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0.8 b 0.8 b b a a b Section (I) Section (II) Two cross-sections ? w (kN/m) L ?max x ?max Fixed Supported Beam with Distributed Load Fig. (1) 1. Deduce the deflection and slope equations for the shown Fixed supported beam shown in Figure (1). 2. Design standard two cross-sections: Solid square (section I), and Box square (Section II) (Use static flexural stress formula only). 3. Calculate the weight of the beam for each cross-section in (kg) and identify the lighter beam weight? 4. Calculate the maximum deflection in (mm) of the loaded beam for each cross-section and identify its position. 5. Comment on the previous results (parts 3 and 4) and state two conclusions. Assume the loaded beam has the following data Young's modulus = 210 GPa, yield strength = 390 MPa and Design factor = 3 w = 5 kN/m, L = 1 m and material's density = 7.8 g/cm³

          0.8 b
0.8 b
b
a
a
b
Section (I)
Section (II)
Two cross-sections
?
w (kN/m)
L
?max
x
?max
Fixed Supported Beam with Distributed Load
Fig. (1)
1. Deduce the deflection and slope equations for the shown Fixed supported beam shown in Figure (1).
2. Design standard two cross-sections: Solid square (section I), and Box square (Section II) (Use static flexural stress formula only).
3. Calculate the weight of the beam for each cross-section in (kg) and identify the lighter beam weight?
4. Calculate the maximum deflection in (mm) of the loaded beam for each cross-section and identify its position.
5. Comment on the previous results (parts 3 and 4) and state two conclusions.
Assume the loaded beam has the following data
Young's modulus = 210 GPa, yield strength = 390 MPa and Design factor = 3
w = 5 kN/m, L = 1 m and material's density = 7.8 g/cm³

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0.8 b
0.8 b
b
a
a
b
Section (I)
Section (II)
Two cross-sections
?
w (kN/m)
L
?max
x
?max
Fixed Supported Beam with Distributed Load
Fig. (1)
1. Deduce the deflection and slope equations for the shown Fixed supported beam shown in Figure (1).
2. Design standard two cross-sections: Solid square (section I), and Box square (Section II) (Use static flexural stress formula only).
3. Calculate the weight of the beam for each cross-section in (kg) and identify the lighter beam weight?
4. Calculate the maximum deflection in (mm) of the loaded beam for each cross-section and identify its position.
5. Comment on the previous results (parts 3 and 4) and state two conclusions.
Assume the loaded beam has the following data
Young's modulus = 210 GPa, yield strength = 390 MPa and Design factor = 3
w = 5 kN/m, L = 1 m and material's density = 7.8 g/cm³

Added by Jessica C.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Supreeta N

08/11/2023

Step 1

For a fixed supported beam with a uniformly distributed load \( w \) over its length \( L \), the deflection \( \delta \) and slope \( \theta \) equations are given by: \[ \delta(x) = \frac{wx^2}{24EI}(L^3 - 2Lx^2 + x^3) \] \[ \theta(x) = \frac{wx}{6EI}(L^3 - Show more…

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Transcript

00:01 Hello students, it is given that the bending moment at a distance x from the fixed end a is given by mx is equal to minus w by 2 into l minus x the whole square.

00:20 So we want to determine the slope that is dy by dx and also the deflection y equations for a beam under a distributed load where l is the length of the beam and w is the load intensity.

00:46 So here in the first step we can equate the given bending moment equation with the bending moment expression from the flexural formula.

00:54 Therefore mx is equal to ei d square y by dx square.

01:03 This gives us the equations that is ei d square y by dx square is equal to minus of w by 2 into l minus x the whole square.

01:21 So let's mark this as equation number one.

01:25 Now in the step two of the equation we can integrate equation one with respect to with respect to x to get the slope dy dx in terms of x.

01:46 So for that ei into dy dx is equal to minus of integral w by 2 into l minus x the whole square dx.

02:03 So here solving the integral and adding the integration constant c1 will get ei dy by dx is equal to w by 6 l minus x the whole cube plus c1 and let's mark this as equation number two and now integrating the equation number two with respect to x to get the deflection equation y in terms of x we can write ei into y will be equal to minus of integral w by 6 into l minus x the whole cube into dx plus c1 x plus c2.

02:46 Therefore solving at the integral and adding the integration constant c2 we can write ei y will be equal to minus omega sorry minus w by 24 into l minus x the whole raised to 4 plus c1 by 2 x plus c2 and let's mark this as equation number three...

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