"SOLID MECHANICS1. IntroductionThe main objective of this project is to analyze the ‘real-life' structure/element by usingSolid Mechanics. The beam can be classified into many types. Among them the suitable beamwill be identified and it should support a significant load. For the implementation process, SSB(Simply Support Beam) will be chosen. The snapshot of the beam element will be taken and thenthe verified the dimensions of the beam element. The external loads of the chosen beam will beestimated and then the conditions which are supported by the beam will be analyzed. The SFD(Shear Force Diagram) and the BMD (Bending Moment Diagram) will be drawn for the entirebeam. The elastic bending stress distribution over the critical section will be calculated. Then theShear stress distribution over the critical section will be calculated. These calculations are basedon the constructed SFD and BMD diagrams. The maximum deflection will be calculated for thechosen beam by using both Macaulay method and the Elastic curve method.In the above figure shows that the uniformly distributed with load 26.847Kn/M. And thedenotedwith the W. We can see that the simply supported beam span is 13m.and which is divided asthree part. First part span is 1m and second part span is 11m.last and the third part of the simplysupported beam span is again 1m. In this SSB is carried a continuous beam. R , R R andR areA B, C A the supports reaction at A, B, C, and Drespectively. 1 SolutionLet M is fixed BM at AA Let M is fixed BM at BB Let M is fixed BM at CC Let M is fixed BM at DD Span BC2 Span CD 3 Now the B.M diagram due to vertical loads taking each span as simply supported beams asshown above figure.Bending Moment diagram due to support momentsLet’s M M M and M are the supports moments at A, B, C and D respectively. But the endA, B, C, D support of a simply supported beam are not subjected to any beam bending. Hence the supportmoments at A and D will be zero.M M 0A= D= To find thesupport moments at B and C, clapeyron’s equation of the three moments is apply forABC and BCDSpan for AB and BC from equation of three moments, we haveM (L ) + 2 M (L ) +L ) + M (L ) = [6 a x / L ]+ [6 a x / L ]A 1 B 1 2 A 2 1 1 1 2 2 1 Deflection of span ABArea of the Span AB2 M = WL /8A = 2/3* M *XAB 1 AB 1 2 = 26.847*1/8 A = 2.237 mm1 M = 3.356 KN/mAB Span of BCArea of the Span AB2 2 M = WL /8A = 2/3* M *XBC 2 BC 2 2 2 = 26.847*11 /8 = 2/3*406.06*112 M = 406.06 KN/mA = 2977.33mmBC 2 4 Span of CDArea of the Span CD2 2 M = WL /8A = 2/3* M *XCD 3 CD 3 2 2 = 26.847*1 /8 = 2/3*3.356*12 M = 3.356 KN/mA = 2.237 mmCD 3 X =L /2=1/2= 0.51 1 X =L /2=11/2=5.52 2 X =L /2=1/2= 0.53 3 M = 0 Because of SSB not subjected to any beam bendingA2 M (1+1) + M (1) = [6*2.237*0.5/1] + [6*2977.33*5.5/11]B C24M + M = 8931.99 + 6.711…………. (1)B C For span BC and CD from equation of three moments, we haveM (L ) + 2 M (L ) +L ) + M (L ) = [6 a x / L ]+ [6 a x / L ] (* M = 0)B 2 C 2 3 D 3 2 2 2 3 3 3 D M * 11 + 2 M (11+1) + 0 = [6*2977.33*5.5/11] + [6*2.237*0.5/1]B C 11M 24 M = 8938.701…………… (2)B + CSolving the equation (1) and (2), we getM =363.84KN-mBM value subjected to equation (1) or (2)B 11(363.84) + 24M = 8938.01c24M =4935.77cM =205.65 KN-mc5 Support Reactions Let’s R R R andR are the support reactions at A, B, C and D respectively.A, B, C DFor the span AB, Taking moment about B, we getM = R *L – W*L *L /2 (*M = -363.84)B A 1 1 1 B -363.84= R * 1 -26.876*1*1/2A R = -350.402 KNAFor the span CD. Taking moment about C, we getM = R *L – W*L *L /2 (*M = -205.65)C D 3 3 3 c -205.65= R * 1 -26.876*1*1/2DR = -192.212 KNDFor the span BC. Taking moment about B, we getM = R *L – W*L *L /2 (*M = -363.84)B C 2 2 2 B -363.84= R * 11 -26.876*11*5.5/2C R = 40.83 KNCTotal load of ABCD= R + R +R + R A B C D26.876= -350.402-192.212+40.83+R BR = -528.66 KNB 6 26.876 KN/mFBDRD R RA C RB V V2 3 V1 SFDV4 X 1 X2 X3 00BMD11 m 1 m1 m7 Bending stress of the beam3 4 3 yEI = [WL X / 12] - [WX / 24] - [WL X / 12]1 1 2 3 3 3 4= [26.876*1*0.5 / 12] – [26.876*6 /24] – [26.876*1*0.5/12] = - 1452.14Maximum deflection of Beam calculation using Macaulay method2 dmax= 5M L / 48EI maxM - MaximumBending moment N-Mmax L- Length of the beam M2 E-Modulus of Elasticity N/M2 I=Area moment of inertia MM2 M = WL /8max 2=26.876*13 / 8 =567.75 KN-mShear force3 3 EI d y/ dx = WL/2 – W*X= 26.876*13/2-26.876*6 =188.138 KN8 ConclusionThere are different types of beams available in Solid Mechanics concept. These types arestudied. To analyze the ‘real-life’ structure/element of this project, Solid Mechanics is applied.The suitable beam is chosen for this project which supports a significant load. The name of thechosen beam is ‘SSB (Simply Support Beam)’ UDL over the full length of the span. Then thedimensions of the beam are identified and the picture of the drawn beam is taken. The externalloading capacity of the SSM beam is analyzed and it is estimated according to the dimensions ofthe beam. The conditions which are supported by the beam are analyzed. For the whole beam,the BMD (Bending Moment Diagram) and SFD (Shear Force Diagram) are drawn. Thecalculations for elastic bending stress distribution and Shear stress distribution over the criticalsection are done based on the SFD and BMD diagrams which are constructed at earlier. By usingtwo methods such as Macaulay and Elastic curve method, the maximum deflection is calculated.9 "