The bending moment M of the beam is given by dM/dx= -w(l-x) where w is a constant
Determine of M. Given M = ½ wt^2 when x=0
dM/dx = -w(l-x)
M = w/2 (l-x)^2 + C
now plug in M(0) = 1/2 wt^2 to find C
To determine the bending moment M, we need to integrate the differential equation dM/dx = -w(l - x) with the given initial condition M = ½ wt^2 at x = 0.
Let's start by integrating dM/dx with respect to x:
∫dM/dx dx = ∫-w(l - x) dx
Integrating the right side gives:
M = ∫(-w(l - x)) dx
M = -w∫(l - x) dx
Now, we can solve this integral:
M = -w∫l dx + w∫x dx
M = -wlx + ½ wx^2 + C
Here, C is the constant of integration.
Now, we can substitute the given condition M = ½ wt^2 when x = 0:
½ wt^2 = -w(0)l + ½ w(0)^2 + C
½ wt^2 = C
So, C = ½ wt^2.
Substituting this back into the equation for M, we get:
M = -wlx + ½ wx^2 + ½ wt^2
Therefore, the bending moment M is given by M = -wlx + ½ wx^2 + ½ wt^2.
To determine the expression for M, we need to integrate the expression dM/dx = -w(l-x) with respect to x.
Let's start by integrating both sides of the equation:
∫ dM/dx dx = ∫ -w(l-x) dx
Integrating the left side of the equation will yield just M:
M = ∫ -w(l-x) dx
Now, integrate the right side of the equation:
M = -w ∫ (l-x) dx
Using the power rule of integration, we get:
M = -w [ (lx - (x^2/2) ) ] + C
where C is the constant of integration.
Given that M = 1/2 wt^2 when x = 0, we can substitute these values into the equation and solve for C:
1/2 wt^2 = -w [ (l(0) - (0^2/2) ) ] + C
1/2 wt^2 = -w (0 - 0) + C
1/2 wt^2 = C
Now, substitute this value of C back into the equation for M:
M = -w (lx - (x^2/2)) + (1/2 wt^2)
Simplifying further, we get:
M = -wlx + wx^2/2 + 1/2 wt^2
Therefore, the expression for M is:
M = -wlx + wx^2/2 + 1/2 wt^2