Week - 2 - Explicit and Implicit Analysis

given: f(u) = F(u) = u3+9u2+4u

Explicit Analysis

To differentiate the function F(u) = u^3 + 9u^2 + 4u with respect to u,

dF(u)/du = d(u^3)/du + d(9u^2)/du + d(4u)/du

d(u^3)/du = 3u^2

d(9u^2)/du = 18u

d(4u)/du = 4

Therefore:

dF(u)/du = 3u^2 + 18u + 4

Stiffness K(U)=df/du= 3u^2 + 18u + 4

This is the derivative of F(u) with respect to u.

△">△△u=incremental displacement,
△">△△F=Incremental external applied forces

using △">△△F=K△">△△u, tolerance 10^-2

The stiffness K(U) is the derivative of the force F(u) with respect to the displacement u. In this case, we have:

dF(u)/du = 3u^2 + 18u + 4

Therefore, the stiffness K(U) is:

K(U) = dF(u)/du = 3u^2 + 18u + 4

Note that the stiffness is a function of the displacement u, so it may vary with u.

let

1. u=0 if we substitute u = 0 in the expression for K(U), we get:

   K(0) = dF(0)/du

   = 3(0)^2 + 18(0) + 4

   = 4

   Therefore, when u = 0, the stiffness K(U) is 4.

   Assuming that ΔF represents the change in external force applied to the system, and Δk(u) represents the change in stiffness due to the displacement change ΔU1, then the calculation

   ΔU1 = ΔF/Δk(u)

   = 1/4

   = 0.25

   Using this value, the new displacement u1 can be found as

   u1 = U0 + ΔU1

   = 0 + 0.25

   = 0.25, where U0 represents the initial displacement.

   Finally,

   substituting u1 = 0.25 into the given expression for internal force Fint, we get:

   F1int = (0.25^3) + 9(0.25^2)(0.025) + 4(0.25)

   = 1.578125

   Therefore, the internal force F1int at the new displacement u1 is approximately 1.578125 units.

2. u1 = 0.25 into this expression, we get:

   k(u1) = 3(0.25)^2 + 18(0.25) + 4 = 5.9375 units/mm

   Using this value, we can calculate the change in displacement Δu from u1 to u2 as:

   Δu = ΔF / k(u1) = 1 / 5.9375 = 0.168

   Therefore, the new displacement u2 is:

   u2 = u1 + Δu = 0.25 + 0.168 = 0.418

   To calculate the internal force F2int at the new displacement u2, we can substitute u2 = 0.418 into the expression for Fint:

   F2int = (0.418^3) + 9(0.418^2)(0.025) + 4(0.418) = 1.706 units

   Therefore, the internal force F2int at the new displacement u2 is approximately 1.706 units.

3. u2=0.3651

   Δu = ΔF / k(u2) = 1 / (3u2^2 + 18u2 + 4)

   Substituting u2 = 0.418 into this expression, we get:

   Δu = 1 / (3(0.418)^2 + 18(0.418) + 4) = 0.0522

   Therefore, the new displacement u3 is:

   u3 = u2 + Δu = 0.418 + 0.0522 = 0.4702

   To calculate the internal force F3int at the new displacement u3, we can substitute u3 = 0.4702 into the expression for Fint:

   F3int = (0.4702^3) + 9(0.4702^2)(0.025) + 4(0.4702) = 1.902 units

   Therefore, the internal force F3int at the new displacement u3 is approximately 1.902 units.

now To calculate the total internal force Fint at the final displacement u3, we need to sum up the internal forces at each displacement. So, we have:

Fint = F1int + F2int + F3int

Substituting the values we calculated earlier, we get:

Fint = 1.578125 + 1.706 + 1.902 = 5.186125 units

Therefore, the total internal force Fint at the final displacement u3 is approximately 5.186125 units.

From the calculations we made earlier, we obtained the following values for the internal forces at each displacement:

• F0int = 0 units
• F1int = 1.578125 units
• F2int = 1.706 units
• F3int = 1.902 units

Therefore, the total force at each displacement can be calculated as:

• F0 = F0int + ∆F1 + ∆F2 + ∆F3 = 0 + 1 + 1 + 1 = 3 units

Therefore, at each displacement, the total force acting on the system is a combination of the internal forces and the external forces applied up to that point.

step	Δ">ΔΔF	U	Fext	Fint	residual = Fint-Fext
1	1	0.	1	1.5781	0.5781
2	1	0.25	2	1.706	0.706
3	1	0.3651	3	1.902	0.902

It is evident from the results that external and internal forces are NOT in equilibrium .

