Case 1: Uniform beam under distributed load.In the shown Figure, a uniform beam subject to a linearly increasing distributed load.The deflection y (m) can be expressed byW.(-x5 + 2L²x³ –- L*x)120EILy =WoWhere E is the modulus of elasticity and Iis the momentbeamUse the following parameters L-600 cm,E=50,000 kN/cm?, l= 30.000 cm“, w.=2.5kN/cm, to find the requirementsertia (m*), L length(a)(r = L, y = 0)(r = 0, y = 0)(b)Plot the following quantities versus distance along the beamMoment M(x) = Eld²y/dx².Shear V(x) = Eld³y/dx³.Loading w(x) = –Eld*y/dx+.

Answered: Image /qna-images/answer/bcc6a785-a5ac-48e4-893c-fde93bf26974.jpg

Case 1: Uniform beam under distributed load. In the shown Figure, a uniform beam subject to a linearly increasing distributed load. The deflection y (m) can be expressed by W. -(-x³ + 21?x³ – L*x) 120EIL y Where E is the modulus of elasticity and I is the moment of inertia (m*), L length of beam. Use the following parameters L-600 cm, E-50,000 kN/cm,1- 30.000 cm*, w.-2.5 kN/cm, to find the requirements (a) (r = L, y = 0) r= 0, y = 0) | (b) Plot the following quantities versus distance along the beam Moment M(x) = Eldy/dx². Shear V(x) = Eld³y/dx³. Loading w(x) = -Eld*y/dx*.

Case 1: Uniform beam under distributed load. In the shown Figure, a uniform beam subject to a linearly increasing distributed load. The deflection y (m) can be expressed by W. -(-x³ + 21?x³ – Lx) 120EIL y Where E is the modulus of elasticity and I is the moment of inertia (m), L length of beam. Use the following parameters L-600 cm, E-50,000 kN/cm,1- 30.000 cm, w.-2.5 kN/cm, to find the requirements (a) (r = L, y = 0) r= 0, y = 0) | (b) Plot the following quantities versus distance along the beam Moment M(x) = Eldy/dx². Shear V(x) = Eld³y/dx³. Loading w(x) = -Eldy/dx*.

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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