Vertical deflection, stress, and strain Help

AI Thread Summary
The discussion revolves around calculating the vertical deflection, stress, and strain of an aluminum nail supporting a raincoat. The shear modulus of aluminum is given as 2.4 × 10^10 N/m², and the calculations involve determining the vertical deflection using shear deformation formulas. Participants express confusion over the lack of an elastic modulus (E) in the problem, which complicates the distinction between shear and bending stresses and strains. The correct approach for shear strain involves using the shear modulus, while bending calculations would require the elastic modulus. Overall, the problem highlights the need for clarity in specifying whether to use shear or bending parameters for accurate calculations.
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The shear modulus of aluminum is 2.4 × 1010 N/m2. An aluminum nail of radius
7.5 × 10–4 m projects 0.035 m horizontally outward from a wall. A man hangs a wet raincoat of weight 25.5 N from the end of the nail.
Assuming the wall holds its end of the nail, what is the vertical deflection of the other end of the nail?

What is the stress for this situation?

What is the strain?





Can someone work this out? I am super stuck! Thanks in advance
 
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I'm assuming you are looking for shear stress and shear deflection for this problem, and not the bending stress and deflection due to bending. Have you not been given a formula for determining deflections and stresses? You'll have to try to work it out first.
 
So I did some more digging, this is what i got, is it correct?

E= 7.0 x 10^10 N/m^2 (youngs modulus of aluminum)
A= 1.77 x 10 ^-6 N/m^2

vertical deflection: deltaY= (F/A)(L/G) (25N/(1.77 x 10 ^-6 N/m^2))/(0.035m/(2.4×10^10 N/m^2)) = 2.06 x10^-5 m

stress= F/A = 25N/(1.77 x 10 ^-6 N/m^2)= 1.4 x 10^7 N/m^2

E = stress/strain ---> strain = stress/E
(1.4 x 10^7 N/m^2)/(7.0 x 10^10 N/m^2)= 2.0x10^-4 = strain

?? good?
 
ambnj99 said:
So I did some more digging, this is what i got, is it correct?

E= 7.0 x 10^10 N/m^2 (youngs modulus of aluminum)
A= 1.77 x 10 ^-6 N/m^2

vertical deflection: deltaY= (F/A)(L/G) (25N/(1.77 x 10 ^-6 N/m^2))/(0.035m/(2.4×10^10 N/m^2)) = 2.06 x10^-5 m

stress= F/A = 25N/(1.77 x 10 ^-6 N/m^2)= 1.4 x 10^7 N/m^2

E = stress/strain ---> strain = stress/E
(1.4 x 10^7 N/m^2)/(7.0 x 10^10 N/m^2)= 2.0x10^-4 = strain

?? good?
I am not sure of the question. It asks for vertical deflection, stress, and strain, but it gives only a shear modulus, G. Shear deformations and strains are a function of G, whereas bending deformations and strains are a function of E, the elastic modulus. I don't know why the problem does not give an E value...you shouldn't have to look it up. Unfortunately, I never use the metric system in engineering, so i don't have a good feel for the numbers..too many darn decimals in SI, so i don't have the will to crank out the numbers to see if bending deformations are more significant than shear deformations for this case, or vice versa. In any case, don't confuse shear stresses and shear strain with bending stresses and bending strains. Where the shear deformation can be calculated (approximated)as you have stated (delta = FL/AG), the bending deformation is FL^3/3EI. And while the the shear stress is F/A, the bending stress is Mc/I, and where the the shear strain is delta_shear/L ,the bending strain is bending stress/E. So where are we now, i don't know, I've even managed to confuse myself :frown:. Please restate the problem as written.
 
PhanthomJay said:
I am not sure of the question. It asks for vertical deflection, stress, and strain, but it gives only a shear modulus, G. Shear deformations and strains are a function of G, whereas bending deformations and strains are a function of E, the elastic modulus. I don't know why the problem does not give an E value...you shouldn't have to look it up. Please restate the problem as written.
I wasn't sure what to due for calculating the strain, so i used the eq. tht included E...
the question is word for word:

The shear modulus of aluminum is 2.4 × 1010 N/m2. An aluminum nail of radius
7.5 × 10–4 m projects 0.035 m horizontally outward from a wall. A man hangs a wet raincoat of weight 25.5 N from the end of the nail.
Assuming the wall holds its end of the nail, what is the vertical deflection of the other end of the nail?

What is the stress for this situation?

What is the strain?
 
Is anyone certain of how to calculate the strain?
 
ambnj99 said:
I wasn't sure what to due for calculating the strain, so i used the eq. tht included E...
the question is word for word:

The shear modulus of aluminum is 2.4 × 1010 N/m2. An aluminum nail of radius
7.5 × 10–4 m projects 0.035 m horizontally outward from a wall. A man hangs a wet raincoat of weight 25.5 N from the end of the nail.
Assuming the wall holds its end of the nail, what is the vertical deflection of the other end of the nail?

What is the stress for this situation?

What is the strain?
This problem is poorly worded. They give you the shear modulus, G, but not the elastic modulus, E. An then they ask you to calculate deflection, stress, and strain, when they shuuld have specified whether they were looking for deflection due to shear or bending , and shear stress vs. bending stresses, and shear strain vs. bending strain, or both. I cranked out some real rough numbers, and it looks like shear stresses and deflections pale in comparison to bending stresses and deflection at the free end as caused by bending moments. But since they gave you G and not E, i guess they are asking for shear strains and stresses and deflections, which are extremely small. Have you studied these equations? They are usually not covered as well as the bending stresses and strains and deflections.
 
ambnj99 said:
So I did some more digging, this is what i got, is it correct?

E= 7.0 x 10^10 N/m^2 (youngs modulus of aluminum)
A= 1.77 x 10 ^-6 N/m^2

vertical deflection due to shear only [/color]: deltaY= (F/A)(L/G) (25N/(1.77 x 10 ^-6 N/m^2))/(0.035m/(2.4×10^10 N/m^2)) = 2.06 x10^-5 m

shear [/color]stress= F/A = 25N/(1.77 x 10 ^-6 N/m^2)= 1.4 x 10^7 N/m^2

E = stress/strain ---> strain = stress/E
(1.4 x 10^7 N/m^2)/(7.0 x 10^10 N/m^2)= 2.0x10^-4 = strain let's stick with shear strain, shall we ??, forget about E right now, since we can't figure what the problem is asking for [/color]

?? good?
see comments above
 
yes, I too have been looking into it,


The above strain is wrong, as you stated with E value my book states shear strain to be:

strain= ΔL= (1/G)(F/A)

=(1/2.4×10^10 N/m^2)(25.5N/(1.77 x 10 ^-6 N/m^2) =2.10 x 10^-5 m
so I am going with this for my solution for strain.
 
  • #10
ambnj99 said:
yes, I too have been looking into it,


The above strain is wrong, as you stated with E value my book states shear strain to be:

strain= ΔL= (1/G)(F/A)

=(1/2.4×10^10 N/m^2)(25.5N/(1.77 x 10 ^-6 N/m^2) =2.10 x 10^-5 m
so I am going with this for my solution for strain.
Correct answer, wrong terminology for shear strain. If you are trying to find shear strain, you must use G, not E. Shear strain = shear stress/G, or , alternatively, shear strain = shear deformation/L , or shear deformation = (FL/AG)/L = F/AG, same result. Do not call shear strain delta L; delta L has units of length;strain has no dimension (it is a dimensionless quantity).
 
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