Contents

## 1. Introduction

## 2. Parts

### 2.1. Cover Flange

### 2.2. Body Flange

### 2.3. Bolt

- Hex bolts
- Eye bolts
- Lifting lug with threaded rod

### 2.4. Pin

### 2.5. Lug

#### 2.5.1. Lug Properties

#### 2.5.2. Bending Stress

#### 2.5.3. Shear Stress

### 2.6. Lug Weld

**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.**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}**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}**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}**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}**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}**Calculate the normal stress parallel to the weld direction**\sigma_{\|} (Insert something)**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}**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

**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

### 3.1. Lug Properties

### 3.2. Bending Stress

### 3.3. Shear Stress

### 3.4. Lug Weld

**Estimate the weld throat size**a We will use a throat size of 5 mm as a first guess.**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}**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}**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}**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}**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}**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}**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}