Shear Force and Bending Moment Diagram of Beam Span
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Added on 2023/06/14
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This article explains how to calculate shear force and bending moment diagram of beam span with solved examples. It covers the concept of uniformly distributed load and point loads. It also discusses the slenderness ratio of I section beam.
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Task: 1 Shear force and bending moment diagram of beam span. ∑Fx=0 HB=0 ∑MA=0 Taking anti close wise moment at positive and clockwise as negative. RA+RB=170 P1x0.5−40x3x1.5+RBx3−P2x3.5=0 7.5−180+RBx3−122.5=¿ RB=98.33 RA=170−98.33=71.67 Shear force: Force, Q(x1=0.5)=−P1 Q1(x1=0)=−15kN Q1(x1=0.5)=−15kN Bending moment at CA span, M(x1)=−Loadxperpendiculardistance M(xc)=−15x0=0 M(xA)=−15x0.5=−7.5 Shear force and bending moment at AB span. Q(x2)=−P1+RA−q1x(x2−0.5) ¿−15+71.67−40x(0.5−0.5) Q(x2=0.5)=56.67kN Q1(x1=3.5)=−15+71.67−40x(3.5−0.5) Q1(x1=3.5)=−63.33kN Bending moment at mid-span: 1
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M(x2)=−P1xx2+RAx(x2−0.5)−q1 (x2−0.5)2 2 M(x2=0.5)=−15x0.5+71.67x(0.5−0.5)−40x(0.5−0.5)2 2=−7.5kNm M(x2=0.5)=−15x3.5+71.67x(3.5−0.5)−40x(3.5−0.5)2 2=−17.5kNm Extreme point, M=Wl2 8=40x3x3 8=45kNm For BD span, Shear force and bending moment. Shear force, Q(x3)=−P1+RA−q1x(3.5−0.5)+98.33 Q(x3=3.5)=−15+71.67−40x(3.5−0.5)+98.33=35 Q(x3=4)=−15+71.67−40x(4−0.5)+98.33=35 Bending moment, M(x3)=−P1+RA−q1x(3.5−0.5)x[(x3−3.5)+(3.5−0.5) 2]+RBx(x3−3.5) M(3.5)=−15+3.5−71.67x(3.5−0.5)−40x3x(1.5)+98.33x(3.5−3.5)=−17.50kNm M(4)=−15+4−71.67x(4−0.5)−40x3x(2)+98.33x(4−3.5)=0 2
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Task: 2 As per the given condition, the serial size 203 x 203 x 52 which have specific dimension and depth of section 206.2mm,width 204.3 mm, flange thickness 7.9mm and web thickness 12.5. A uniformly distributed load is applied over the span which additional two point loads. For the point loads the bending stress, SF=25kN And the bending moment, BMxx=225x2−25x2=0 BMyy=12.5xX−25X+50 FindingdM dx, dM dx=12.5−25=−12.5 Bending moment for shear force for the uniform distributed load, BMxx=−WxX2 2 Apply boundary condition and loading condition, x = 0 and x = 4m, BMx=4=−Wx42 2=−40x16 2=−320kNm2 Now,dm dx=−Wx=−40x4=−160 Bending moment at uniformly distributed load, BMx=WL 2x−Wx2 2=40Lx 2−40x2 2=20(4x−x2) dBMxx dx=20(4−2x) 20(4−2x)=0 4
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x=2m The point of contra-flexure obtained by, −63.33−15(x+0.5)−35xXxX 2 Differential above equation by x, dm dx=15x−7.5−17.5x2 15−35x=0 x=2.33m As per shown in diagram, Y=A1Y1+A2Y2+A3Y3 A1+A2+A3 ¿ 204.3x12.5x(106.8+12.5+12.5 2)+(106.8)(7.9)(160.8 2+12.5)+(204.3x12.5x6.25) 2x204.3x12.5+160.8x12.5 Y=83.24mm I=INA+Ah2 Iy1=2x[1 12(204.3)(12.5)3+(204.3x12.5)(86.65)3 ]+1 12(160.8)(7.9)3=¿ ¿3322942056.0909375+6606.7226 ¿3322948662.8135375 Iy1=3.322x109mm4 M I=σ Y σ=MY I=45x83.24x106 3.322x109=1127N/mm2 5
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Task: 3 Slenderness ratiok=l r=185.8 5.18=35.86 Value of slenderness is less than 200 which is safe for the design point of view. If the slenderness value greater than 200 so that there are higher chance to get failed. This I section beam is short because of safe operation with respect to y-directional length. 7