{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW3 - 1.39 Member ABC which is supported by a pin and...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1.39 Member ABC, which is supported by a pin and bracket at C and a cable BD, was designed to support the 4-kip load P as shown. Knowing that the ultimate load for cable ED is 25 kips, determine the factor of safety with respect to cable failure. PROBLEM 1.39 SOLUTION P 1‘ Use member ABC as a. A T ¥v~ee book, and “0+3 1 15m. chul' m-em ber 80 is a +wc> — ‘Force. mem lo if L______’ 12in.—-l ZMC = O 18 in. <9 M, Mtaom + LP swowss») — we» snow -03» SM 3o‘)(,|2 tn) = o _ 32.0.3 _ 32.c23)(t+) _ . F» - m? - W - 5°87" kw Fu‘l’or ol' sa‘Fe‘l‘y 'Fbl‘ cwlale 30 PS. = TEES-it: = 3:82:75— : 3.G9 ‘ 1.40 Knowing that the ultimate load for cable ED is 25 kips and that a factor of safety of 3.2 with respect to cable failure is required, determine the magnitude of the largest force P which can be safely applied as shown to member ABC PROBLEM 1.40 SOLUTION Us: member ABC as a. Free. boar anA no‘l'? 'th‘l‘ member 8D is G. ﬁnale-Force. men he». ZMcto (Peas 90°)(3O l“) +(P5M 40°)(15;..) 7', (Fan Cos 30°)(l5in3 " (Fm, Sin 30°)(l2 [va : o _ \8.QQo _ P — ”37213? FED _ 0.58216 F” .' . ,, i‘, F _ _E‘)£’- - as. ‘ kn Ap/owabie penal ‘POI‘ member 80 IS 31, -- F5 — ET_ 7.8)25 ups Nﬂouab/e local P = (0.582IC )(7.3l25') =' 4.55 kips PROBLEM 1.41 1.41 Members AB and AC of the truss shown consist of bars of square cross section made of the same alloy. It is known that a 20- mm- square bar of the same alloy was tested to failure and that an ultimate load of 120 kN was recorded. If bar AB has a 15- mm— square cross section, determine (a) the factor of safety for bar AB, (b) the dimensions of the cross section of bar AC if it is to have the same factor of safety as bar AB. 0 .4m . B—L SOLUTION Lena-Hi a? member AB 9M3 ’-' _/B752+ 62f: = 0-85 m use en+fre +V‘uss as a ‘Free Boaiy A. TAAY DEMcr-o m A“ - (0.7:)(283 = 0 Ar 15 W \$2520 A,—22=o A1=22uN 3 Use Joia‘l‘ A as "Free boaly 2‘ lcN Ax A, LZFﬁo ﬂag—Ape c c FAB AB :(ocﬁ'gls) = ,7 kN F“ HEP-71:0 Ara-35:3; For H‘e +es+ loam A r (0.02.0)a T QOOYID" m" Pa = WONGa N - - P - Mow/03 __ s For He mdenaﬂ 6” - A‘u - W5 — 300x") PG. (a) Foo- bar A8 E3: f; - GLA = soowo‘ (0.015),.I ﬁe ' Fm a7x/o = 3.77 .. (b3 For bar AC ES. : E"— : M ._. 6230.1 5" FA: ﬁt. a}: \$5255 ._. (3.97)§20xlo‘) = 24:47 “0-; M: 6?; aooxlo‘ (1.: 16.2%1/0'3 m ...
View Full Document

{[ snackBarMessage ]}