2.5 Threaded Fasteners, Welded Connections & Mechanical Joints

Threaded Fasteners, Welded Connections & Mechanical Joints

Structural integrity in mechanical assemblies depends entirely on the design of mechanical joints. Threaded fasteners and welded connections transfer static and fluctuating multiaxial loads across structural interfaces. Understanding bolt preloading physics, elastic load-sharing stiffness constants ($C$), joint separation prevention, and weld group throat shear vectors is mandatory for machine design engineering.

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1. Threaded Fasteners, Bolt Grades & Proof Strength

+-----------------------------------------------------------------------------+
|                      THREADED FASTENER DEFINITIONS                          |
|                                                                             |
|   Tensile Stress Area:    A_t = \frac{\pi}{4}\left(d - 0.938194 p\right)^2    |
|                                                                             |
|   Proof Strength (S_p):   \approx 0.85 S_y \quad (\text{Stress without permanent set}) |
|                                                                             |
|   Bolt Preload (F_i):     F_i = 0.75 A_t S_p \quad (\text{Reusable joints})   |
|                           F_i = 0.90 A_t S_p \quad (\text{Permanent joints})  |
|                                                                             |
|   Tightening Torque:      T = K F_i d                                       |
|                           (K \approx 0.20 \text{ as-received, } 0.15 \text{ lubricated}) |
+-----------------------------------------------------------------------------+

Common Bolt Property Classes & Grades

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2. Bolted Joint Elastic Interaction & Joint Constant $C$

A preloaded bolted joint acts as two parallel springs: the bolt is stretched in tension (stiffness $k_b$), while the clamped joint members are squeezed in compression (stiffness $k_m$).

                       Bolt (Tension Spring k_b)
                          +---/\/\/\/\/\---+   
                          |                |   
   External Load P <======+  CLAMPED ZONE  +======> External Load P
                          |                |   
                          +---/\/\/\/\/\---+   
                     Members (Compression Spring k_m)

+-----------------------------------------------------------------------------+
|                        BOLTED JOINT MECHANICS                               |
|                                                                             |
|   Joint Stiffness Constant:      C = \frac{k_b}{k_b + k_m}                  |
|                                  (Typical range: 0.15 \le C \le 0.30)        |
|                                                                             |
|   Total Bolt Tensile Load:       F_b = F_i + C P                            |
|                                                                             |
|   Total Member Clamping Load:    F_m = F_i - (1 - C) P                      |
|                                                                             |
|   Joint Separation Load:         P_{\text{sep}} = \frac{F_i}{1 - C}          |
|                                                                             |
|   Proof Safety Factor:           n_p = \frac{S_p A_t}{F_b} = \frac{S_p A_t}{F_i + C P} |
|                                                                             |
|   Joint Separation Safety Factor: n_{\text{sep}} = \frac{P_{\text{sep}}}{P} = \frac{F_i}{P (1 - C)} |
+-----------------------------------------------------------------------------+

[!NOTE] Because member stiffness $k_m$ is typically 3 to 5 times larger than bolt stiffness $k_b$, the joint constant $C$ is small ($C \approx 0.20$). When an external tensile force $P$ is applied, the bolt feels only 20% of the load, while 80% goes into decompressing the clamped flange members!

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3. Bolt Fatigue under Fluctuating Tensile Load

Consider an external tensile load cycling repeatedly from $0$ to $P$:

• Bolt Minimum Force: $F_{b, \text{min}} = F_i$
• Bolt Maximum Force: $F_{b, \text{max}} = F_i + C P$

$\text{Alternating Force: } F_a = \frac{F_{b, \text{max}} - F_{b, \text{min}}}{2} = \frac{C P}{2} \implies \sigma_a = \frac{C P}{2 A_t}$ $\text{Mean Force: } F_m = \frac{F_{b, \text{max}} + F_{b, \text{min}}}{2} = F_i + \frac{C P}{2} \implies \sigma_m = \frac{F_i + C P / 2}{A_t}$

Modified Goodman Fatigue Safety Factor: $n_f = \frac{S_e (S_{ut} - \sigma_i)}{\sigma_a (S_{ut} + S_e)}$ Where $\sigma_i = F_i / A_t$ is initial preload stress.

