Table 6-3: Typical values at failure for plates

3 AISI 304 2 0.03 6 0.07 3.4 0.1 3.2 4 AISI 1010 _{2 0.07 7.6 0.13 5.0 0.2 3.7} 5 AISI 304 3 0.15 8.2 0.45 7.6 0.5 6.8 6 AISI 1010 3 0.3 9.9 0.6 8.5 0.9 9

Table 6-2: Typical values at yield for plates.

1

# Material Support

Type

2 mm Plate

5 mm Plate

6 mm Plate

Pressure value at failure Maximum deflection before rupture Pressure value at failure Maximum deflection before rupture Pressure value at failure Maximum deflection before rupture

(Bar) (mm) (Bar) (mm) (Bar) (mm)

1 AISI 304 1 18.4 195 48 192 56 188 2 AISI 1010 _{1 8 87 18.9 83 21 76} 3 AISI 304 2 1.9 52.7 4.8 50.6 6.3 50 4 AISI 1010 2 0.6 19 1.4 18 1.9 17 5 AISI 304 3 5 61.5 13.5 63 15.2 59 6 AISI 1010 _{3 2.4 24 4.7 21 5.5 20.7}

Table 6-3: Typical values at failure for plates.

2

The following general observations can be made from Table 6-2 and Table 6-3: 3

• A simply-supported plate can withstand higher pressure than a bolted plate because of a rotational degree 4

of freedom at boundaries, which transfers the critical failure stress to the geometric center of plate and 5

hence reduces high localised stress on the support. 6

• A plate with a higher number of bolts can withstand higher pressure than a plate with a lower number of 7

bolts. A plate with a smaller number of bolts fails due to high localised stress around these bolts. 8

• A thicker plate will deform less and will withstand higher pressure. 9

• The ductility of material AISI 304 helps it in withstanding higher overpressures even though it has low yield 10

strength. 11

This section provided the basic understanding of the effects of different types of material and support on the 12

mechanical strength of metal plates. A basic assumption is made that the load acting on the plates is static, which 13

means it is applied slowly so that dynamic effects can be neglected, but it has provided a guideline for the strength 14

of plates. 15

In the next section, the rate of rise of overpressure will be taken into account when estimating the mechanical 16

stresses during an internal arc. 17

6.2.2 Calculation results for the failure of metal plates due to dynamic

1

overpressure

2

Since the internal arc overpressure is a short duration time-varying load, the inertial effect has to be considered 3

when calculating deformation and stresses. The structure behaves differently if the force is applied slowly 4

compared to when it is applied in an instant, like hitting the wall with a hammer. 5

Mathematical models are created for dynamic loading using a simplified mass-spring representation for the 6

structure. This can be used to estimate the deformation for simplified geometry in the elastic and plastic regimes

.

7

6.2.2.1 ELASTIC DEFORMATION REGIME

8

The method is based on converting a structure into a spring mass system and solving it for arbitrary excitation. 9

Equation (6-1) is the governing equation. 10

𝑑

2

_𝑥

𝑑𝑡

2

+ 2𝜁𝜔

0

𝑑𝑥

𝑑𝑡 + 𝜔

02

𝑥 =

𝐹(𝑡)

𝑚

(6-1)

11

Where 𝑥 is deformation, 𝑡 is time, 𝑚 is mass, 𝐹(𝑡) is force, 𝜔₀ is the natural frequency, and 𝜁 is the damping ratio.

12

The equation can be solved analytically or numerically for an arbitrarily-varying load for small time steps to yield 13 displacement at time 𝑡_𝑖 14

𝑦(𝑡

𝑖

) =𝑒

−𝜁𝜔𝑡𝑖

𝑚𝜔

𝐷

{𝐴

𝐷

(𝑡

𝑖

) sin 𝜔

𝐷

𝑡

𝑖

− 𝐵

𝐷

(𝑡

𝑖

) cos 𝜔

𝐷

𝑡

𝑖

}

(6-2)

where 15

𝐴

𝐷

(𝑡

𝑖

) = 𝐴

𝐷

(𝑡

𝑖−1

) + � 𝐹(𝜏)𝑒

𝜁𝜔𝜏

cos 𝜔

𝐷

𝜏𝑑𝜏

𝑡𝑖 𝑡𝑖−1

(6-3)

𝐵

𝐷

(𝑡

𝑖

) = 𝐵

𝐷

(𝑡

𝑖−1

) + � 𝐹(𝜏)𝑒

𝜁𝜔𝜏

sin ω

D

𝜏𝑑𝜏

𝑡𝑖 𝑡𝑖−1

(6-4)

where 𝜔_𝐷 is the damped natural frequency and 𝐶_𝑖 is the damping.

