Question

Statically Indeterminate Propped Cantilevered Beam Reaction and Deflection Derivation 1. Determine the reactions R4, Rg, and M, and the elastic equation for the section of the beam between the wall and the load P. 2. Note: It will take 3 solutions to solve for the elastic equations for the entire beam: 0<x<d, d<x<s, and s SXSL 1. The derivation of the elastic equation for the section between the wall and the load (0 <x<d) is derived above. 2. The assignment for this lab is to: a. Derive the deflection curve for the section between the load and the support (d<x<s) and (s <x<L) using the variables d, L, P, s, and x and simplify it similar to that shown above. b. Solve for the reactions R, Ry, My, the angle, and the deflections V, and V, for both of the cases provided using an Excel spreadsheet. The spreadsheet must fit on one 8.5" x 11" page. c. You may derive the equations by hand (show all work) but place the final simplified elastic equations and the equation for angle, in Equation Writer and insert it into the Excel spread sheet that provides the solutions. You may use the Excel form provided as a template if you desire. d. There are 13 parts to the assignment, the reactions RA, Rg, Mw, the angle 0. V and V, for both cases (% each) and the derived simplified equations (28%). Hint: All of the V, deflections are between 3 and 5 mm e. You may submit your solution up to three times by 7/27, each will be graded and returned with each part being marked as either right or wrong; i.e. no partial credit. The highest grade will be applied to the assignment f. The beam dimensions are 20.09 mm wide by 3.01 mm high. The beam is 6061-T6 aluminum; use E = 68.9 GPa.

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Answer 1

PAGE) ANSWER -) GIVEN CONSIDERATIONS canklevend Beam Reaction statically intermediate Propped and deplecion derivation X2 L Free Body diagram & A d B р S 2 Mw RA RB By wing {fy PA + RB. =P -O T o 1 EMA 20 . Mw a Roxs pxd Emozo Mw & P (sad) 2 Raad CS

วนง + P (s-d) = Raxd RAUER RA - P-RB the equation in this equation By substituting we get a 2 and 3 Mwt RBS - Pd Mw + P (s-ot) = (P-RB) * Mw+Ps - Pd Pd Rod Mw tes apd & Rod so 4 Mot RBS Pd By substiterting u and 2 equations we get : Ps - RBS. Pd + Rod p (sad) RB (sd) = 0 - RB Р RA zo in equations By substituting RBIRA Mw tPs - Pd pd-s) Mw wer As (a-s) co Mw - -PCs-d) Therefore in Mw is clockwise ditelion. CS

ܝܘ PAGE 3 by considering sections clockwise 5 force 112 SECTON AC osaed v (3) M(x) = P (sad) SECTION danes V (*) -P 3 M(= P(sd) P(x-d) Gin SECTION BD 3 veu) - -P+P -o M(s) = P(S-d) P(ted) + P(4-5) the equations ie the 1,2,3,415, 6 ore elastic equations of the system ie for osaed d sxes seas.l. respectively. please upvote its necessary thanking you at CSC