   Key Engineering Principle:
   Alternating stress \sigma_a depends ONLY on (C * P) and is INDEPENDENT of preload F_i.
   Higher preload does NOT increase fatigue alternating amplitude, but prevents joint separation!

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4. Welded Connections: Fillet & Groove Welds

+-----------------------------------------------------------------------------+
|                        FILLET WELD THROAT MECHANICS                         |
|                                                                             |
|   Leg Size:                     h (or w)                                    |
|                                                                             |
|   Effective Throat:             t = h \cos 45^\circ = 0.707 h               |
|                                                                             |
|   Effective Throat Area:        A_w = 0.707 h L                             |
|                                                                             |
|   Allowable Shear Stress (AWS): \tau_{\text{allow}} = 0.30 S_{ut, \text{electrode}} |
|                                 (E70XX: S_{ut}=70 kpsi \implies \tau_{\text{allow}}=21 kpsi = 145 MPa) |
+-----------------------------------------------------------------------------+

        Plate 1
     +------------+
     |            |\ 
     |            | \  <--- Fillet Weld (Leg h)
     |            |  \ 
     +------------+---+-------------------+
                  |   |  Throat t = 0.707h
                  |   |                   |
                  +---+-------------------+
                           Plate 2

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5. Eccentric Loading on Weld Groups (Direct Shear + Torsion)

When a load $V$ acts eccentrically at distance $e$ from the center of gravity (centroid $G$) of a weld group, it induces simultaneous primary direct shear ($\tau'$) and secondary torsional shear ($\tau''$):

+-----------------------------------------------------------------------------+
|                     ECCENTRIC WELD GROUP FORMULATION                        |
|                                                                             |
|   1. Primary Direct Shear:    \tau' = \frac{V}{A_w} = \frac{V}{0.707 h \sum L_i} |
|                                                                             |
|   2. Secondary Torsional Shear: \tau'' = \frac{M r}{J_w} = \frac{(V e) r}{0.707 h J_u} |
|      (Where J_u = I_{ux} + I_{uy} is unit polar moment of inertia of weld lines) |
|                                                                             |
|   3. Vectorial Superposition: \tau_{\text{total}} = \sqrt{(\tau'_x + \tau''_x)^2 + (\tau'_y + \tau''_y)^2} \le \tau_{\text{allow}} |
+-----------------------------------------------------------------------------+

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6. Step-by-Step Worked Engineering Problem

Problem Statement

A rigid steel bracket is attached to a vertical column using two identical vertical fillet welds of length $L = 120\text{ mm}$ spaced $b = 100\text{ mm}$ apart. An eccentric downward shear load $V = 30\text{ kN}$ is applied at an eccentricity $e = 150\text{ mm}$ from the weld group centroid. The welds are made using E70XX electrodes ($\tau_{\text{allow}} = 145\text{ MPa}$).

Determine the minimum required weld leg size $h$.

                     |<--- e = 150 mm --->|
                     |                    |
      Column         |                    v  V = 30 kN
     +----+    +-----+--------------------+
     |    |==| |                          |
     |    |  | |                          |
     |    |  | |   G (Centroid)           |
     |    |==| |                          |
     +----+    +--------------------------+
          |<b=100>|
          Length L = 120 mm each weld

Step 1: Weld Group Geometry & Centroid

• Two vertical lines of length $L = 120\text{ mm}$ at $x = \pm 50\text{ mm}$.
• Centroid $G$ is at $(0, 0)$. Total length $\sum L = 2(120) = 240\text{ mm}$.
• Critical point is the outermost corner: $x = 50\text{ mm}, y = 60\text{ mm}$. $r = \sqrt{x^2 + y^2} = \sqrt{50^2 + 60^2} = \sqrt{2500 + 3600} = 78.10\text{ mm}$

