Answer:
a) 65.139 MPa
b) 17490 GPa
Explanation:
a) calculate the flexural strength
we now that for a three-point bending test the bending moment will occur at the midsection
Max stress ( flexural strength ) = MY / I
Max stress can be expressed as ; 3/2 * ( F*l / b*d^2 ) ------- ( 1 )
F ( load at fracture ) = 283
l (distance between support point ) = 50
b = 11.6
d = 5.3
substitute given data into equation 1 above
= 3/2 * (( 283 * 50) /( 11.6 * 5.3^2) )
= 3/2 * ( 14150 / 325.844 )
= 3/2 * 43.425 = 65.139 MPa
B ) Determine deflection ( Δ y )
load = 265 N
elastic modulus = 66 GPa
elastic modulus = tensile stress / tensile strain
Tensile stress = Δ y
strain = load = 66 GPa
therefore Δ y = 265 * 66 = 17490 GPa
(a) The flexural strength will be "65.139 MPa".
(b) The deflection will be "17490 GPa".
According to the question,
Load at fracture,
Distance b/t support points,
Width,
Height,
(a)
The flexural strength will be:
= [tex]\frac{MY}{I}[/tex]
or,
= [tex]\frac{3}{2}\times (\frac{(F\times I)}{b\times d^2} )[/tex]
By substituting the values, we get
= [tex]\frac{3}{2}\times (\frac{(283\times 50)}{11.6\times 5.3^2} )[/tex]
= [tex]\frac{1}{2}\times (\frac{14150}{325.844} )[/tex]
= [tex]\frac{3}{2}\times 43.425[/tex]
= [tex]65.139 \ MPa[/tex]
(b)
We know,
Load,
Elastic modulus,
Now,
→ [tex]Elastic \ modulus = \frac{Stress}{Strain}[/tex]
here, Stress = Δy
Now,
→ [tex]\Delta y = 265\times 66[/tex]
[tex]= 17490 \ GPa[/tex]
Thus the above answers are correct.
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