Question :
The cantilever beam AB of length L carries a uniformly distributed W load , which includes the weight of the beam.
(Note that the left end end A is free , while right end has a fixed support)
(i) Derive the equation of the elastic curve.
(ii) Compute the maximum displacement of the beam.
My work :
First I considered a section of beam of length x from left end A and derived an equation for bending moment(M) which gave,
M + (Wx/L)(x/2) = 0
M = -(W*(x^2))/2L) which is equal to EIy' = -(W*(x^2))/2L)
==> EIy' = -(W*(x^3))/6L) + C1
==> EIy = -(W*(x^4))/24L) + C1L + C2 , where C1, C2 are arbitrary constants and y' & y' denote second & first derivatives of y, respectively.
EI is constant.
I found C1 = (w*(L^2))/6 and C2 = -(w*(L^3))/8
==> EI*y' = -(W*(x^3))/6L) + (w*(L^2))/6
==> EI*y = -(W*(x^4))/24L) + (w*(L^3))/6 + -(w*(L^3))/8
For the 2nd part,
I tried calculating max. displacement manually by equating y' = 0 and substituting the x value I got, to y.
y' = -(W*(x^3))/6L) + (w*(L^2))/6 = 0
==> -(W*(x^3))/6L) + (w*(L^3))/6L = 0
==> (w*(L^3))/6L = (W*(x^3))/6L)
==> (w*(L^3)) = (W*(x^3))
==> L^3 = x^3 ==> x^3 - L^3 = 0
==> (x-L)(x^2 + Lx + L^2) = 0
==> x = L or x^2 + Lx + L^2 = 0
But at x=L, y= 0
Hence, x^2 + Lx + L^2 = 0
==> x =[ -L + /- sqrt(L^2 - 4L^2)]/2
==> x = [ -L + /- sqrt(-3L^2)]/2 ] , which are not real values
Following a different process, then I differentiated y w.r.t.x and equated it to 0
EI*y = -(W*(x^4))/24L) + (w*(L^3))/6 + -(w*(L^3))/8
y' = w/24EI(4L^2 - ((3x^3)/L) )
and equating y' = 0 to find x such that max.y can be found,
w/24EI(4L^2 - ((3x^3)/L) ) = 0
4L^2 - ((3x^3)/L) ) = 0
3x^3 = 4L^3 x^3 = 4/3(L^3) ,
which also doesn't seem to give the expected answer of x = 0
Could you help find my mistake?
Thanks!
Based on your calculations, it seems that you have made a mistake in deriving the equation for the bending moment.
Let's start fresh and go step-by-step:
1. Start by considering a small section of the beam of length dx at a distance x from the free end.
2. The bending moment at this section can be given as M = -Wx, where W is the uniformly distributed load. The negative sign indicates that the bending moment is in the opposite direction to the load.
3. The bending moment is also related to the curvature of the beam through the equation M = EIy'', where E is the modulus of elasticity and I is the moment of inertia of the beam's cross-sectional area.
4. Since the beam is subjected to a uniformly distributed load, the curvature can be assumed to be linear, i.e., y'' = kx, where k is a constant.
5. Substituting the values of M and y'' in the bending moment equation, we have -Wx = EI(kx), which simplifies to: k = -W/(EI).
6. Integrating the curvature twice gives us the equation for the deflection (elastic curve) of the beam: y = (-W/(24EI))x^3 + C1x + C2, where C1 and C2 are integration constants.
7. Applying the boundary conditions of a cantilever beam (free end and fixed support), we can determine the slope (dy/dx) and deflection (y) at each end.
- At the free end (x = 0), y = 0 (no deflection).
- At the fixed support (x = L), dy/dx = 0 (no vertical displacement).
8. Using the boundary conditions, we can determine the values of C1 and C2:
- At x = 0, y = 0:
0 = (-W/(24EI))(0^3) + C1(0) + C2
C2 = 0
- At x = L, dy/dx = 0:
0 = (-W/(24EI))(L^3) + C1
C1 = (W/(24EI))(L^3)
9. Substituting the values of C1 and C2 back into the deflection equation, we have:
y = (-W/(24EI))x^3 + (W/(24EI))(L^3)x
10. To find the maximum deflection, we can differentiate the deflection equation with respect to x and set it equal to zero:
dy/dx = (-W/(8EI))x^2 + (W/(24EI))(L^3) = 0
11. Solving for x, we get:
x^2 = (L^3)/3
x = sqrt((L^3)/3) = (L/√3)
Therefore, the maximum deflection occurs at x = (L/√3), and the maximum deflection value can be calculated by substituting this value into the deflection equation:
y_max = (-W/(24EI))((L/√3)^3) + (W/(24EI))(L^3)(L/√3)
You can simplify this expression further if needed.
Looking at your calculations, it seems that you made a mistake in the equation for the bending moment. Let's go through the derivation step by step to identify the error.
Given:
M + (Wx/L)(x/2) = 0
To find the bending moment equation, we need to integrate the distributed load equation twice.
First integration:
∫[M + (Wx/L)(x/2)] dx = 0
∫M dx + ∫(Wx^2/2L) dx = 0
The integral of M dx gives us the equation of the elastic curve (y). The integral of (Wx^2/2L) dx gives us the equation for the deflection angle (θ).
Second integration:
∫(∫M dx + ∫(Wx^2/2L) dx) dx = 0
∫(∫M dx) dx + ∫(∫(Wx^2/2L) dx) dx = 0
The first term on the right-hand side, ∫(∫M dx) dx, gives us the equation for y. The second term, ∫(∫(Wx^2/2L) dx) dx, gives us the equation for θ.
Now let's integrate each term separately.
For the first term, ∫(∫M dx) dx:
∫(∫M dx) dx = ∫(C + ∫(Wx/L)(x/2) dx) dx
= Cx + ∫(Wx^3/2L^2) dx
= Cx + (W/2L^2) * ∫(x^3) dx
= Cx + (W/2L^2) * (x^4/4)
= Cx + (Wx^4/8L^2)
Now, let's differentiate this equation with respect to x to find y.
dy/dx = C + (Wx^3/2L^2)
Now let's integrate the second term, ∫(∫(Wx^2/2L) dx) dx:
∫(∫(Wx^2/2L) dx) dx = ∫((W/2L) * ∫(x^2) dx) dx
= ∫((W/2L) * (x^3/3)) dx
= (W/2L) * (x^4/12)
= (Wx^4/24L)
Differentiating this equation with respect to x gives us the equation for θ.
d^2y/dx^2 = (Wx^3/6L)
Now we have the equations for y and θ. Let's combine them to find the complete equation for the elastic curve.
EI * d^2y/dx^2 = M = (Wx^3/6L)
Now, we can integrate this equation to solve for y.
Let's assume the integration constant is zero for simplicity.
Integrating EI * d^2y/dx^2 = (Wx^3/6L) with respect to x gives us
EI * d^2y/dx^2 = (Wx^4/24L) + C
Integrating again gives us
EI * dy/dx = (Wx^5/120L) + Cx + D
Integrating one last time gives us the equation of the elastic curve:
EI * y = (Wx^6/720L) + (Cx^2)/2 + Dx + E
Now that we have the correct equation for the elastic curve, we can proceed to find the maximum displacement by equating y' = 0 and solving for x.