1- write the equation 2- name of hgures 3-graphs .. standard part b plus the graphs and standards

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1- write the equation 2- name of hgures 3-graphs .. standard part b plus the graphs and standards

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1- write the equation
2- name of hgures
3-graphs .. standard
1 Write The Equation 2 Name Of Hgures 3 Graphs Standard Part B Plus The Graphs And Standards 1
1 Write The Equation 2 Name Of Hgures 3 Graphs Standard Part B Plus The Graphs And Standards 1 (61.22 KiB) Viewed 10 times
1 Write The Equation 2 Name Of Hgures 3 Graphs Standard Part B Plus The Graphs And Standards 2
1 Write The Equation 2 Name Of Hgures 3 Graphs Standard Part B Plus The Graphs And Standards 2 (61.22 KiB) Viewed 10 times
part b plus the graphs and standards
(8-51) 1 S, 0 +0. which is equivalent to Eq. (8-28). EXAMPLE 8-5 H.W Solution Figure 8-21 shows a connection using cap screws. The joint is subjected to a fluctu- ating force whose maximum value is 5 kip per screw. The required data are: cap screw, 5/8 in 11 NC, SAE 5: hardened-steel washer, le = t in thick: steel cover plate, 11 = in, E, = 30 Mpsi; and cast-iron base, 12 = in. Es = 16 Mpsi. (a) Find kt. km. and C using the assumptions given in the caption of Fig. 8-21. (b) Find all factors of safety and explain what they mean. (a) For the symbols of Figs. 8-15 and 8-21. h=1+=0.6875 in. I = h+d/2 = 1 in, and D2 = 1.5d = 0.9375 in. The joint is composed of three frusta; the upper two frusta are steel and the lower one is cast iron. For the upper frustum: 1 = 1/2 = 0.5 in, D = 0.9375 in, and E = 30 Mpsi. Using these values in Eq. (8-20) gives k, = 46.46 Mibt/in. For the middle frustum: 1 = h -1/2 = 0.1875 in and D=0.9375 +21 - h) tan 30º = 1.298 in. With these and E. = 30 Mpsi, Eq. (8-20) gives k = 197.43 Mibffin. The lower frustum has D = 0.9375 in, 1 =1-h=0.3125 in, and Ed = 16 Mpsi. The same equation yields ks = 32.39 Mibt/in. Substituting these three stiffnesses into Eq. (8-18) gives ks = 17.40 Mlbf/in. The cap screw is short and threaded all the way. Using 1 = 1 in for the grip and A = 0.226 in? from Table 8-2, we find the stiffness to be ks = A, E/1 = 6.78 Mlbf/in. Thus the joint constant is ko 6.78 0.280 kotka 6.78 + 17.40 Answer CE Figure 8-21 Pressure cone frustum member model for a cap screw. For this model the significant sizes are Jh+17/2 In+d/2 D = dy + I tan a 1.50 +0.5771 D = d, 1.50 where I = effective prip: The solutions are for a = 30 and do 1.5d.
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