Question

Problem 2: (50 points) The beam has a rectangular cross section as shown below. If the intensity of the distributed load w=800N/m, Determine: a) The absolute maximum shear stress and its location. b) The maximum compressive stress and its location. 50 mm V | 150 mm 2m + 2m + 2m

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Answer 1

Page 1 Ho Sol (2) y = 800N Idolgomm Lº vì ly A k i I'm k am MAROON b = Somm SOF Diagram SOON - 16oon B-M Diagram -goon - 16oox-m - 160w-m * Consider the beam as overhanging beam with equal overtars of am from each support at Bfc. First we need to find the son Max shear force 4 Max Bending Momat to find absolute max. shear stress & Max. compressive stress. * To do so we xleed to find the Rxn at supports B&C → son Rod Rc Rang at supports B&C resp. • PB + Rc - 800x6 (verótical force balance) PB+ Rc - 4800) - (i) Now, for the equilibrium of the beame it should bhane 'O' B Mabout any point, so taking moment about it - 800x4x4 + RB X 2 + 800x2x2 = 0

Pago-a which gives RA - 9400 N/ Now from 6 we get Re=9400 N a). Absoute Max. shear stress. & Now do we have to calculations for sof at various locations : SF b/w Af B => -Wn SFA = -WXo = 0 4 convention is This sign taken tue. S.FR = - 800x9= -/boon Now, as at this concenhaled toad is acting cut 'B point, ech we will calculate other so there will be gst also also by consideorghe. Sif bh Bfc => - WH + RB S.Fe = -foox 9 +9 40o = 800N. SC = - 800 x 4 + 940 = - 800N i) S.F 6/0 c&D - - Wn+RB+ Re S.F at .C - 8000 4+24oot 9460= 16oon S.F = - Sooy 6 + 9400 + 9400 = ON s. Födiagam is drawn at start of que. to determine absolute shear stress > TA How

Page 3 here, F= max shear force from S.FD lielfoon T 1600 50x1508106 = 0.913x106 N/m 2 » 0.913 1984 but this is avg. shear stress (Tavg). We was know a Tmax = 3 tang for rectangular X-sle that too at the lehtral axis as → Ermax This diagrom shear stress variation along depth for reclay -ular beam. Tmax = 3 x 0.913 Tmax = 0.3195 MPa (b). Now we will find Born varsation along beam Bith bow A&B - W22 BOMA = 0 (as, n:o) Bote - soo xul.-1600 Nem Bernb/w Bhl – Wh2 + RBX (1-2) B.Me= - Soox y2 + gurox (4-0) = -16000-m Now, as ren sof is charging signal na b/w Bf C so, wel need to find that both at that point → Birst we need to find Hat point (E) Taking S.FD & Appelying similarity of triangles for

AMNE & APOE → page-4 I = sto gol sto l=lm N=3 for boint { → B. Mę = -800x3 +9400*(3-5) = -190011-m & Birn blw c&D - wn? + Rg(n-g) + Rc (2-4) 9 Bong = - 800x6² +9400x (6-8) + 94ool 6-4) a onim B. n.D is drawn at starting of que. Ssom Bormid we get Maximom Bending moment = 1600mm Now, taking Bending formula 1 - 1 . I = Moment of Ivestia of Beam about Nechal Axis - 1600 = 15x10-3 50x1503x10-12 T = 8.53 MPa

* Location of max. Comp stress Page 5 The maximum comb stress will be located at a those points where o Bonis Maximum liez 16oon-m son at u= amd um → Max Comp stress. & This compressive shels will act on lower fibre of beam. max. comp. y This fibre will exbesience shen.