# Swing Bolt Closure Design

## 1. Introduction

A swing bolt is usually used in pressure vessels that have covers that need to be opened many times in a realativelly short period (let’s say, twice a month, for example). This is the case of filters, in which you have to open it to change the filtration element. You can also use this kind of clousure when you want to avoid thick plate for the body flange.The procedures described in this article consider the design of the swing bolt closure using ASME standard code and equations from strength of materials, which can be solved by hand or with softwares like Excel. If the designer whichs to design a more reliable and cheap clousure, it’s recommended to use a finite element analysis (FEA) software like Ansys or SolidWorks Simulation. The design using FEA softwares will be covered in another article.

## 2. Parts

The swing bolt closure can be separated in five parts: cover flange, body flange, eye bolt, pin and lugs. The calculation of the lug, pin and eye bolt will be based in the strength of materials, the body flange will be calculated using ASME BPVC Section VIII Division 1 Appendix 2 (Mandatory Appendix 2 – Rules for Bolted Flange Connections with Ring Type Gasket) with some considerations described in the Section 2.2 and for the cover flange the designer can either calculate it using Appendix 2 or use an ASME B16.5 (Mandatory Appendix 2 – Rules for Bolted Flange Connections with Ring Type Gaskets) flange and cut the holes.The allowable stress, minimum tensile strength, minimum yield strength and modulus of elasticity of the material used in the calculations can be found in the ASME BPVC Section II.

### 2.1. Cover Flange

As mentioned above, the cover flange can either be calculated using Appendix 2 or made from an ASME B16.5 flange. If you use the B16.5 flange, just cut the holes as shown in the Figure 2.The design of bolted flanged according to the Appendix 2 will not be described here, as there’s already an article about it in the site: Bolted Flange Design.If the calculated flange is too thick, you can use a compound cover, made of a dished head, a body flange (see Section  2.2) and lugs (see Section 2.5), as shwon in Figure X.

### 2.2. Body Flange

The body flange is the part where the lugs will be welded. It will be designed according to the Appendix 2, with the exception that the flange outside diameter will be the outside diameter of the rolled plate and the flange thickness will be the same as the flange hub length. With these changes it is considered that the lugs don’t contribute to the strength of the flange, even though it isn’t all true, but it’s necessary because the Appendix 2 don’t consider welded lugs in the design of the flanges.

### 2.3. Bolt

The size and quantity of the bolts are selected when designing the body and cover flanges. If you choose to use standard bolts no further action is required.There are three types of bolts that is usually used with swing bolts clousure:
1. Hex bolts
2. Eye bolts
3. Lifting lug with threaded rod

### 2.4. Pin

The calculation of the pin will be use to find the maximum

### 2.5. Lug

There are three types of lugs used in swing bolt closures:

#### 2.5.1. Lug Properties

For the calculation of the stresses acting in the lug, three important properties must be defined: moment of inertia ($I_{lug}$), first moment of the area ($Q_{lug}$) and center of gravity ($y_{c}$).Consider the cross-section shown in the Figure X, where $a$ is the lug height, $c$ the lug width and $b$ the lug thickness. The force will act along the $y$ axis, so we are interested in finding the moment of inertia and center of gravity related to it. The radii due to the shape bending will not be considerede as it’ll not influency that much in the results and will simplify our equations.

Figure X – Lug cross-section

First, find the center of gravity using Equation X: \begin{aligned} \tag{1} y_c = \cfrac{2 \cdot a^2 + 4 \cdot a \cdot b + c \cdot b}{4 \cdot a + 2 \cdot c}\end{aligned} Then, find the moment of inertia using Equation X: \begin{aligned} \tag{2} I_{lug} &= 2 \cdot a \cdot b \cdot \left(\cfrac{a}{2} + b - y_c \right)^2 + c \cdot b \cdot \left(y_c - \cfrac{b}{2} \right)^2 \\&+ \cfrac{2 \cdot b \cdot a^3 + c \cdot b^3}{12} \end{aligned} And finally find the first moment of the area above the center line ($y_c$) using Equation X: \begin{aligned} \tag{3} Q_{lug} = b \cdot \left(a - y_c \right)^2 \end{aligned}

#### 2.5.2. Bending Stress

T0 calculate the bend stress in the lug we’ll consider it as a beam as shown in Figure X.

Figure X – Lug length and force

The maximum bending stress is calculated using Equation X: \begin{aligned} \tag{1} \sigma_{lug} = \cfrac{M \cdot c}{I_{lug}}\end{aligned}Where $M$ is the bending moment ($M = L \cdot F$), $c$ is the distance from the centroid to the extreme fiber ($c = a / 2$) and $I_{lug}$ is the moment of inertia, which is calculated using the Equation X.

#### 2.5.3. Shear Stress

To calculate the shear stress in the lug we use the Equation X: \begin{aligned} \tag{1} \tau_{lug} = \cfrac{F \cdot (a - y_c)^2}{2 \cdot I_{lug}}\end{aligned} Where $F$ is the force acting in the lug, $a$ is the lug height, $y_c$ is the distance from the bottom of the lug to its center of gravity (see Figure X) and $I_{lug}$ is the moment of inertia.

