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 – 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*.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text: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.
(-x5 + 2L²x³ –- L*x)
120EIL
y =
Wo
Where E is the modulus of elasticity and I
is the moment
beam
Use the following parameters L-600 cm,
E=50,000 kN/cm?, l= 30.000 cm“, w.=2.5
kN/cm, to find the requirements
ertia (m*), L length
(a)
(r = L, y = 0)
(r = 0, y = 0)
(b)
Plot the following quantities versus distance along the beam
Moment M(x) = Eld²y/dx².
Shear V(x) = Eld³y/dx³.
Loading w(x) = –Eld*y/dx+.
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