Implicit Methods

To solve the equation using implicit method, we need to discretize the equation in time and space. This leads to the following equation:

[ u(n+1) - u(n) ] / ∆t^2 = [ f(n+1) - f(n) + f_int(n+1) - f_int(n) ] / m

where u(n) and f(n) are the displacement and external force at time step n, f_int(n) is the internal force at time step n, m is the mass of the system, and ∆t is the time step.

u(1) = 0

We can rewrite this equation as:

u(n+1) = 2u(n) - u(n-1) + (∆t^2/m) * [ f(n+1) + f_int(n+1) - f(n) - f_int(n) ]

We can apply this equation to our problem with the following values:

m = 1 k(u) = 3u^2 + 18u + 4 u(0) = 0 u(1) = 0.25 f(0) = 0 f(1) = 1

To calculate u(2), we need to use the values of u(1) and u(0):

u(2) = 2u(1) - u(0) + (∆t^2/m) * [ f(2) + f_int(2) - f(1) - f_int(1) ]

To calculate f_int(1) and f_int(2), we need to use the values of u(1) and u(2):

f_int(1) = k(u(1)) * u(1) = 1.640625 f_int(2) = k(u(2)) * u(2) = 2.575233

Substituting these values in the equation, we get:

u(2) = 2(0.25) - 0 + (0.1^2/1) * [ 1 + 2.575233 - 1 - 1.640625 ] = 0.330259

Therefore, the displacement at time step 2 is u(2) = 0.330259.

To find the displacement at time step 3, we can use the equation derived in the implicit method:

(2 + 12𝑡 + 36𝑡^2)u(3) − (1 + 6𝑡 + 12𝑡^2)u(2) + (1 − 6𝑡 + 12𝑡^2)u(1) = 4𝑡^2 + 2

We know the values of u(1) and u(2) from the previous calculations:

u(1) = 0.25 u(2) = 0.3651

We also know that the time step size is 0.1, so t = 0.3.

Substituting these values into the equation, we get:

(2 + 12(0.3) + 36(0.3)^2)u(3) − (1 + 6(0.3) + 12(0.3)^2)(0.3651) + (1 − 6(0.3) + 12(0.3)^2)(0.25) = 4(0.3)^2 + 2

Simplifying and solving for u(3), we get:

u(3) = 0.4267

Therefore, the displacement at time step 3 is approximately 0.4267 units.

similarly we know,

u(1) = 0

u(2) = 0.3651

u(3) = 0.4267

We also know that the time step size is 0.1, so t = 0.4.

Substituting these values into the equation, we get:

(2 + 12(0.4) + 36(0.4)^2)u(4) − (1 + 6(0.4) + 12(0.4)^2)(0.4267) + (1 − 6(0.4) + 12(0.4)^2)(0.3651) = 4(0.4)^2 + 2

Simplifying and solving for u(4), we get:

u(4) = 0.4837

Therefore, the displacement at time step 4 is approximately 0.4837 units.

Fint(u) = u^3 + 9(0.025)u^2 + 4u

Using this equation and the values of the displacements, we can calculate the internal forces

k(u) = 3u^2 + 18(0.025)u + 4

Using this equation, we can calculate the stiffness at each displacement as follows:

1. k(u1) = 3(0.25)^2 + 18(0.025)(0.25) + 4 = 3.96875
2. k(u2) = 3(0.3651)^2 + 18(0.025)(0.3651) + 4 = 5.39771
3. k(u3) = 3(0.4267)^2 + 18(0.025)(0.4267) + 4 = 6.03913
4. k(u4) = 3(0.4837)^2 + 18(0.025)(0.4837) + 4 = 6.84721

Now, we can calculate the internal forces at each time step using the equation

1. Fint(u1) = (0.25)^3 + 9(0.025)(0.25)^2 + 4(0.25) = 1.578125
2. Fint(u2) = (0.3651)^3 + 9(0.025)(0.3651)^2 + 4(0.3651) = 2.557385
3. Fint(u3) = (0.4267)^3 + 9(0.025)(0.4267)^2 + 4(0.4267) = 3.246396
4. Fint(u4) = (0.4837)^3 + 9(0.025)(0.4837)^2 + 4(0.4837) = 4.053496

To calculate the external forces at each time step, we can use the formula:

Fext = k(u)*Δu

1. Fext(1) = k(u1)Δu1 = 3.968750.1151 = 0.457364
2. Fext(2) = k(u2)Δu2 = 5.397710.0911 = 0.491999
3. Fext(3) = k(u3)Δu3 = 6.039130.0736 = 0.445555
4. Fext(4) = k(u4)Δu4 = 6.847210.0636 = 0.436082

step	∆F	u	f	R=internal-external
1	1	0.33	2.5	2
2	1	0.4	3.2	2.7
3	1	0.42	4.0	3.5

conclusion:

1. The main difference between these two methods is how they handle the time-dependent terms in the equations and methods are numerical techniques used to solve differential equations.
2. Explicit methods use the information from the current time step to calculate the solution at the next time step.
3. Implicit methods, on the other hand, use the information from the current and next time step to calculate the solution at the next time step.
4. Explicit methods are generally less accurate than implicit methods, especially for problems with rapidly varying solutions or where the time step is large.
5. Implicit methods, on the other hand, are generally more accurate than explicit methods, especially for problems with complex or nonlinear dynamics.