{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

568_Mechanics Homework Mechanics of Materials Solution

568_Mechanics Homework Mechanics of Materials Solution -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: LH—i' 5.85 and 5.86 Knowing that the allowable normal stress for the steel used is 160 PROBLEM 5-35 MIPa,.selocl the most economical metric S—shapc beam to support the loading shown. 75 kN SOLUTION 40 thu +52Mo=o ~3.GA +(2.1Y1sh(o.fi)(|.s¥uo3 = o A = 7125 mm 021%,: o 3.6 D ~(o.q)6753-C2.73(I-8)C4°) = O D = 72. 75 kt: Shear? A +98 V = 74.25 klU B+DC V='H.2§-75=-o.7sku v, -. - ans-(1.9040) .— -72.75 M} Areas unoler SLCQJN oln‘aal‘au A in 13 SW: - (OAKWLQS) .~ 66.825 Wm [3 +0 c 5va = (o.qu(.o.7s)= .—_ om: kN-m C l» 1) (flag Long-42.75) = .. sews w. m Bending Momen+5= MA =0 A 3 C D M3: o+ 66.325 = £6325ka ,L: se.325— 0.675" = €6.15 kN-m Mo" QQJS— 66.!5‘ = o Mm3num IMI = 66.825 W-M e €6.225'xto‘ Hm 6.1? \80 ”pa. = IEOVIO‘ 13¢ S ‘ lMI 66.815xlo’ ---— r ——-—----= n|7.7x:o“m’ = 41172103 m‘ 6'41 £60 x10 ‘ S (Io’m?) S 380‘ G“ qjl Lljk+es+ S"5€€+I‘DU\ ‘0 l“.3 ‘73 2 2350152 is: 8 310.413 9 47.3 lea/u 4 .-[.-, ...
View Full Document

{[ snackBarMessage ]}