A tapered beam is shown below, the dimensions of the beam vary along the length. The beam has a non-uniform distributed load applied to the upper surface. L X+ w(x) = f(x) The length of the beam, L, is 1,000.0 mm. The function f(x) equal to 2.000sin (7) N/mm. The cross-section of the beam is square, with the side length of a. The allowable strength, σ, of the material is 250.0 MPa. Determine the side length, a, of the cross-section when the value of x is equal to 450.0 mm.
A tapered beam is shown below, the dimensions of the beam vary along the length. The beam has a non-uniform distributed load applied to the upper surface. L X+ w(x) = f(x) The length of the beam, L, is 1,000.0 mm. The function f(x) equal to 2.000sin (7) N/mm. The cross-section of the beam is square, with the side length of a. The allowable strength, σ, of the material is 250.0 MPa. Determine the side length, a, of the cross-section when the value of x is equal to 450.0 mm.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:A tapered beam is shown below, the dimensions of the beam vary along the length.
The beam has a non-uniform distributed load applied to the upper surface.
·a
L
a
X+
The length of the beam, L, is 1,000.0 mm.
The function f(x) equal to 2.000sin (7) N/mm.
The cross-section of the beam is square, with the side length of a. The allowable
strength, O, of the material is 250.0 MPa.
w(x) = f(x)
Determine the side length, a, of the cross-section when the value of x is equal to
450.0 mm.
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