (given figure for steel is 210 x 10^9 pa)
I is second moment of area about the neutral axis (taken from Bs4 and converted to m^4)
M is bending moment (worked out to be 262.5 x 10^3 Nm)
And
H = R-√R^2-m^2
Were H is deflection in m (then converted to mm for purpose of the exercise)
R is radius
m is the half the length of the original beam (20/2)
Using the beam with dimensions 1016 x 305 x 314, we get the following:
R=E.I/m
R= 210 x 10^9Pa x 6442 x 10^-6 m^4/262.5 x 10^3Nm
R=5153.6
So to get the moment of deflection we can use the formula
H=R√R^2-m^2
H=5153.6-√5153.6^2-10^2
Multiplying answer by 1000 to convert from m to mm we get the figure
9.701 mm
This answer seems to satisfy the requirements.
But since
R=y.E/s
Were y is distance from the neutral axis to edge or surface of the beam (m) (0.5m)
s(should be sigma but I am using ipad and don't have access to Greek symbols)=stress due to bending at distance y from the neutral axis (given as 80 MPa)
I am now getting the following
R=0.5 x 210x10^9/80x10^6
R=1312.5 this figure is drastically different to original 5153.6
Entering this figure into deflection formula we have
H=1312.5-√1312.5^2-10^2
H=38mm
No longer satisfactory
Do these figures and calculations look correct and if so, how come they are so drastically different. Is the second set of equations taking into account degradation of the beam/wear and tear? And as an engineer which figures/equation would you use... Obviously you would go for the worst case scenario, but what would the reasons be
I know it's a long winded question, thank you for taking the time out to read it
Regards Joe
R=E.I/M
We're R is radius
E is Elastic modulus (given figure for steel is 210 x 10^9 pa)
I is second moment of area about the neutral axis (taken from Bs4 and converted to m^4)
M is bending moment (worked out to be 262.5 x 10^3 Nm)
And
H = R-√R^2-m^2
Were H is deflection in m (then converted to mm for purpose of the exercise)
R is radius
m is the half the length of the original beam (20/2)
Using the beam with dimensions 1016 x 305 x 314, we get the following:
R=E.I/m
R= 210 x 10^9Pa x 6442 x 10^-6 m^4/262.5 x 10^3Nm
R=5153.6
So to get the moment of deflection we can use the formula
H=R√R^2-m^2
H=5153.6-√5153.6^2-10^2
Multiplying answer by 1000 to convert from m to mm we get the figure
9.701 mm
This answer seems to satisfy the requirements.
But since
R=y.E/s
Were y is distance from the neutral axis to edge or surface of the beam (m) (0.5m)
s(should be sigma but I am using ipad and don't have access to Greek symbols)=stress due to bending at distance y from the neutral axis (given as 80 MPa)
I am now getting the following
R=0.5 x 210x10^9/80x10^6
R=1312.5 this figure is drastically different to original 5153.6
Entering this figure into deflection formula we have
H=1312.5-√1312.5^2-10^2
H=38mm
No longer satisfactory
Do these figures and calculations look correct and if so, how come they are so drastically different. Is the second set of equations taking into account degradation of the beam/wear and tear? And as an engineer which figures/equation would you use... Obviously you would go for the worst case scenario, but what would the reasons be
I know it's a long winded question, thank you for taking the time out to read it
Regards Joe