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Mechanics of Structure Assignment

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Added on  2020-04-07

Mechanics of Structure Assignment

   Added on 2020-04-07

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Mechanics of Structure 1MECHANICS OF STRUCTURENameCourseProfessorUniversityCity/stateDate
Mechanics of Structure Assignment_1
Mechanics of Structure 2Beam deflection 01Linearly distributed load, w0 = 3 kN/mE = 270 GPa, b = 200 mm and h = 400 mmMoment of inertia, I = bh³12=200x400³12 = 1.0667 x 109 mm4[ CITATION Thend \l 1033 ]EI = 270 GPa x 1.0667 x 109 mm4 = 270 x 103 N/mm2 x 1.0667 x 109 mm4 = 2.88 x 1014 Nmm2Taking moments at B to the left hand side:∑MB=0 → -RAL + (woL2¿(L3) = 0RAL = woL²6 → RA = woL6; and RB = w0L3Taking moment at a point x from support A to the left hand side:∑Mx=0(wox²2L¿(x3)−woLx6+M=0 → M = woLx6−wox³6LBut M = EId²ydx² CITATION Sel14 \l 1033 (Selvam & Bindhu, 2014)
Mechanics of Structure Assignment_2
Mechanics of Structure 3ThereforeEId²ydx²=woLx6−wox³6L ..................................................................... (1)Integrating equation 1 with respect to x givesEIdydx=woLx²12−wox424L+C1 ......................................................... (2)Integrating equation 2 with respect to x givesEIy=woLx³36−wox5120L+C1x+C2 .............................................. (3)When x = 0, y = 0; substituting this in equation 3 gives:0 = 0 – 0 + 0 + C2 → C2 = 0When x = L, y = 0; substituting this in equation 3 gives:0=woL⁴36−woL4120+C1L+0 → C1L=woL4120−woL⁴36; C1 = −7woL³360Substituting the value of C1 in equation 3 and making y the subject gives:EIy=woLx³36−wox5120L−7woL³x360y=woEI[Lx336−x5120L−7L3x360] ....................................................... (4)Equation 4 is the equation of the elastic curveSlope of the curve is determined by differentiating equation 4 with respect to x, which gives:
Mechanics of Structure Assignment_3
Mechanics of Structure 4dydx=woEI[Lx212−x424L−7L3360] ...................................................... (5) At maximum deflection, slope, dydx=0 (substituting this in equation 5)0 = 30L2x2 – 15x4 – 7L4 (where x is the point at maximum deflection)15x4 – 30L2x2 + 7L4 = 0 Solving this equation gives x2 = 0.2697L2 → x = 0.5193LThus maximum deflection is:y=woEI[L(0.5193L)³36−(0.5193L)⁵120L−7L3(0.5193L)360]y=woEI[0.14L⁴36−0.0378L⁴120−3.6352L⁴360]y=woEI[1.4L4−0.1134L4−3.6352L⁴360]y=woEI[−2.3486L4360]y=woEI[−0.006524L⁴]y=0.006524woL⁴EI (ignore the negative sign) ............................................ (6)Equation 6 is the maximum deflection of the beamSubstituting the values of E, I, w0 and L in equation 6EI = 2.88 x 1014 Nmm2; w0 = 3 kN/m = 3 N/mm; L = L mm
Mechanics of Structure Assignment_4

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