11.1 Threaded Fasteners, Rivets & Welded Joint Analysis

11.1 Threaded Fasteners, Rivets & Welded Joint Analysis

Mechanical joints form the foundational load-bearing links in machine design, structural frameworks, and pressure vessels. Fasteners and joint assemblies must be designed to withstand static loads, cyclic fatigue, shear forces, and bending moments without suffering catastrophic separation or structural yield. This section examines the engineering principles governing threaded fasteners, riveted joints, and structural welds, featuring step-by-step mathematical models and worked calculations required for the PRC Mechanical Engineering Licensure Examination.

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1. Threaded Fasteners & Bolt Mechanics

Threaded fasteners convert rotational torque into clamping force (preload) to secure mechanical assemblies. Fastener geometries are standardized under ISO Metric and Unified National (UNC/UNF) thread systems.

Thread Geometry & Tensile Stress Area

Thread pitch ($p$) is the axial distance between adjacent thread crests. The pitch diameter ($d_2$) and minor diameter ($d_3$) define the internal geometry of external threads. However, under axial tensile loading, thread roots create stress concentrations. The effective load-carrying area of a threaded fastener is defined by the tensile stress area ($A_t$):

$$A_t = \frac{\pi}{4} \left( d - 0.938194 p \right)^2$$

where $d$ is the nominal major diameter (mm or in) and $p$ is the thread pitch (mm or 1/TPI in inches).

Bolt Preload & Tightening Torque

To prevent joint separation and improve fatigue life, bolts are tightened to a initial clamping force called preload ($F_i$). The recommended preload for reusable joints (subject to disassembly) is $75%$ of the bolt's proof load ($F_p$), whereas permanent joints are preloaded up to $90%$ of $F_p$:

$$F_p = A_t S_p$$ $$F_i = 0.75 F_p = 0.75 A_t S_p \quad \text{(reusable joints)}$$

where $S_p$ is the bolt proof strength (typically $85%$ to $90%$ of yield strength $S_y$).

The relation between applied tightening torque ($T$) and resulting bolt preload ($F_i$) is expressed by the standard torque equation:

$$T = K F_i d$$

where:

• $T$ = tightening torque (N·m or lb·in)
• $K$ = torque coefficient (empirical factor accounting for thread friction and collar friction)
  • $K \approx 0.20$ for unlubricated/as-received steel bolts
  • $K \approx 0.15$ for lubricated threads
  • $K \approx 0.12$ for cadmium-plated or PTFE-coated fasteners
• $d$ = nominal bolt major diameter (m or in)

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2. Bolted Joint Stiffness & External Load Distribution

When an external tensile load ($P$) is applied to a preloaded bolted joint, the load is partitioned between the bolt and the clamped members according to their relative elastic stiffnesses.

Joint Stiffness Constant ($C$)

The bolt acts as a tension spring with spring rate $K_b$, while the clamped members act as a compression spring with combined spring rate $K_m$:

$$K_b = \frac{A_b E_b}{L_b}$$

where $A_b$ is the unthreaded shank area, $E_b$ is the modulus of elasticity, and $L_b$ is the grip length. The joint stiffness constant ($C$) represents the fraction of external load transferred to the bolt:

$$C = \frac{K_b}{K_b + K_m}$$

Typically, $C$ ranges from $0.15$ to $0.35$ for standard metal-to-metal joints without soft gaskets, meaning clamped members absorb $65%$ to $85%$ of any external tensile force.

