Week - 2 - Explicit and Implicit Analysis

AIM :

To solve a non linear static problem by both explicit and implicit analysis.

Given Parameters :

A simple bar in tension as shown in Figure 1. Suppose that the force in the bar is a nonlinear function of the displacement F(u) = u³+9u²+4u

Three equal load increments (or steps) of ∆F = 1 unit each are used to load the tension bar. The variables used are u for displacement, f for internal force in the bar, F for external force applied to the bar and k for stiffness. Increemental displacements or incremental externally applied forces are represented as ∆u or ∆F, respectively. Use is made of the relationship ∆F = k∆u.

⇒k(u) = df/du = 3u²+18u+4

Explicit analysis :

Step 1 :

u₀ = 0, ⇒ k(u₀) = 4, ∆F = 1, ∆u₁ = ∆F/k(u₀) = 0.25, u₁ = u₀ + ∆u₁ = 0.25

Step 2 :

⇒ k(u₁) = 3*(0.25)² + 18*0.25+4 = 0.1875+4.5+4 = 8.6875, ∆F = 1, ∆u₂ = ∆F/k(u₁) = 0.1151, u₂ = u₁ + ∆u₂ = 0.3651

Step 3 :

⇒ k(u₂) = 3*(0.3651)²+18*0.3651+4 = 10.7178, ∆F = 1, ∆u₃ = ∆F/k(u₂) = 1/10.7178, u₃ = u₂+∆u₃= 0.4584

Finally, from the three steps the following results are obtained.

(F_int)₁ = u₁³+9u₁²+4*u₁ = 0.25³+9*0.25²+4*0.25 = 1.5781

(F_int)₂ = u₂³+9u₂²+4*u₂ = 0.3651³+9*0.0.3651²+4*0.3651 = 2.7087

(F_int)₃ = u₃³+9u₃²+4*u₃ = 0.4584³+9*0.4584²+4*0.4584 = 3.821

Step i	∆Fi	∆ui	ui	(Fext)i	(fint)i	fint − Fext = R
1	1	0.25	0.25	1	1.5781	0.5781
2	1	0.1151	0.3651	2	2.7087	0.7087
3	1	0.0933	0.4584	3	3.821	0.821

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

Implicit Analysis :

An implicit analysis is the same as the explicit analysis, except that at the end of each step Newton-Raphson iterations are used to enforce equilibrium before moving to the next step. Basically, an incremental force is applied to advance the solution forward at the beginning of a step. However, internal forces and external forces will not be in equilibrium unless the stiffness is linear for the given step. Hence, in order to achieve equilibrium, corrections must be made to the displacement. This is accomplished by using Newton-Raphson iterations to minimize the residual, R_(u) = f_int − F_ext. Expanding the residual as a Taylor series about the current displacement u^j gives

R^j+1 = R^j(u^j) + dR(u^j)/du *δu^j+1 + ... = R^j + k(u^j)δu^j+1 + ...

By neglecting higher order terms in the series, setting R^j+1 = 0 and solving for δu^j+1 the

following correction is obtained

δu^j+1 = −[k(u^j)]⁻¹R^j

Notice that the iteration variable is j and it may take several iterations for the norm of the residual, ||R||, to be reduced below the chosen tolerance criteria (for the example below a tolerance of 10⁻² is chosen).

Step 1 :

u₀ = 0, ⇒ k(u₀) = 4, ∆F = 1, ∆u₁ = ∆F/k(u₀) = 0.25, u₁ = u₀ + ∆u₁ = 0.25

(F_int)₁ = u₁³+9u₁²+4*u₁ = 0.25³+9*0.25²+4*0.25 = 1.5781

R⁽⁰⁾ = f_int − F_ext = 1.5781 – 1 = 0.5781

||R⁽⁰⁾|| > 10⁻² ⇒ Newton-Raphson iterations is necessary.

the correction is δu⁽¹⁾ = −[k(u₁⁽⁰⁾)]⁻¹ * R⁽⁰⁾ = -[ 3*0.25²+18*0.25+4 ]^-1 * 0.5781 = -0.0665

⇒u₁¹=u₁⁰+∂u¹ = 0.25-0.0665 =0.1835

(F_int)₁ = u₁³+9u₁²+4*u₁ = 1.0432 > 10^-2 ⇒ Newton-Raphson iterations is necessary.

