johnson9883 Posted November 6, 2016 Share Posted November 6, 2016 for this question , I'm having problem with the shear stress at point E and shear stress at centorid. normally , the shear stress at the centoid will be maximum , right ? But , in my working , I found that the shear stress at the centroid is smaller than the shear stress at E. What's wrong with the working ? i get y coordinates of centorid = 66.7mm For Ixx , i get (5.00x10^-5)(m^4) , For V(shear force ) , I use (437.5x10^3) For shear stress at centroid , i use formula of $$\tau = V(Q) / It $$ so at centroid , Q = (66.67x10^-3)(160x10^-3)(66.67x10^-3 / 2 ) = 3.56x10^-4 so $$\tau $$= (437.5x10^3)(3.56x10^-4) / (5.00x10^-5)(160x10^-3) = 1.9x 10^7 Pa at E , Q = Ay = (40x10^-3)(80x10^-3)(53.33x10^-3)**(2)** = 3.41x10^-4 so , $$\tau $$= (437.5x10^3)(3.41x10^-4) / (5.00x10^-5)(80x10^-3) = 3.6x 10^7 Pa For Q at E i have labelled it with the orange part , for Q at centroid , i have labelled it with the green part .... Link to comment Share on other sites More sharing options...
studiot Posted November 6, 2016 Share Posted November 6, 2016 That's a poorly worded question and you have cut off the bottom of fig 1(b). So what is the dimension across the soffit of the trough? 1 Link to comment Share on other sites More sharing options...
johnson9883 Posted November 6, 2016 Author Share Posted November 6, 2016 The part 2 has a length of 80mm , so total length of base = 40 +80 + 80 =160 Link to comment Share on other sites More sharing options...
johnson9883 Posted November 8, 2016 Author Share Posted November 8, 2016 bump Link to comment Share on other sites More sharing options...
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