Question 13.1: A SDOF elasto-plastic system has a mass 2.5 kg, tangent stif...
A SDOF elasto-plastic system has a mass 2.5 kg, tangent stiffness 250 N/m and viscous damping of 5%. Its yield strength is 5 N. It was subjected to a dynamic loading and its response at 2.20 sec and 2.34 sec was as shown in the table. The step size was 0.02 sec and load increment was 0.1 N. Find the response at (i) 2.22 sec and (ii) 2.36 sec using the Newmark β method.
| Time | Displacement | Velocity | Acceleration | Spring Force | |
| Sec | m | m/s | m/s² | N | N |
| 2.2 | 0.01825 | 0.2205 | −1.8455 | 4.5625 | 0.5 |
| 2.34 | 0.0346 | 0.0147 | −3.3147 | 5 | −3.25 |
tural frequency ω = \sqrt{\frac{250}{2.5} } = 10 rad/sec
Period = 0.628 sec, step size = 0.02 sec
Damping coefficient c = 2mωξ = 2 × 2.5 × 10 = 5.2 Ns/m
Factors β =1/4 and γ = ½.
\overline{k} =k+\frac{\gamma }{\beta \Delta t} c+\frac{1}{\beta (\Delta t)^{2}} m
\overline{k} =250+\frac{0.5}{0.25\times 0.02} c+\frac{1}{0.25\times (0.02)^{2}} m
\overline{k} = 250 + 100c + 1000m = 25500 N/m
Effective load increment is given by
\Delta \overline{p} _{i}=\Delta p_{i}+\left\lgroup\frac{1}{\beta \Delta t}m+\frac{\gamma }{\beta }c \right\rgroup \dot{x} _{i}+\left[\frac{1}{2\beta }m+\Delta t \left\lgroup\frac{\gamma }{2\beta }-1 \right\rgroup c \right] \ddot{x} _{i}
\Delta \overline{p} =\Delta p\left\lgroup\frac{2.5}{0.25\times 0.02}+\frac{0.5\times 2.5}{0.25} \right\rgroup \dot{x} +\left\lgroup\frac{2.5}{2\times 0.25}+0.05\left\lgroup\frac{0.5}{0.5}-1 \right\rgroup 2.5 \right\rgroup \ddot{x}
\Delta \overline{p} =\Delta p+ 505 \dot{x} + 5 \ddot{x}
(i) At time step 2.20 sec,
\Delta \overline{p} =\Delta p+ 505 \dot{x} + 5 \ddot{x} =102.225
\Delta x_{i}=\frac{\Delta \overline{p} }{\overline{k} } =\frac{102.225}{25500} =0.004009
Velocity increment is given by
\Delta \dot{x} =\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}
\Delta \dot{x} =\frac{0.50}{0.25\times 0.02} \Delta x-\frac{0.50}{0.25} \dot{x} +0=100\Delta x-2\dot{x} =-0.0401
Spring force is given by
f_{i+1}=f_{i}+\Delta f= 4.5625+250\times 0.004009= 5.5648 \ngtr 5 N
It shows that between the load step 2.20 and 2.22 sec, the spring has yielded. There is a need to step back and find the displacement corresponding to the yield force 5 N.
