1. The problem statement, all variables and given/known data member AD of the timber truss shown (attached) is framed into the 100 X 50 -mm bottom chord ABC as shown (attached - detailed view). determine the dimension 'a' that must be used if the average shearing stress parallel to the grain at the ends of the chord ABC is not to exceed 2.25 MPa 2. Relevant equations shearing stress, tau = V/A where V is shear force, A is cross sectional area --> average 3. The attempt at a solution is my free body diagram correct? if so, do i first solve for the forces in each member using the method of joints. do i consider that the chord in question is also the same at where the roller is located? once i have found all the forces in each member based on the free body diagram do i just use what i found to be the force at C as the value for V in the shearing stress formula? do i then just solve for A aka 'a'? please tell me if i am on the right track, if not, any assistance appreciated
Your FBD is not correct. There can be no horizontal external A_x force at the pin A, because there are no other external forces in that direction to balance it. The reaction loads at supports A and C are both vertical only. Determine what they are, and solve for the tensile force in the bottom chord. The shear plane is not the cross section area of the lower chord, it is the plane parallel to the grain of length 'a' and 50mm in width. Your figure is a bit confusing, I think the chord is 100mm deep and 50mm into the page (a 100 X 50mm timber).
when i solve for the tensile force in the bottom chord, is the best way to do this is: - treat the bottom chord as the sum of the two 2m pieces is the average shearing stress a component of an angle, there is none given or am i meant to deduce the angle from finding out the reactions etc?
when i solve for the tensile force in the bottom chord, is the best way to do this is: - treat the bottom chord as the sum of the two 2m pieces --> find reactions/tensile in each 2m portion and then sum them is the average shearing stress a component of an angle, there is none given or am i meant to deduce the angle from finding out the reactions etc?
No, whether they are 2 pieces or one piece does not matter. They do not add. The tension in AB is the same as the tension in BC. If you isolate joint B, you should see that in the x direction, the tension in AB must be equal to the tension in BC (AB tension points left, BC tension points right). But first, in order to arrive at that tension value, you should determine the vertical reactions and then isolate joint A and solve for the member forces at that joint using the method of joints. Have you determined the vertical reactions? Are you familiar with the method of joints? Note it is given that member AD has a slope of 1.5/2. This problem is a bit tricky, in that usually in a pure truss one deals with axial tension or compression forces only. In this case, you have a tensile force in AB due to the horizontal component of AD, that is pushing the member AB outward (to the left). That introduces a longitudinal shear in the wood timber over the length 'a' and 50mm width, trying to split and tear the wood apart in shear across that plane. To calculate the distance 'a', note that the shear stress is equal to the tensile force in AB divided by the shear plane area, that is parallel to the load, of (.050)(a).
after checking the solutions, the dimension for a should be 53.3mm, i am getting 80mm. after doing the method of joints here are my values for the reactions F_BA = F_BC = 9kN F_BD = 18kN F_DA = F_DC = F_CD = 12kN Cy = Ay = -7.94kN see attachment for corrected free body diagram and angles calculated from given dimensions ultimately solving for a using tau = V/A 2.25MPa = 9kN/0.05a --> a = 0.08m = 80mm
A few comments. Your calculation for theta is wrong, you used the sine function when you should have used the tangent. But beyond that, your member forces and reactions are incorrect in any case, except for F_BD. The vertical reactions at Cy and Ay must add to 18, per Newton 1, so I don't know how you got -7.94 instead of +9(upward load on the joint). Make that correction, then isolate joint A. The 9kN vert load must be balanced by a 9kN vert component downward in AD. The 9kN vert component in AD implies a horizontal component of 12 kN in AD {(2/1.5)(9) = 12}. So what must be the tension in AB? (And BTW, when I do the calculation for a, I get a value of 106.6mm, which is twice the book answer, so it appears that they are using a shear area of .1(a) instead of .5(a), maybe I'm misreading your sketch which is unclear as to what is the depth and what is the width of the bottom chord).