Question 13.3: Solve Example 13.2 again using Newmark method with Newton–Ra...

The calculations were carried out in an Excel sheet using the Newmark β method with iterations as discussed. Whenever there is a change in state from elastic to plastic, Newton–Raphson iterations were carried out within the time step until the unbalanced load was zero. The results are shown in Table 13.2.
At 0.3 sec, the elastic force reaches yield level. The tangent stiffness needs to be set equal to nil and iterations are required between 0.3 sec and 0.4 sec time step. The details are shown as follows.
Initialize

x_{i+1}^{(0)}=x_{i} = 0.0115,      f_{s}^{(0)} = 230.2010

\Delta R^{(1)}=\Delta \overline{p}_i=2309.8684, \ \overline{k}_T=226320.0

Cycle 1

\Delta x^{(1)}=0.0102

x_{i+1}^{(1)}=x_{i+1}^{(0)}+\Delta x^{(1)} = 0.0217

\Delta f^{(1)}=f_{s}^{(1)}-f_{s}^{(0)}+(\overline{k} _{T}-k_{T})\Delta x^{(1)}

\Delta f^{(1)}=f_{s}^{(1)}-f_{s}^{(0)}+\left\lgroup\frac{1}{\beta (\Delta t)^{2}}m+\frac{\gamma }{\beta \Delta t} c \right\rgroup \Delta x^{(1)}=2125.5433

\Delta R^{(2)}=\Delta R^{(1)}-\Delta f^{(1)}=184.3251

Table 13.2 Newmark β Method With Newton–Raphson Iteration

Cycle 2

\Delta x^{(2)}=0.0008

x_{i+1}^{(2)}=x_{i+1}^{(1)}+\Delta x^{(1)} = 0.0225

\Delta f^{(2)}=f_{s}^{2}-f_{s}^{(1)}+(\overline{k} _{T}-k_{T})\Delta x^{(2)}=168.0362

\Delta R^{(3)}=\Delta R^{(2)}-\Delta f^{(2)}=16.2889

Cycle 3

\Delta x^{(3)}=0.0001

x_{i+1}^{(3)}=x_{i+1}^{(2)}+\Delta x^{(3)} = 0.0226

\Delta f^{(3)}=f_{s}^{3}-f_{s}^{(2)}+(\overline{k} _{T}-k_{T})\Delta x^{(3)}=14.8494

\Delta R^{(4)}=\Delta R^{(3)}-\Delta f^{(3)}=1.4395

Cycle 4

\Delta x^{(4)}=0.00001

x_{i+1}^{(4)}=x_{i+1}^{(3)}+\Delta x^{(4)} = 0.0226

\Delta f^{(4)}=f_{s}^{4}-f_{s}^{(3)}+(\overline{k} _{T}-k_{T})\Delta x^{(4)}=1.3123

\Delta R^{(5)}=\Delta R^{(4)}-\Delta f^{(4)}=1.4395

Cycle 5

\Delta x^{(5)}=0.000001

x_{i+1}^{(5)}=x_{i+1}^{(4)}+\Delta x^{(5)} = 0.0226

\Delta f^{(5)}=f_{s}^{5}-f_{s}^{(4)}+(\overline{k} _{T}-k_{T})\Delta x^{(5)}=0.1160

\Delta R^{(6)}=\Delta R^{(5)}-\Delta f^{(5)}=1.4395

These results are substituted in the table below the same time step, and the procedure is continued till another change of state is encountered.
At 0.7 sec
At 0.7 sec, the velocity is positive but at 0.8 sec, the velocity is negative. It means there is a change in state between 0.7 and 0.8 sec. It indicates unloading. The tangent stiffness needs to be restored to 20000 N/m. Here, a different strategy will be required as discussed in Figure 13.18 and Example 13.1, Figure 13.22.
Setting total velocity = 0,

\Delta x=\Delta x_{1}+\Delta x_{2}    numerical

=\Delta x_{1}+\Delta x_{2}  true

\dot{x} _{i+1}=\frac{2}{\Delta t} (\psi \Delta x)-\dot{x} _{i} =0

or,    \frac{2}{0.1}   (ψ(-0.0002)-0.0229) = 0,

ψ = -5.725

\Delta x_{1} = ψΔx = 0.001145 m

also,

\overline{k} \Delta x_{2}=(1-\psi )\Delta \overline{p}

where , k_{T} because unloading is always elastic

∴          \Delta x_{2}=\frac{(1-\psi )\Delta \overline{p} }{\overline{k} } =\frac{(1+5.725)\times (-42.4858)}{226320} =-0.001262 m

Total displacement increment in the time step

Δx = 0.001145− 0.001262 = − 0.000117 m

Displacement at the end of the time step

  x_{i+1} =  x_{i}+ \Delta x = 0.0491 – 0.000117 = 0.04898 m

Velocity increment

 \Delta \dot{x} _{i}=\frac{\gamma }{\beta \Delta t} \Delta x_{i}-\frac{\gamma }{\beta } \dot{x} _{i}+\Delta t\left\lgroup1-\frac{\gamma }{2\beta } \right\rgroup \ddot{x} _{i}

or,                    \Delta \dot{x} _{i}=\frac{0.50}{0.25\times 0.1} \Delta x-2\dot{x}

Velocity at the end of the time step

\dot{x} _{i+1}=\dot{x} _{i}+\Delta \dot{x} =20 \Delta x-\dot{x} _{i}=20\times (-0.000117)-0.0229=-0.02524

Similarly, acceleration at the end of the time step can be computed from the equation of motion.
Resisting force at the end of the time step

f_{i+1}=f_{i}+k_{T}\Delta x_{2} = 250+ 20000 × (-0.001262) = 247.66 N

Acceleration at the end of the time step

\ddot{x} _{i+1}=\frac{P_{i+1}-c\dot{x}_{i+1}-(f_{s})_{i+1} }{m}

\ddot{x} _{i+1}=\frac{0-316 \times (-0.02524)-247.66 }{500} = -0.4794 m/s²

Knowing the updated values at 0.8 sec, rest of the calculations can be carried out as usual.