Load Distribution & Joint Separation

Under external tensile load $P$ per bolt:

• Total Bolt Load ($P_b$): $P_b = C P + F_i$
• Total Member Load ($P_m$): $P_m = (1 - C) P - F_i$
• Joint Separation Condition: Joint separation occurs when clamping force in members drops to zero ($P_m = 0$). The external load required to cause joint opening ($P_{\text{sep}}$) is:

$$P_{\text{sep}} = \frac{F_i}{1 - C}$$

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Worked Step-by-Step Calculation: Bolted Joint Analysis

Problem: An M16 x 2.0 Class 8.8 metric bolt ($S_p = 600 \text{ MPa}$) with a tensile stress area $A_t = 157 \text{ mm}^2$ is used in a rigid flange connection. The joint stiffness ratio is $C = 0.25$. Calculate:

1. The recommended preload $F_i$ for a reusable connection.
2. The required tightening torque $T$ assuming an unlubricated condition ($K = 0.20$).
3. The maximum resultant load on the bolt when an external tensile load $P = 30 \text{ kN}$ is applied to the joint.

Step 1: Calculate Proof Load and Preload

$$F_p = A_t S_p = 157 \text{ mm}^2 \times 600 \text{ N/mm}^2 = 94,200 \text{ N} = 94.20 \text{ kN}$$ $$F_i = 0.75 F_p = 0.75 \times 94.20 \text{ kN} = 70.65 \text{ kN} = 70,650 \text{ N}$$

Step 2: Calculate Tightening Torque Convert nominal diameter $d = 16 \text{ mm} = 0.016 \text{ m}$:

$$T = K F_i d = 0.20 \times 70,650 \text{ N} \times 0.016 \text{ m} = 226.08 \text{ N\cdot m}$$

Step 3: Calculate Tensile Load on Bolt under External Load

$$P_b = C P + F_i = (0.25 \times 30 \text{ kN}) + 70.65 \text{ kN} = 7.50 \text{ kN} + 70.65 \text{ kN} = 78.15 \text{ kN}$$

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3. Riveted Joints & Failure Analysis

Riveted joints are permanent mechanical connections widely used in pressure vessels, structural trusses, and aerospace fuselages. Joints are classified as lap joints (overlapping plates) or butt joints (butt-aligned plates connected via single or double cover straps).

Lap Joint (Single Shear):            Butt Joint (Double Strap / Double Shear):
   Plate A ----->                      Strap Top ------> [===]
   ==============                         Plate A =====|===== Plate B
   ==============                      Strap Bottom ---> [===]
   <----- Plate B

Primary Failure Modes of Riveted Joints

For a riveted joint of pitch $p$ (distance between rivet centers), plate thickness $t$, rivet hole diameter $d$, allowable tensile stress $\sigma_t$, allowable shear stress $\tau$, and allowable crushing stress $\sigma_c$:

1. Tearing Failure of Plate Between Rivets ($P_t$):

   $$P_t = (p - d) t \sigma_t$$

2. Shear Failure of Rivets ($P_s$):

   • Single Shear (Lap joint or single-strap butt joint): $$P_s = n \cdot \frac{\pi}{4} d^2 \tau$$
   • Double Shear (Double-strap butt joint, ASME Boiler Code factor $1.875$): $$P_s = 1.875 \cdot n \cdot \frac{\pi}{4} d^2 \tau$$

   (where $n$ is the number of rivets per pitch length)

3. Crushing (Bearing) Failure of Plate or Rivet ($P_c$):

   $$P_c = n \cdot d \cdot t \cdot \sigma_c$$

4. Margin Shearing/Tearing: Prevented by maintaining a margin distance $m \ge 1.5 d$ from the plate edge to rivet center.

Joint Efficiency ($\eta$)

Joint efficiency is the ratio of the minimum joint strength to the strength of an unpunched solid plate of width $p$:

$$P_{\text{solid plate}} = p \cdot t \cdot \sigma_t$$ $$\eta = \frac{\min(P_t, P_s, P_c)}{P_{\text{solid plate}}} \times 100\%$$

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Worked Step-by-Step Calculation: Riveted Joint Efficiency

Problem: A single-riveted lap joint connects two $10 \text{ mm}$ thick steel plates using $20 \text{ mm}$ diameter rivets spaced at a pitch of $60 \text{ mm}$. Allowable stresses are $\sigma_t = 120 \text{ MPa}$, $\tau = 90 \text{ MPa}$, and $\sigma_c = 160 \text{ MPa}$. Determine the joint efficiency.

