Question

Problem 12.47 3o7 The shaft is made of steel and has a diameter of d. 17 mm. The bearings at A and Bexart only vertical reactions on the shaft E -200 GP Figure 1) Part A Determine the maximum deflection of the shaft Express your answer to three significant figures and include the appropriate units. Enter positive value if the deflection is upward and negative value of the deflection is downward. tế HÀ • ? • • Ga Units - Value Figure Submit Request Answer Provide Feedback

Answer 1

Draw the force diagram. AR N Take the net moment about point B. R,x(700)-(250)x(500)-(50) (200)=0 (250)(500) +(50) (200) (700) = 192.86 N Write the balanced force equation. R+Mg = 250+50 192.86+R3 = 250+50 ► Rg =107.14 N Write the general deflection equation.

=-[R_x-(250)(x-0.2)-(50)(x-0.5)] =-Rxx+(250)(x-0.2)+(50)(x-0.5) =-192.86x+(250)(x-0.2)+(50)(x-0.5) dx Integrate the equation. Erd -96.43x +125(x-02 |-25(x-0,5) +C, Integrate the equation. Ely = -32.14x" |+41.67(x-0.25 +8.33(x-0.5)+Cx+C; At x=0 =y=0. EI (0) = -32.14(0)' +G(0)+C2 > C =0 At x=0.7m =y=0. EI 0 =-32.14(0.7)|+41.67(0.7–0.2)* 18.33(0.7–0.5)° +G(0.7)+0 0 =-32.14(0.7)* + 41.67(0.7–0.2)* +8.33(0.7 -0.5)* +G(0.7) _32.14 (0.7) ° - 41.670.7 -0.2) -8.33(0.7 -0.5) (0.7) = 8.2

Write the deflection equation. Ely = -32.14x +41.67(x-0.2) +8.33(x-0.5)° +8.2x+0 = -32.14x° +41.67(x-0.2) +8.33(x-0.5)' +8.2x Calculate the maximum deflection of the shaft (at x = 0.2 m). Ely = -32.14(0.2) +41.67(0.2-0.2)' +8.33(0.2-0.5)' +8.2 (0.2) = 1.15797 (1.15797) y=- (200x10") * -(0.017)") = 0.00141222 m 1.4122 mm