Question

15 kN 2 kN/m B A Roller support 3 m 3 m A beam is fixed at A and supported by a roller at B. The beam has E = 100 GPa and I = 1.0 x 10-5 m4. For the loading shown, complete the following: 1. Calculate the magnitude and direction of all reactions at A. 2. Calculate the magnitude and direction all reactions at B. 3. Calculate the magnitude and direction of the rotation at B.

Please show all work and steps. Any help will be appreciated and thank you for your time!

Answer 1

Given, 15 KN 2 kN/m. 3m 3m E100 GPa = 100x10² mpa 105 M PA : 105 N/mm 105 x (103)4 mm I= 105 my Х EI 105 N/mm² x 105 m4 5 10 N/m2 x 185 m (103)2 ΕΙ 106 N.m² STEP(1) : calculate Propped reaction (@B) 15 2 kN/m قصه 3 3m. 3m. 2 kN/m. 3 2 x 63 te SEI 8 EI + 15 KN t 5W/ 5x15x63 48 EI 48 EI + RB L3 RB (63) 3 EI ЗЕІ RB.

Upward deflection - Downward de flection PB (6) ( 5x15x+63 4EU 31 81 4 75 RB 나 4 1 48 3 RB + 25 16 3 RB 4 25 16. 16 3. 29 X3 RB 16 (2) : RB - 82 16 = 5.44 KN() STEP-2: 15 KN 12 kN/m B (0) it 3m. 3 m 4 RB -5.44 KN RAT RB = 2 (6) + 15 RA 27- 5.44 Il RA 21.56 KN (1) EMER FO.(C+, (7z+ , 9:-) RA (6) -(2 x 6x ý ) 15 (3) - MA = 0. MA - (21.56 x6) - 36-45 MA= 48.36 KN.m (5)

STEP-3 (3) magnitude 8 Direction of the Rotation @B 15 KN • 2 kN/m Зm 3m. ,2 kN/m. 15 KN + + Tih 6m 3m 3m. ARB 2x62 RB Xb EI + 15x6 2EI D EI A Az - diagrams. EI MOMENT-AREA METHOD according to Moment area method, change in (Theorem-1): from Slope section - A to section-B diagram will be EI blw. A & B equal to zie og OA = A + A + A3 + Ą = -1/2 xbxh bh 2 544net + A, = - 1/2 bxh 2x 36 2ΕΙ - 12 -1xBx 3 2 / 3 x x 90 ΕΙ ΕΙ 135 + 16.32 IN EI EI EI - 72 - 135 + 16.32 - 190.68 EI ΕΙ (OA=0).

- 190.68 1 'B EI EI - 106 N.m² 106x103 KN.m² 103 KN.m? - 190.68 - 0.19 B 103 lor) 0.199 (2) B RESULTS: (1) RA 21.56 KN (4) ; MA= 4836 kNmd 3) 5.44 KN (4) (2) RBS (3) 0.19°C OB