the correction is δu⁽²⁾ = −[k(u₁⁽¹⁾)]⁻¹ * R⁽¹⁾ = -[3*0.1835²+18*0.1835+4]^-1 * 0.0432 = -0.0058

u₁²=u₁⁰+u₁¹+∂u² = 0.25-0.0665-0.0058 =0.1777

(F_int)₁² = u₁³+9*u₁²+4*u₁ = 0.1777³ + 9*0.1777²+4*0.1777=1.0006

R⁽²⁾ = f_int − F_ext = 1.0006 – 1 = 0.0006 < 10⁻²

So, (F_int)₁ = 1.0006

Step 2 :

(F_int)₂ = u₂³+9u₂²+4*u₂ = 0.3651³+9*0.0.3651²+4*0.3651 = 2.7087

R⁽⁰⁾ = f_int − F_ext = 2.7087 – 2 = 0.7087

||R⁽⁰⁾|| > 10⁻² ⇒ Newton-Raphson iterations are necessary.

the correction is

δu⁽¹⁾ = −[k(u₂⁽⁰⁾)]⁻¹ * R⁽⁰⁾ = -[ 3*0.3651²+18*0.3651+4 ]^-1 * 0.5781 = -0.0646

u₂¹=u₂⁰+∂u¹ = 0.3651-0.0646 =0.3005

(F_int)₂¹ = u₂³+9*u₂²+4*u₂ = 0.3005³ + 9*0.3005²+4*0.3005=2.037

R⁽¹⁾ = f_int − F_ext = 2.037 – 2 = 0.037> 10^-2

the correction δu⁽²⁾ = −[k(u₂⁽¹⁾)]⁻¹ * R⁽¹⁾ = -[3*0.3005²+18*0.3005+4]^-1 * 0.037 = -0.0038

u₂²=u₂⁰+u₂¹+∂u² = 0.3651-0.0646-0.0038 =0.2967

(F_int)₁² = u₁³+9*u₁²+4*u₁ = 0.2967³ + 9*0.2967²+4*0.2967=2.0052

R⁽²⁾ = f_int − F_ext = 2.0052 – 2 = 0.0052 < 10⁻²

So, (F_int)₂ = 2.0052

Step 3 :

(F_int)₃ = u₃³+9u₃²+4*u₃ = 0.4584³+9*0.4584²+4*0.4584 = 3.821

R⁽⁰⁾ = f_int − F_ext = 3.821– 3 = 0.821

||R⁽⁰⁾|| > 10⁻² ⇒ Newton-Raphson iterations are necessary.

the correction is

δu⁽¹⁾ = −[k(u₃⁽⁰⁾)]⁻¹ * R⁽⁰⁾ = -[ 3*0.4584²+18*0.4584+4 ]^-1 * 0.821 = -0.0637

u₃¹=u₃⁰+∂u¹ = 0.4584-0.0637 =0.3947

(F_int)₃¹ = u₃³+9*u₃²+4*u₃ = 0.3947³ + 9*0.3947²+4*0.3947=3.0424

R⁽¹⁾ = f_int − F_ext = 3.0424 – 3 = 0.0424> 10^-2

the correction δu⁽²⁾ = −[k(u₃⁽¹⁾)]⁻¹ * R⁽¹⁾ = -[3*0.3947²+18*0.3947+4]^-1 * 0.0424 = -0.0037

u₂²=u₂⁰+u₂¹+∂u² = 0.4584-0.0637-0.0037 =0.3911

(F_int)₁² = u₁³+9*u₁²+4*u₁ = 0.3911³ + 9*0.3911²+4*0.3911=3.0009

R⁽³⁾ = f_int − F_ext = 3.0009 – 2 = 0.0009< 10⁻²

So, (F_int)₃ = 3.0009

Step i	∆Fi	ui	(Fext)i	(fint)i	fint − Fext = R
1	1	0.1777	1	1.0006	0.0006
2	1	0.2967	2	2.0052	0.0052
3	1	0.3911	3	3.0009	0.0009

RESULT :

The drift from the exact solution illustrates the lack of equilibrium between the internal and external forces. To correct for this problem an implicit analysis is required.