. Design of a portable pneumatic riveter. ?FT - 7 v//////////////-a S£CT/OW/\A Fig. 11. -27- TABLE VI. Mom. c No Area Mom. Ct -Cc leg h Ah-^ I Tc C 1 1 1 9 : 2 » 5 • 5 6 .65 • 1 • 1 6 Substituting the values given in Table VI in (14) we have sc = St = 2q6|^ = 2360 sq. in. Ss = 21620 = 2250 sq. in. These values are low but a wide section must be used onaccount of clearances. The toggle lever arms will be considered as cantileverbeams uniformly loaded. Hence substituting the known value
. Design of a portable pneumatic riveter. ?FT - 7 v//////////////-a S£CT/OW/\A Fig. 11. -27- TABLE VI. Mom. c No Area Mom. Ct -Cc leg h Ah-^ I Tc C 1 1 1 9 : 2 » 5 • 5 6 .65 • 1 • 1 6 Substituting the values given in Table VI in (14) we have sc = St = 2q6|^ = 2360 sq. in. Ss = 21620 = 2250 sq. in. These values are low but a wide section must be used onaccount of clearances. The toggle lever arms will be considered as cantileverbeams uniformly loaded. Hence substituting the known values ofP and 1 in (15) we have for the maximum bending moment at thepoint of support M = 34700 x = 13000 lb. in. Substituting the known values of M, and d in (14) we have o _ 32 x 13000 _ / , S ~ - 4920 sq. in. which is a safe value to use. TOGGLE BOLT. The toggle bolt will be considered as a beam fixed a: bothends and carrying two concentrated loads as shown in Fig. 12. -28- r. Fig. 12. The formulae for a beam fixed at both ends and carrying oneconcentrated load are Mi = ?*»[*- T + fs](17) *s = -^8f-?]* * (18)Ma = ? Pa2|^ . ia + (ig) where = the bending moment at the left support. Mg = the bending moment at the rightsupport. Ma = the bending mo*rent under the = the load. 1 = the distance between supports,a = the distance of the load from theleft Bupport. Substituting the known values of P,a, and 1 in (17), (18),and (19) we have Mx = - 15150 lb. in. -29- Mg = - 2490 lb. in. M = + 4460 lb. These values give the bending moments due to one load butsince the beam carries two symmetrical loads it is evident that,due to Pg, there will be a bending moment Mg equal to at theright support another bending moment equal to Mg, at the left support. The maximum bending moment will evidently occurat one of the supports and is equal to the algebraic sum of thetwo moments at the support, or, Bince the loads are symmetrical M = M1 + Mg = - 15150 - 2490= - 1764
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Keywords: ., bookcentury1900, bookdecade1910, booksubjecttheses, bookyear1912
