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W, complete following: a. Determine the number of degrees of indeterminaey b. Solve for all reactions using reaction B (a roller) as a redundant e. Draw the shear and moment diagrams ww 4 kN/m 2 m

W, complete following: a. Determine the number of degrees of indeterminaey b. Solve for all reactions using reaction B (a roller) as a redundant e. Draw the shear and moment diagrams ww 4 kN/m 2 m

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Degra 1 inden minacy R- 2 3-2 Taking redundat Fore (Removing For ) fhu radundat 4x 8 2048 (Removing. Load) th extornad 2 + Rx 2 SET 3ES be 2o10 Dylection at 2048 12 bean Fe fora diegr. - Ra-ux ) = ५ 28-4x S
Transcribed Image Text:Degra 1 inden minacy R- 2 3-2 Taking redundat Fore (Removing For ) fhu radundat 4x 8 2048 (Removing. Load) th extornad 2 + Rx 2 SET 3ES be 2o10 Dylection at 2048 12 bean Fe fora diegr. - Ra-ux ) = ५ 28-4x S
1) For the beam shown below, complete the following: Determine the number of degrees of indeterminacy b. Solve for all reactions using reaction B (a roller) as a redundant Draw the shear and moment diagrams a. e. w= 4 kN/m B 2 m Step 2 (b) Taking Reaction B as redundant : Udi The deflection at B is given by. Due to (õB )udi = (6L² – 4Lx + x²) 24 EI X = 8m L = 10m 4C8) 24 ET (6(10)2 - 4 (10) (8) + 82) 3669.3 (ön )udi = Due to reaction point load The deflection at B is given by. (öB )R = R.2 (3a - x) a - x - 8 (ôn R = (3 x 8 - 8) 6 EI 170.667R so. (öB )R = (ön)udi 170.667R 3660.3 R = 21. 5 kN Step 4 135 M 148 10 28 ot
Transcribed Image Text:1) For the beam shown below, complete the following: Determine the number of degrees of indeterminacy b. Solve for all reactions using reaction B (a roller) as a redundant Draw the shear and moment diagrams a. e. w= 4 kN/m B 2 m Step 2 (b) Taking Reaction B as redundant : Udi The deflection at B is given by. Due to (õB )udi = (6L² – 4Lx + x²) 24 EI X = 8m L = 10m 4C8) 24 ET (6(10)2 - 4 (10) (8) + 82) 3669.3 (ön )udi = Due to reaction point load The deflection at B is given by. (öB )R = R.2 (3a - x) a - x - 8 (ôn R = (3 x 8 - 8) 6 EI 170.667R so. (öB )R = (ön)udi 170.667R 3660.3 R = 21. 5 kN Step 4 135 M 148 10 28 ot
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