### 2.6. Lug Weld

An all around fillet weld will be used to join the lugs in the body flange. The weld will only be made in the external face of the lugs, as shown in Figure X, as it’s quite impossible to do a quality weld in the internal face.The calculation is quite simples, just follow the steps:
1. Estimate the weld throat size $a$ As there isn’t a global rule accepted by everyone about the proper weld thickness to be used as a first guess or to avoid calculation, I’ll leave it to the designer to use whatever values he likes.
2. Calculate the weld group area $A_w$ The total area of the weld group is the sum of three separated areas, as shown in the Figure X and calculated using Equation 1: \begin{aligned} \tag{1} A_w = a \cdot (2 \cdot H + B)\end{aligned}
3. Calculate the weld group center of gravity $\hat{Y}$ The center of gravity of the weld group about the y-axis is calculated using the Equation 2: \begin{aligned} \tag{1}\hat{Y} = \cfrac{H^2 - \cfrac{a \cdot B}{2}}{2 \cdot H + B} \end{aligned}
4. Calculate the weld group moment of inertia around x-axis $I_{wx}$ The moment of inertia of the weld group around x-axis is calculated using Equation 3: \begin{aligned} \tag{1}I_{wx} &= \cfrac{a \cdot H^3}{6} +2 \cdot a \cdot H \cdot \left(\cfrac{H}{2}-\hat{Y}\right)^2 \\ &+\cfrac{B \cdot a^3}{12} + a \cdot B \cdot \left(\cfrac{a}{2}+\hat{Y}\right)^2\end{aligned}
5. Calculate the weld group moment of inertia around y-axis $I_{wy}$ The moment of inertia of the weld group around the y-axis is calculated using Equation 4: \begin{aligned} \tag{1}I_{wy} &= \cfrac{H \cdot a^3}{6} + 2 \cdot a \cdot H \\ &\cdot \left(\cfrac{a + B}{2}\right)^2 + \cfrac{a \cdot B^3}{12}\end{aligned}
6. Calculate the normal stress vertical to the weld direction $\sigma_{\bot}$ The normal stress vertical to the weld direction is calculated using Equation X: \begin{aligned} \tag{1}\sigma_{\bot}=\cfrac{M\cdot r_y}{I_{wx}}\end{aligned}
7. Calculate the normal stress parallel to the weld direction $\sigma_{\|}$ (Insert something)
8. Calculate the shear stress parallel to the weld direction $\tau_{\|}$ The shear stress parallel to the weld direction is calculated using Equation 8: \begin{aligned} \tag{1}\tau_{\|}=\tau_{y}=\cfrac{F_y}{A_w}\end{aligned}
9. Calculate the equivalent stress in the weld $S_{w}$ The equivalent stress in the weld is calculated using Equation 9 or 10:, \begin{aligned} \tag{1}S_{w} = \sqrt{\sigma_{\bot}^2+\sigma_{\|}^2-\sigma_{\bot}\cdot \sigma_{\|}+\tau_{\bot}^2+\tau_{\|}^2}\end{aligned} If $\sigma_{\|}=0$ then: \begin{aligned} \tag{1}S_{w} = \sqrt{\sigma_{\bot}^2+\tau_{\bot}^2+\tau_{\|}^2}\end{aligned}

## 3. Example

Consider the flange shown in Figure XX. The body was designed using Appendix 2 of the ASME BPVC Section VIII Division 1, as described in the Section 2.2. I used an integral flange, but you can any of flange type available in the Code.
Input Values
Design Temperature: 60 °C Design Pressure: 1,5 MPa
Output Values
Number of bolts: 12 Bolt size: 1/2″-13UNC Bolt matertial: SA-193 B7 Force per bolt: 15000 N

Figure X – Example of swing bolt closure

### 3.1. Lug Properties

Calculation of the center of gravity using Equation X: \begin{aligned} y_c & = \cfrac{2 \cdot a^2 + 4 \cdot a \cdot b + c \cdot b}{4 \cdot a + 2 \cdot c} \\ & = \cfrac{2 \cdot 103^2 + 4 \cdot 103 \cdot 12,7 + 80 \cdot 12,7}{4 \cdot 103 + 2 \cdot 80} \\ & = 48,02 \ mm \\ \end{aligned}Then, find the moment of inertia using Equation X: \begin{aligned} I_{lug} & = 2 \cdot a \cdot b \cdot \left(\cfrac{a}{2} + b - y_c \right)^2 + c \cdot b \cdot \left(y_c - \cfrac{b}{2} \right)^2 \\ & + \cfrac{2 \cdot b \cdot a^3 + c \cdot b^3}{12} \\ & = 2 \cdot 103 \cdot 12,7 \cdot \left(\cfrac{103}{2} + 12,7 - 48,02 \right)^2 \\ & + 80 \cdot 12,7 \cdot \left(48,02 - \cfrac{12,7}{2} \right)^2 \\ & + \cfrac{2 \cdot 12,7 \cdot 103^3 + 80 \cdot 12,7^3}{12} \\ & = 4775667,10 \ mm^4 \end{aligned}And finally find the first moment of the area above the center line ($y_c$) using Equation X: \begin{aligned} Q_{lug} & = b \cdot \left(a - y_c \right)^2 \\ & = 12,7 \cdot \left(103 - 48,02 \right)^2 \\ & = 38389,57 \ mm^3 \end{aligned}

