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Load, elasticity, strain, strength 
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#1
Dec1708, 08:36 PM

P: 122

1. The problem statement, all variables and given/known data
A continuous and aligned glass reinforced composite consists of 40 volume % of glass fibers, having modulus of elasticity = 10 X 10^6 psi and 60 volume % of polyester resin that when hardened, displays a modulus of elasticity = 5 X 10^5 psi. a) compute modulus of elasticity in the longitudinal direction b) If the crosssectioned area is .4in^2 a stress of 7000psi is applied on the longitudinal direction, compute magnitude of the load by each of the fiber and matrix. c) Distance of strain that is sustained by each phase when stress is applied. d) Assuming tensile strength of 500,000 psi and 10,000 psi, respectively, for glass fibers and resin. Determine the longitudinal strength of this fiber composite. 2. Relevant equations E = elasticity C = volume % P = load A = area F = tensile strength (i think) e = strain subabbreviations: c = composite f = fiber m = matrix 3. The attempt at a solution (this is a lot of reading, sorry. i summarized it on the last sentence) i need help BIG time. i was hoping somebody here would be able to give me the formulas that im missing in order calculate all these, or at least be a pal and help me figure the formulas out, because i need to know it for tomorow's exam but i dont have it cus my professor was difficult to understand (he has a heavy accent and bad handwriting). here's what i have so far though: a) Ec = EfCf + EmCm = (10 x 10^6)(.4) + (5x10^5)(.6) = 4.3 x 10^5 psi b) stress of composite = 7000psi = P/A A = .4 in^2 load = Pc = (7000)(.4) = 2800 lbs ...now how do i figure out the load on each? do i just multiply the volume % by total load? (continue reading, i beleive i have part of the answer further down) c) no idea how to do this. the professor had random calculations all over the board, i couldnt follow it cus once again i cant understand what hes saying and it takes me ten minutes to figure out what each letter is cus his handwriting is terrible. d) same thing, no idea, just a bunch of giberish that i wrote down on the board. (im sure i found the answer further down) _______________________________________________________________________ Solving this is like trying to put puzzle peices togeather. here's some calculations that i have written down: these looked like F's but probably werent: Ff/Fm = 13.3 Ff = 13.3Fm Fc = (Ac)(stress) = (.4)(7000) = 2800lbs (this i think is part of part b)) Fc = Ff + Fm = 13.3Fm + Fm = 14.3Fm Fm = 2800/14.3 = 195 lbs Ff = 2800  195 = 2605 lbs (further down, it appears these calculations are for part b) ...ok i have no idea where the 13.3 came from and im not sure what numbers where supposed to be in place of Ff/Fm so if anyone can figure that out id appreciate it. so here is some more calculations which im not sure what part they are for: Am = VmAc = (.6)(.4) = .24in^2 Af = VfAf = (.4)(.4) = .16 in^2 stressm = Fm/Am = 195/.24 = 812.5 psi stressf = Fc/Af = 2605/.16 = 16281 psi ok i HIGHLY appreciate it if you are still trying to keep up. it would really make a difference on my grade if you could help me figure this out, so thanks so far.. lol. now given these peices of the puzzle, i came up with this conclusion: the formula is supposed to be stress = P/A. so therefore i think the values Fm (195) and Fc (2605) are the values of the loads. maybe he had bad handwriting and it should have been: Pm = 195 lbs Pf = 2605 lbs and this answers part b. now some more calculations: em = stressm/Em = (812.5)/(5x10^5) = 1.63 x 10^3 in/in ef = stressf/Ef = (16281)/(10x10^6) = 1.63 x 10^3 in/in Ec = stressc/ec = stressc/em = (7000)/(1.63 x 10^3) = 4.3 x 10^6 ok this makes sense because it goes along with the formula and em = ef. im not sure why the units are in/in. maybe it was something else and i misread his handwriting? anyways, id like to know which question these calculations are apart of... lol. where these calculations for part c? when they ask for "distance of strain" do they just want the strain?? moving on to the last calculation: (10,0000psi)(.6) + (500,000)(.4) = 206,000 psi this appears to be the tensile strength for part d). so i guess: F = 206,000 psi if you're unclear on ANYTHING let me know, i know its confusing. i just have random calculations and im trying to put this togeather so i can figure out how to do it correctly to summarize it: my main concern is i'd like to know how to answer part C and i'd like to know where the 13.3 came from (in part b). i highlighted the 13.3 part in red and the part that i think is for part c in blue 


#2
Dec1708, 09:01 PM

P: 122

AHA!!!! i figured out the 13.333...
Ff/Fm = (Cf)(Ef)/(Cm)(Em) = (.4)(10x10^6)/[(.6)(5x10^5)] = 13.3333 now i would just like somebody to confirm if im right about the strain being the same as the distance of strain 


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