Question

A cantilever beam supports the loads shown. The cross-sectional dimensions of the shape are also shown. Assume mm, by - 85 mm, 5 mm, 9 mm. Determine - 0.5 m, P. - 4.0 kN, Pg - 7.5 kN, Pe-2.0 kN, -85 (a) the maximum vertical shear stress. (b) the maximum compression bending stress. (c) the maximum tension bending stress. See the coordinate system for the beam in the problem figure with the origin of the x axis at the feed support. Consider four points along the beam's axis Point A ato Point Batx 0.5 m Point Cat x = 1.0 m Point Datx=1.5 m break the beam into three segments: AB, C, and CD. Enter the shear force in each segment with its correction based on the coordinate system in the problem figure and the sign convention for shear forces in Chapter 7.

Answer 1

7-5KN 4KN D 0.Sm 0.Sm 0 Sm Shea FoTu (F) at dittesend Poirt S FA 4KN F6-4+7S 3.5fn c 4+15+2 =5.5 kn Ves tical Shea foru 5.5 kN The maximum Bemdin moments (m) at dibbedent POínt S MA O MB = yX0 S 2KN.m M c 4x1-7.5X05 0.25 KN.m 4XS 7.SX1-aXoS-2.5 KNm Mo at Pofnt D momemt maximum bmding Mmax 5 KNm

85mm SmM A 8Smm 80 Given Sectiom is SymmttTY about anis So lentod Jies om a11ta2 = a4+ar ai80x9 70 mm =40m m a285 x5 425mm 3 80 t82s o (70x40) (425 K 82:5) 7o t125) - 55.76 mm moment 06 Imes tia of Secti on about N.A IN,A INAI + J NA 2 -bidt b2d + +A2 hi 12 12 bi ndi80mM a1 1a0mmh - 1518 m b 85mm d Smm a1 415mh 2-9 36-2 mm

9x803 +(12oy 1518) 85(195 2 12 =0.563 x lo t 0-304 Xlo 0-86X0 Simpl bemdimg equation R tmax y-85-9 A X C max NA bucd camt Juv beam the dayeas below the Im the dayers above A SUbiecttd compessive stess omd Stess to A) maximum Vestcal Shea stess max. shea foou Vmax ara of kction 55xlo3 N 4.8 Nmm (70tY25)

maximum Compressive bendfng Stess is at bottom must Jayer ob the Section = 5578mm J= distank blw N.A omd bottom Jayes Mmax ymax Cmax as1 (sS18 Mm 35xNmm X SS.78 0.867 160.84 1mm CMax maxmu tensiom bendfmg StsesS is at tor must Jayer ot the SecHom 3= (85-5518) im 25xlo x (8s-55-18) tmax 0.867x lo ৪५.1१४ |mn otm