16 17

So, for a given time-varying overpressure the deformation of a simplified plate can be calculated in the elastic 18

regime with equation (6-2). The stresses are calculated from the resulting displacement and compared against the 19

mechanical failure properties of the material. 20

It is important to remember that equation (6-2) is valid for displacement in the elastic regime only when there is no 21

permanent deformation. 22

6.2.2.2 ELASTO-PLASTIC REGIME

1

When the load is so high that the structure undergoes plastic deformation, the above mentioned equation (6-2) is 2

not valid, as it doesn’t account for changes in the stiffness of the structure in the plastic region. 3

So equation (6-2) is rewritten for a given instant of time for the next incremental change in loading. This is then 4

solved for a different stiffness in the plastic region. For simple cases and low plastic deformation, the restoring force 5

is considered constant after the yield point. However, a separate non-linear stiffness function can be used for 6

elastic and plastic parts if needed. 7

The dynamics equation for a nonlinear single degree of freedom system becomes incremental at a given time 𝑡_𝑖 8

𝑚𝛥𝑦̈

𝑖

+ 𝑐

𝑖

𝛥𝑦̇

𝑖

+ 𝑘

𝑖

𝛥𝑦

𝑖

= 𝛥𝐹

𝑖

(6-5)

where 𝑦 is displacement , 𝑦̇ is velocity and 𝑦̈ is acceleration for 𝐹, which is the applied force, Δ represents the 9

change in value during an infinitesimal time increment at the 𝑖th time step, 𝑚 is the mass , 𝑐 is damping, and 𝑘 is 10

stiffness. 11

The solution of equation (6-5) yields 12

𝑘�

𝑖

Δ𝑦

𝑖

= Δ𝐹�

𝑖

(6-6)

where 𝑘�, 𝐹� are effective stiffness and force: 13

𝑘�

𝑖

= 𝑘

𝑖

+6𝑚_Δ𝑡

₂

+3𝑐_Δ𝑡

𝑖

(6-7)

Δ𝐹�

𝑖

= Δ𝐹

𝑖

+ 𝑚 �_{Δ𝑡 𝑦̇}6

𝑖

+ 3𝑦̈

𝑖

� + 𝑐

𝑖

�3𝑦̇

𝑖

+Δ𝑡_{2 𝑦̈}

𝑖

�

(6-8)

The stiffness for the 𝑖th time step is taken from the actual force versus displacement graph of the material or it can 14

be assumed constant for the elastic region and ‘0’ for plastic region. The above equations can be solved for each 15

time step starting from the given initial condition. The stresses are calculated from the deformation plot of the 16

above equations. 17

As mentioned earlier the applicability of above equation becomes difficult for complex structures and high plastic 18

deformations. For example, take the case of a plate which is deformed by a given time-varying load and undergoes 19

plastic deformation. However this plastic deformation is not uniform throughout the plate. A part of the plate 20

undergoes plastic deformation whereas the other areas remain elastically deformed. This presents a problem for 21

stiffness being used in the above-mentioned formula at a given time step. 22

So the next logical step is to divide the plate into small discrete subsections and define the stiffness of these 23

subsections based on an elastic or plastic regime. The division of the structure into discrete elements can be 24

handled efficiently with Finite Element Analysis (FEA). The finite element method is a numerical method for solving 25

partial differential equations in physical systems. The solution of standard differential equations exists for simplified 26

cases. When complicated geometry or non-linearity are involved, an approximate numerical solution is the right 27

way to proceed. The finite element formulation results in simultaneous algebraic equations for the degrees of 28

freedom. These algebraic equations provide a solution for the primary variable at a discrete number of points rather 29

than during the continuous solution. The mechanical body is modeled as the union of smaller sub-bodies called 1

finite elements. The response of these finite elements is formulated by the governing differential equation. The 2

assembly of these elements provides the solution for the complete structure. So the finite element method 3

formulates the equations for finite elements and combines them to get the solution for the complete body. 4

6.2.3 Application

5

As stated earlier certain areas suffer plastic deformation and others elastic deformation. In order to account for this 6

non-uniform stiffness, finite difference/element methods are used. One case from the previous internal arc tests is 7

taken to check the validity of the method (Case A in Table 2-3). The before and after test pictures of the switch tank 8

are shown in Figure 6-6. FEA analysis was used to calculate the von Mises stress and the static mechanical 9

deformation of this tank under two different conditions; in the first simulation the tank was filled with SF6 gas. In the

10

second simulation the tank was filled with air (see Table 6-4). 11

Simulation no. 1 Simulation no. 2

Type of Gas SF6 Air

Construction: Welded Welded

Steel: AISI, 1010, 6 mm thick AISI 1010, 6 mm thick Maximum Pressure: 13.8 bars in 70 ms (from

measurement)

13.8 bars in 12 ms (from simulations, using basic model)

Table 6-4: Two cases that are used for the FEA analysis of the mechanical deformation and von Mises

In document TOOLS FOR THE SIMULATION OF EFFECTS OF THE INTERNAL ARC IN TRANSMISSION AND DISTRIBUTION SWITCHGEAR (Page 80-83)