Change of State
Consider Figure 13.21
\Delta x=\Delta x_{1}+\Delta x_{2}
Let
\Delta x_{1}=\psi \Delta x, and f_{i}+bc=f_{y}
bc=\psi \Delta xk_{T},
∴ \psi =\frac{(f_{y}-f_{i})}{\Delta x k_{T}}
\psi =\frac{5-4.5625}{0.004009\times 250} = 0.4365,
∴ Δ x_{1} = 0.001750
Again in Figure 13.21,
ab = ac – bc = Δ f_{i}- ψ Δf_{i}=(1 – ψ) Δf_{i}
Also,
\Delta x_{0}=\Delta x_{1}+\Delta x_{p}
The plastic deformation \Delta x_{p} in the element is given by
\overline{k} \Delta x_{p}=(1-ψ)\times\Delta \overline{p} similar to Equation (13.10)
\overline{k}_{T} \Delta x_{0}=\Delta p_{0} (13.10)
where \overline{k} = 25250 because k_{T} = 0 at yield
or,
\overline{k} \Delta x_{p}=(1-0.4365)\times 102.225 or \Delta x_{p}= 0.002281
Total displacement at the end of this step
x_{i+1}=x_{i}+\Delta x_{1}+\Delta x_{p} = 0.01825 + 0.001750 +0.002281 = 0.02228 m
Velocity increment is given by
\Delta \dot{x} =100\Delta x-2\dot{x} = −0.379 m/s
Velocity at the end of the time step is given by
\dot{x} _{i+1}=\dot{x} _{i}+\Delta \dot{x} = 0.1826 m/s
Acceleration at the end of the time step can be found from the equation of motion, that is
\ddot{x} _{i+1}=\frac{p_{i+1}-c\dot{x}_{i+1}-(f_{s})_{i+1} }{m}
\ddot{x} _{i+1}=\frac{0.6-2.5×0.1826-5 }{2.5} = 1.942 m/s²
(ii) At time step 2.34 sec
Given, x = 0.0346 m , \dot{x} = 0.0147 m/s, and \ddot{x} = -3.3147 m/s²
Since, f_{s} = 5 N = f_{y} , stiffness k_{T} = 0
\Delta \overline{p} =\Delta p+505\dot{x} +5\ddot{x} = -0.1 + 505 × 0.0147 + 5 × (-3.3147) = -9.25
\Delta x_{i}=\frac{\Delta \overline{p} }{\overline{k} } =\frac{-9.25}{25250} =-0.0003663 m
Velocity increment is given by
Δ\dot{x} =100 Δx – 2\dot{x} = -0.0660 m/s
Total velocity is given by
\dot{x}_{i+1} = \dot{x}_{i} + Δ \dot{x} = -0.05133 m/s
Since velocity is negative, it means somewhere in between this step, the velocity has become positive to zero and the direction of loading has changed from loading to unloading.
Change of State
Consider Figure 13.22
Setting total velocity = 0,
Δx = Δx_{1} +Δx_{2} numerical
= Δx_{1} + Δx_{2} true
\dot{x}_{i+1} =\frac{2}{\Delta t} (\psi \Delta x)-\dot{x} _{i} = 0 or ψ = −0.4013
Δ x_{1} = ψΔx = 0.000147 m
also,
\overline{k} \Delta x_{2}=(1-\psi )\Delta \overline{p} similar to Equation (13.10)
where, k_{T}= 250 because unloading is always elastic
∴ \Delta x_{2}=\frac{(1-\psi )\Delta \overline{p} }{\overline{k} } =\frac{(1+0.4013)\times (-9.25)}{25500} = -0.0005083 m
Total displacement increment in the time step
Δx = 0.000147 − 0.0005083 = − 0.0003613 m
Displacement at the end of the time step
x_{i+1} = x_{i}+Δx = 0.0346 – 0.0003613 = 0.03424 m
Velocity at the end of the time step
\dot{x} _{i+1}=\dot{x}_{i} +\Delta \dot{x} = 100\Delta x-\dot{x} _{i} = 100 × (-0.0003613) – 0.0147 = -0.05083 m/s
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} = 5 + 250 × (-0.0005083) = 4.8729 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{-3.15-2.5\times (-0.05083)-4.8729 }{2.5} = -3.1583 m/s²
Comments
In any hysteresis model such as those shown in Figures 13.2, 13.9, 13.10 and 13.11, a series of control points are identified at the end of each loading or unloading branch. These control points essentially represent a state of transition. The stiffness at each control point is based on experimental data. A specific computer code can be written for each control point to model loading and unloading states and compute displacement and force increments as illustrated in this example