### 3.2. Bending Stress

To calculate the maximum bending stress we use the Equation X: \begin{aligned} \sigma_{lug} & =\cfrac{M \cdot c}{I_{lug}} \\ & = \cfrac{L \cdot F \cdot a}{2 \cdot I_{lug}} \\ & = \cfrac{120 \cdot 1250 \cdot 103}{2 \cdot 4775667,10} \\ & = 1,62 \ MPa \end{aligned}

### 3.3. Shear Stress

To calculate the shear stress we use the Equation X: \begin{aligned} \tau_{lug} & = \cfrac{F \cdot (a - y_c)^2}{2 \cdot I_{lug}} \\ & = \cfrac{1250 \cdot (103 - 48,02)^2}{2 \cdot 4775667,10} \\ & = 0,40 \ MPa \end{aligned}

### 3.4. Lug Weld

Weld stresses calculation:
1. Estimate the weld throat size $a$ We will use a throat size of 5 mm as a first guess.
2. Calculate the weld group area $A_w$ The weld group are will be calculated using the Equation X: \begin{aligned} A_w & = a \cdot (2 \cdot H + B) \\ & = 5 \cdot (2 \cdot 103 + 80) \\ & = 1430 \ mm^2 \end{aligned}
3. Calculate the weld group center of gravity $\hat{Y}$ The center of gravity of the weld group about the y-axis will be calculated using the Equation X: \begin{aligned} \hat{Y} & = \cfrac{H^2 - \cfrac{a \cdot B}{2}}{2 \cdot H + B} \\ & = \cfrac{103^2 - \cfrac{5 \cdot 80}{2}}{2 \cdot 103 + 80} \\ & = 36,40 \ mm \end{aligned}
4. Calculate the weld group moment of inertia around x-axis $I_{wx}$ The moment of inertia of the weld group around x-axis will be calculated using Equation X: \begin{aligned} I_{wx} & = \cfrac{a \cdot H^3}{6} +2 \cdot a \cdot H \cdot \left(\cfrac{H}{2}-\hat{Y}\right)^2 \\ &+\cfrac{B \cdot a^3}{12} + a \cdot B \cdot \left(\cfrac{a}{2}+\hat{Y}\right)^2 \\ & = \cfrac{5 \cdot 103^3}{6} + 2 \cdot 5 \cdot 103 \cdot \left(\cfrac{103}{2} - 36,40\right)^2 \\ & + \cfrac{80 \cdot 5^3}{12} + 5 \cdot 80 \cdot \left(\cfrac{5}{2} + 36,40 \right)^2 \\ & = 1751573,47 \ mm^4 \end{aligned}
5. Calculate the weld group moment of inertia around y-axis $I_{wy}$ The moment of inertia of the weld group around the y-axis will be calculated using Equation X: \begin{aligned} I_{wy} & = \cfrac{H \cdot a^3}{6} + 2 \cdot a \cdot H \\ &\cdot \left(\cfrac{a + B}{2}\right)^2 + \cfrac{a \cdot B^3}{12} \\ & = \cfrac{103 \cdot 5^3}{6} + 2 \cdot 5 \cdot 103 \\ & \cdot \left(\cfrac{5 + 80}{2} \right)^2 + \cfrac{5 \cdot 80^3}{12} \\ & = 2075916,67 \ mm^4 \end{aligned}
6. Calculate the normal stress vertical to the weld direction $\sigma_{\bot}$ The normal stress vertical to the weld direction will be calculated using Equation X: \begin{aligned} \sigma_{\bot} & =\cfrac{M\cdot r_y}{I_{wx}} \\ & = \cfrac{F \cdot L \cdot (H - \hat{Y})}{I_{wx}} \\ & = \cfrac{1250 \cdot 120 \cdot (103 - 36,40)}{1751573,47} \\ & = 5,70 \ MPa \end{aligned}
7. Calculate the shear stress parallel to the weld direction $\tau_{\|}$ The shear stress parallel to the weld direction will be calculated using Equation X: \begin{aligned} \tau_{\|} & = \cfrac{F_y}{A_w} \\ & = \cfrac{1250}{1430} \\ & = 0,87 \ MPa \end{aligned}
8. Calculate the equivalent stress in the weld $S_{w}$ The equivalent stress in the weld will be calculated using Equation X: \begin{aligned} S_{w} & = \sqrt{\sigma_{\bot}^2+\tau_{\|}^2} \\ & = \sqrt{5,70^2 + 0,87^2} \\ & = 5,77 \ MPa \end{aligned}
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