Step 2: Unit Polar Moment of Inertia ($J_u$)

For two vertical parallel lines of length $L$ separated by distance $b$: $I_{ux} = 2 \left(\frac{L^3}{12}\right) = \frac{2(120)^3}{12} = 288,000\text{ mm}^3$ $I_{uy} = 2 \left(L \left(\frac{b}{2}\right)^2\right) = 2 (120)(50)^2 = 600,000\text{ mm}^3$ $J_u = I_{ux} + I_{uy} = 288,000 + 600,000 = 888,000\text{ mm}^3$

Step 3: Primary Direct Shear Stress Vector ($\tau'$)

Direct downward load: $V_y = -30\text{ kN}$. $\tau'_y = \frac{V}{0.707 h \sum L} = \frac{30,000}{0.707 h (240)} = \frac{176.81}{h}\text{ MPa (Downward)}$ $\tau'_x = 0$

Step 4: Secondary Torsional Shear Stress Vector ($\tau''$)

Clockwise moment: $M = V e = 30,000\text{ N} \times 150\text{ mm} = 4.50 \times 10^6\text{ N}\cdot\text{mm}$. $\tau'' = \frac{M r}{0.707 h J_u} = \frac{(4.50 \times 10^6)(78.10)}{0.707 h (888,000)} = \frac{559.88}{h}\text{ MPa}$

Resolving $\tau''$ into orthogonal components at top-right corner $(+50, +60)$: $\tau''_x = \tau'' \sin \theta = \tau'' \left(\frac{y}{r}\right) = \frac{559.88}{h} \left(\frac{60}{78.10}\right) = \frac{430.13}{h}\text{ MPa (Rightward)}$ $\tau''_y = -\tau'' \cos \theta = -\tau'' \left(\frac{x}{r}\right) = -\frac{559.88}{h} \left(\frac{50}{78.10}\right) = -\frac{358.44}{h}\text{ MPa (Downward)}$

Step 5: Vectorial Superposition & Sizing

$\tau_{\text{total}, x} = \tau'_x + \tau''_x = 0 + \frac{430.13}{h} = \frac{430.13}{h}$ $\tau_{\text{total}, y} = \tau'_y + \tau''_y = -\frac{176.81}{h} - \frac{358.44}{h} = -\frac{535.25}{h}$

$\tau_{\text{resultant}} = \sqrt{\tau_x^2 + \tau_y^2} = \frac{\sqrt{(430.13)^2 + (-535.25)^2}}{h} = \frac{\sqrt{185011.8 + 286492.6}}{h} = \frac{686.66}{h}\text{ MPa}$

Setting $\tau_{\text{resultant}} \le \tau_{\text{allow}} = 145\text{ MPa}$: $h \ge \frac{686.66}{145} = 4.74\text{ mm} \implies \text{Specify standard } h = 5.0\text{ mm or } 6.0\text{ mm leg size}$

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7. Common Exam Traps & PE Pro-Tips

• Trap 1 — Direct Addition of Primary and Secondary Weld Shears: Primary and secondary shear stresses act in different angular directions. Always break $\tau''$ into $x$ and $y$ vector components before adding to $\tau'$.
• Trap 2 — Torque-Tension Unit Mismatch: In $T = K F_i d$, nominal diameter $d$ must be in meters if torque is in $\text{N}\cdot\text{m}$, or inches if torque is in $\text{lb}\cdot\text{in}$.
• Trap 3 — Misunderstanding Preload in Fatigue: Preload does not increase cyclic stress amplitude $\sigma_a$. In fact, higher preload prevents joint separation, keeping the effective joint stiffness constant $C$ active.