Question

For an extruded beam having the cross section shown, determine: (a) the location of the shear center o (b) the distribution of the shearing stresses caused by a vertical 2.5-kN force V applied at 0. 80-

Answer 1

3 Taks. I AB I au = t x 80 x 67 80x6x552 - 1.453448168 mm Toebe *8*1632 -2.736 2m vom Parts BD and th: ABD = lah= √60²+252 = 65mm sind = 25 , losa = 60/5. ABD = AFG = 658 - 52omma São - Jorna 55180 - 67.5mm IBO - Isha 1 / 2 & 128 3 27.083x10 mm LBD - Ifu= 520x6.5 + 27.083 X10 - 2.39633110 mm 6 I = {I = 10.43024 10 mm By Symmetry, fs = F, , fu = F2 A (s) = LAB(s) a(s) = SARKAB (1) 9 (3) ¥ Q 13) . SABE ABS fase q (3) ds- & TAB tas tasas V HAR TABIAB- I. LAB I 2 I. Six 6x802 2x10.43x166 = 0.10/244V hty

st point B: ds = AB KAB= do = 55X67 5o = 26.4x10 mm Calculation of tz and fy: in pant BO 0(3) = toos 513) - You & I J Sina. Q(s) = OB + A(3) Ý (5) - dattoos AB tfs sind) - dar tuis TAB 3 + ₂ s² sina. q(8) - V0 (3) - (Og+tan TAB 3 th Bosh sind) Fr = 1 - 7 / 2 (as "Byt 3 to y es lo 7 toplo sina) - 10.4 302466 ( 26.4.8 x 65 + 12 48 x 55 x 65 + £ x8x65 (42) = 0.267144 V ta , Fu = 0.267144 v Ja At point D oo 26.4x10 3 + 320x67.5= 61.5x1033 & 3 need not be calculated since its moment about E is zero S&Me - JEME by statics, Ve - 556, + sofa cosa + sofy cosa +55fs - llof, + 16ot2 =(110x 0. 101244 + 16 +0. 267141x600) - 50.6V. Dividing by v e = 50.6 mm

At point & QE = dat ADE S DE = 61.5x10 3 + 8x gonno = 87.1x103 mm3. strearstoess: -64 V= 2.58 10 N, 1 =10.4302810 m . At At points A and H B and G Z=0. = 26.4x10 ms. o qaro - 2.54103 x 26.48158 - = 6.3278 AB 10.4302x10-6 9. 3 6. 327X10 on the right : t = 6x103 m, - q- t 6x10-3 =losux loopa. a losu mpa on the left , t= 8x10²m. 6.3278x102 = 0.791x10 Pa = 0.791 mpa 872103 3 m At points and F: 10 - bl.sX16 m t = 8 . - vo - 2.5x105x6lisnios 10.4302118-6 I t = 1.843 xoopa = 1.843 mpa At point & vo Z - 2.5x103 x 83. 1810-6 10.43022156 x 8x10 3 Ir 2.6180 Pa 2.61 mpa