Question

Q1. For the cantilever beam and loading shown with circular section of 60 mm diameter and E = 200 GPa, use Double Integration Method to determine the value of the first arbitrary constant C1. Unit of force must be in KN and unit of length must be in m. Express your answer in three decimal places. Q2. For the cantilever beam and loading shown with circular section of 60 mm diameter and E = 200 GPa, use Double Integration Method to determine the value of the second arbitrary constant C2. Unit of force must be in KN and unit of length must be in m. Express your answer in four decimal places. Q3. For the cantilever beam and loading shown with circular section of 60 mm diameter and E = 200 GPa, use Double Integration Method to determine the value of the slope at the free end in degree measure. Express your answer in four decimal places. Q4. For the cantilever beam and loading shown with circular section of 60 mm diameter and E = 200 GPa, use Double Integration Method to determine the value of the deflection at the free end in mm. Express your final answer in absolute value and use two decimal places. 2 kN/m 1 KN B A 0.40 m 0.60 m ta 1.00 m

Answer 1

an2 - Mai &095: IKN .alealm A B coum k Samsung с -0,6m. Mars - 1X2 - 29-4) (no) 2 - (n=472 By Double Integration method : -n'- CM-0.472 EI 224 on integration! EI dy (n-0:43 +C1 AQ Again integrating EI Y5 (n-0.474 -23 +C19 +62 + 12 using end conditions ; At a=2m, dy = $c = 0 dn .: from eancis EI хо - - 14-04).° +C1 C1 = 0.572 ()

OA = 0:2575 degree Also, ad nofm o y = Sco from egniig EI XO 5 01- (3-0-474 + 0.572x17 ca 12 (2) Ca=-0.3945 E = 200 G Pa = 200X106 Knim 2 Is I *74 => 64 T 64 *(60X10-334 I = 6-36 *10-7 74 EI = 2X108 6-36x 10-7 EI: 127.23 kN-m? (3) Slope (dy J al poree end dy dr BIC-22- (2–0-4)3 + 0.572) 3 content => 127:23 [0-0 +0.572) 2150 OA = 4.49 *10-3 radion OA = 4.49x10-3 x 180 degree T

4) Deflection of free end y = SA= ja con -213_(4-0-4) 4 +0.5720-03945 12 Sach-o) [0-0+0 -0.3945) 127-23

SA= -3.100 x10-3m SAE - 3.1 mm (downward )