A cantilever beam shown in the figure below is commonly used to support a load acting on a balcony.The left side of a horizontal beam is connected to a vertical wall. Eight equally spaced vertical arrows start above the beam and point downward, toward the beam. The collection of arrows is labeled w. The leftmost arrow is near the left end of the beam at the wall, and the rightmost arrow is at the right end of the beam. A point near the left end of the beam is a distance x to the right of the wall, and the length of the beam is L > x.The deflection of the centerline of the beam is given by the following equationy = −wx224EI(x2 − 4Lx + 6L2),where we have the following.y = deflection at a given location (m)w = distributed loadE = modulus of elasticity (N/m2)I = second moment of area (m4)x = distance from the support as shown (m)L = length of the beam (m)What is the appropriate unit for w if the preceding equation is to be homogeneous in units? Show all steps of your work.Starting with the equation y = −wx224EI(x2 − 4Lx + 6L2), the units of y are m, the units of the product E I are , and the units of x2, Lx, and L2, are all equal to . Rearranging the equation above to solve for w, we obtain the following. (Use any variable or symbol stated above as necessary.)w = Then, the units of w are N · m3 = .
Question
A cantilever beam shown in the figure below is commonly used to support a load acting on a balcony.The left side of a horizontal beam is connected to a vertical wall. Eight equally spaced vertical arrows start above the beam and point downward, toward the beam. The collection of arrows is labeled w. The leftmost arrow is near the left end of the beam at the wall, and the rightmost arrow is at the right end of the beam. A point near the left end of the beam is a distance x to the right of the wall, and the length of the beam is L > x.The deflection of the centerline of the beam is given by the following equationy = −wx224EI(x2 − 4Lx + 6L2),where we have the following.y = deflection at a given location (m)w = distributed loadE = modulus of elasticity (N/m2)I = second moment of area (m4)x = distance from the support as shown (m)L = length of the beam (m)What is the appropriate unit for w if the preceding equation is to be homogeneous in units? Show all steps of your work.Starting with the equation y = −wx224EI(x2 − 4Lx + 6L2), the units of y are m, the units of the product E I are , and the units of x2, Lx, and L2, are all equal to . Rearranging the equation above to solve for w, we obtain the following. (Use any variable or symbol stated above as necessary.)w = Then, the units of w are N · m3 = .
Solution
The equation given is y = -wx^2/24EI * (x^2 - 4Lx + 6L^2).
The units of y are meters (m), the units of the product EI are Newton meters squared (N*m^2), and the units of x^2, Lx, and L^2 are all meters squared (m^2).
We can rearrange the equation to solve for w:
w = -y * 24EI / (x^2 * (x^2 - 4Lx + 6L^2))
The units of y are meters (m), the units of 24EI are Newton meters squared (N*m^2), and the units of the denominator are meters to the fourth power (m^4).
Therefore, the units of w are:
N*m^2 / m^4 = N/m^2
So, the appropriate unit for w, the distributed load, is Newton per square meter (N/m^2).
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