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15) (10 Points) Consider a horizontal beam of length L. with uniform cross-section and made out of uniform material. It is resting on the x-axis, with one end at the origin. It is acted upon by a vertical force it's own weight in this simple version). The deflection of the beam at any point x,for 0 <=<L.is given by Ely) = w, where E, I, ware constants. E is the Young's modulus of elasticity of the material, and / is the beam's moment of inertia about the horizontal axis, and wis the force per unit length on the beam from it's own weight. a) Find the general solution in terms of Elw, b) Using the general solution you just found, solve the Boundary Value Problem (0) - '(0) - (L) - (L.) -0. These condition imply that the ends of the beam are securely fixed c) Determine where the beam experiences the maximum (in absolute value) defection, and find the value of this deflection d) Suppose the beam is a cylinder 2 meters long with a radius of 3 cm. Look up Young's Modulus for aluminum in m the moment of inertia of a solid cylinder, and the density of aluminum (you will need to use the density to find the mass and then w, which is the weight per unit length). Write the equation of the deflection (2) for 0<=< 2 for this piece of aluminum. e) Use the information above to draw a detailed graph of (+) for 0 << 2.

Answer 1

solotion ( 0) (0) () dx =wx3 da = wx ay +603 a, we have dux) EIX dx ETY duc = woxtc, da) El drux) = with GXt2 El dux) tot (2X+(3 1) EIX U(X) = +(₂1 tc (2) b, Apply Boundary conditions at 1-0, due to and at IL, da o from =no; 20+0+0+ (3 (3=0 And o- WL with t Ch 2 +G. 20 – (3) Also, at a=0 ; u(x)=0 and at x=L; u(x220; O=Cu Cu =0 + G2² E+CL And +GL3 GL² + toto

뺨당 1월 0 from =n @ ando 4 + L W - - 뽑 명왕왔 뽐 -8 모니 GD-OL 빵는 -WL² +COL 6 tOL3 6 - ge 1. Erulex) = Log - Golt tolian CDum) - tor _ LOL3 Woo 0 쁨 받기적왕자이 =0&가스 발부 x = " I JulgItegy 3 발 5 유로

3 - We tal 2= 4WL QLOL 689 x=L & Į 1. ulo at 2:0, 14, L since at =0 and at, 461)=0; .. morrimum deflection will occur at u .. umu) - 139.604 63 +Coc" 13" GL u(max) = 5WLY ΩΣΕΤ d for alumineum, E= 7o upa INA II (6)4 = 63.617 cmH = 63-6178 108 mH I. EI = 70 x 109 x 6 3.6174108 44453190 m Also , unit-weight of aluminium = a110 kglmz we 2710x9 8 X TT X (3X10-2) W = 75.091 N/m

E :. Also ,L=am for Aluminium Bow; u(x) = 1 [ 15. 1 4 75 12 4 3 76.1 x 2xx 445311 UN ) )41 X10-4 4-5.6240-4434 1.12X10-302 e, at noo ,u=0; atazam, uzo at naimi ucman) = 7.7980 Um = 0 779mm im 10.779mm
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