Cantilever beam point load analysis with diagram

[Jason24]

Homework Statement

Require various values from cantilever beam analysis including
Bending moment
K stiffness
2nd moment of inertia
Max stress

Homework Equations

Shown below

The Attempt at a Solution

Here is my attempt at this question but I am unsure about units and therefore my answers?Bending moment

Image attached below

Calculated as follows:
L=0.24M
L2=0.02M
P=-0.0049KN

FX=0
FY=0

RAx=0 (No horizontal forces)
+Ray-0.0049=0

M=0
- (0.0049) (0.25) +Ma=0
Ma=1.176x10-3

Second moment on inertia

Ix=bh^3/12 = (30) (260) ^3/12 = 43.94x10^6 (439.4mm^4)Neutral axis (Y)

X axis = 6mm hence Y=3mm

Flexural formula

To find max stress
M/I = Sigma/Y
M=1.176x10-3
I=43.94x10^6
y=3mm
Stress= My/I (flexural formula)
Stress = (1.176x10-3) (0.003) = 8.04x10^-14
(43.94x10^6)

K stiffness

K=3EI = (3)(190x10^9)(43.94x10^6)/0.26^3 = 1.425x10^13
l^3

Young’s modulus formula

E= f
e
Ee=f
Strain= Stress/Young’s modulus
Strain= 6.68/190x10^9= 3.51Stress

Stress = W/Z (l-x)

Stress = 4.9/14.64x10-3 (.26 -.24) = 6.69N

E= stress/strain
E=190x10^9
Strain = Stress/E = 6.69x190x10^9 = 3.52x10^-11
Stress = E x Strain = 190x10^9 x 3.52 x 10^-4 = 6.68

Any help is greatly appreciated!

Homework Statement

Homework Equations

The Attempt at a Solution

 

[SteamKing]

I see a bunch of numbers which may or may not be correct.

Without an intelligible problem statement, who knows what you have done?

 

[Jason24]

Im essentially trying to gain a value for the maximum stress recorded from a cantilever beam so that this can be compared to that of an FEA model. However to do this I need to obtain values for bending moment, stiffness, strain etc beforehand.

I have no direct question for this problem other than are the values for these problems right with the information that I have provided?

Once I know they are right I can compare the max stress value here to the values from the rig using voltage to strain conversion

This of any help?

 

[SteamKing]

All right, here goes:

1. L of the beam = 0.24m Why is your moment = P * 0.25m?

2. Neutral axis (Y): X axis = 6mm hence Y=3mm
What does this mean?

3. Second moment on inertia

Ix=bh^3/12 = (30) (260) ^3/12 = 43.94x10^6 (439.4mm^4)

It is technically the second moment of area (or moment of inertia)

The dimensions of your beam cross section are not given in your problem statement
It appears you are saying the beam is 260 mm deep by 30 mm wide.
This beam has very unusual proportions, in that its width is approximately equal to its length.

All of your other calculations depend on the accuracy of the results above.

 

[Jason24]

Its point load so its 0.24m from fixed end to load (P) then 0.02 after load, giving a total length of 0.26m. The load at 0.24m is 4.9N

Y from the bending formula is distance from neutral fibres to base and so neutral axis is 3mm as total thickness is 6mm.

Second moment of area I've now got as 4.394x10-5 m4

dimensions of beam are as follows
b=30mm
h=260mm
so its a long thin rectangular beam as usual

thanks for your help

 

[SteamKing]

Let's try this again:

M = 4.9N * 0.24m = 1.176 N-m

If the beam is 6 mm deep and 30 mm wide, then

I = (1/12)*(30)*(6^3) = 540 mm^4

Since 1 m = 1000 mm, then 1 m^4 = (1000 mm)^4 = 10^12 mm^4

so I = 540 /10^12 = 5.4 * 10^-10 mm^4

 

[SteamKing]

Check the I calc:

I = 540 mm^4 = 5.4 * 10^-10 m^4

 

[Jason24]

I understood for a rectangular beam

Ix = b*h3 / 12

where

b = width

h = height

its not the thickness but the height of the beam that is entered into the formula?

 

[SteamKing]

The moment of inertia is calculated for the cross section of the beam.
The cross section is taken transversely, not longitudinally.

Your initial calculation had b = 30 mm and h = 260 mm, yet you said the NA was 3 mm from the outer fiber of the beam because the depth = 6 mm