Sol HW1 - flag/u] 1:41: we, 6/14 / 1.1 T Id 1' ‘ Problem...

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

View Full Document Right Arrow Icon
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: flag/u] 1:41: we, 6/14 / 1.1 T Id 1' ‘ Problem 1 ‘1 wo so 1 cy 1ndncal nods AB and BC are welded together at B and loaded as shown. Knowing that d} = 50 mm and d2 : 30 mm, find average normal stress at the ‘ mldsection of (a) rod AB, (b) rod BC. (OJ Real AB P.— |+o+30=7o M} :70xlOSN _ L 3 "~_ " ‘3 1 A = w’df: fi’Cs‘o) —— {.9655X10W" - “$3940 ““ 3 61a "A: = “ 35-7”); 7)“ 6A6 T 3527 MPA ‘ (m Rad! BC P = 30 w = 30xlo3N A : {59:2 r 70;.86 mmL-L- 70G-gé’d0—c MK = 30:13.1. 2 42,4;40‘ Pa 709.3010" 65‘ :- d e 2% Problem 1 .1 2 1.12 The rigid bar EFG is supported by the truss system shown. Detexmine the cross-sectional area of member AE for which the normal stress in the member is 15 ksi. 3ft h Usrn‘i Porln‘ov‘ EFGCB 0-5 fir?“ L"JAY ' +T7Fj=o: gHE—géoo :0 i— 4 360011) FA? 2 6000 fl” = Q'00 W5 4ft 4ft—>|<-4ft Sirees Em Inae-nlned“ Al; (37.; = ,5 k5“ He 6A6 ’ A” Fm: $.00 . 1 AM = 6’” .— [S : voo w. 4 Probll)em 1_31 1,31 The 1.4 kip load P is supported by two wooden members of uniform cross 50 in section that are joined by the simple glued scarf splice shown. Determine the normal 8/ and shearing stresses in the glued Splice. } " P: 14002:: errro°—éo‘ = 30“ A; (5.0mm T 15 cn‘ g : 125:2 : A; L5 5: 70..0F=‘L‘ ‘41 ,Z. : P3». 28 : (Hools-‘n 60° 1; ’Z’: 5/99. 49:23! 2A.; (23053 / ’ Problem 1.9 1.9 Each of the four vertical links has an 8 X 36— mm uniform rectangular cross section and each of the four pins has a 16— mm diameter, Determine the maximum value of the average normal stress in the links connecting (a) points B and D, (17) points C and E. Use. bar G5 a. "Free heel)“ 20 in 0.025 Ala—0.040 ———~i FM“ 8 C ., FED RE 2Mc = o Lamar-1n — (0,025+0.ovo)(20x1033 = 0 Fee: = 32.52“)! N Link 8D is m +W3'W‘ 2MB=O -(0.0'-lo)F-m —(o.025)(20x/03):o F“,- " ‘ lQuSI XIO‘ N Link CE: :5 l4 prrez‘smw Ne‘l‘ area o‘lz One [pink 'POV‘ 'l‘evASloM = (0.008)(b.036-0.0|€) ' I60 X10.G m2. For +0110 Paulie? ,l’rnks And: 32,0 >410” w. 2. Tensile s-lwess in link BD 3 :8 La) 6‘,» = {3}?” = = Ichsen/o or 101.6 MP4 4 «if ‘ Area. 43¢» one ,ptmk in acupressth = (0.008)(0-036) = 223 x Jo" m‘, Fw +9» Parafl’efl in!“ A = 57g ~10? mi _ - 3 _ (b) 665 = 12-5 = 3%%§r= -2:.7owo‘ or -2.I.7MPa. 4 ...
View Full Document

This note was uploaded on 02/20/2009 for the course CE MAE 373 taught by Professor N/a during the Spring '09 term at CSU Long Beach.

Page1 / 2

Sol HW1 - flag/u] 1:41: we, 6/14 / 1.1 T Id 1' ‘ Problem...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online