Step 1: Calculate Solid Plate Strength per Pitch

$$P_{\text{solid}} = p \cdot t \cdot \sigma_t = 60 \text{ mm} \times 10 \text{ mm} \times 120 \text{ N/mm}^2 = 72,000 \text{ N}$$

Step 2: Calculate Tearing Strength ($P_t$)

$$P_t = (p - d) t \sigma_t = (60 - 20) \text{ mm} \times 10 \text{ mm} \times 120 \text{ N/mm}^2 = 48,000 \text{ N}$$

Step 3: Calculate Rivet Shear Strength ($P_s$, Single Shear)

$$P_s = 1 \times \frac{\pi}{4} (20)^2 \text{ mm}^2 \times 90 \text{ N/mm}^2 = 314.16 \times 90 = 28,274 \text{ N}$$

Step 4: Calculate Crushing Strength ($P_c$)

$$P_c = 1 \times 20 \text{ mm} \times 10 \text{ mm} \times 160 \text{ N/mm}^2 = 32,000 \text{ N}$$

Step 5: Determine Minimum Strength and Efficiency The governing (lowest) failure load is rivet shear: $P_{\min} = P_s = 28,274 \text{ N}$.

$$\eta = \frac{28,274 \text{ N}}{72,000 \text{ N}} \times 100\% = 39.27\%$$

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4. Welded Joints & Throat Area Stress Analysis

Structural welds fall into two primary types: butt welds (groove welds) and fillet welds.

Butt Welds

For a full-penetration butt weld of plate thickness $t$ and length $L$, the effective throat is equal to plate thickness $t$. Tensile stress is:

$$\sigma = \frac{P}{t L}$$

Fillet Welds

Fillet welds transfer loads through shear along the minimum cross-section known as the effective throat ($t$). For a standard equal-leg fillet weld with leg size $h$:

$$t = h \sin 45^\circ = h \cos 45^\circ = 0.707 h$$ $$\text{Effective Throat Area } A_w = 0.707 h L$$

where $L$ is the total length of the fillet weld.

Fillet Weld Geometry:
     |\
     | \
 Leg |  \  Hypotenuse
  h  |   \ 
     |____\  <--- Effective Throat t = 0.707 h
       Leg h

Primary & Secondary Shear in Eccentric Welded Joints

When an eccentric load $P$ acts at distance $e$ from the centroid of a weld group, it generates a direct shear force $P$ and a twisting moment $M = P e$.

1. Primary Shear Stress ($\tau'$):

   $$\tau' = \frac{P}{A_w} = \frac{P}{0.707 h L_{\text{total}}}$$

2. Secondary Shear Stress ($\tau''$):

   $$\tau'' = \frac{M r}{J_w}$$

   where $r$ is the distance from the weld group centroid to the critical weld point, and $J_w$ is the polar moment of inertia of the weld throat group ($J_w = 0.707 h \cdot J_{\text{unit}}$).

3. Resultant Maximum Shear Stress ($\tau_{\max}$):

   $$\tau_{\max} = \sqrt{(\tau')^2 + (\tau'')^2 + 2 \tau' \tau'' \cos \theta}$$

   where $\theta$ is the angle between the primary and secondary shear stress vectors at the critical location.

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Worked Step-by-Step Calculation: Parallel Fillet Weld Sizing

Problem: A steel plate is welded to a rigid frame using two parallel fillet welds, each of length $L = 100 \text{ mm}$. The joint transmits a static transverse tensile load $P = 50 \text{ kN}$. If the allowable shear stress of the weld metal is $\tau_{\text{allow}} = 80 \text{ MPa}$, determine the required weld leg size $h$.

Step 1: Express Total Throat Area Total length $L_{\text{total}} = 2 \times 100 \text{ mm} = 200 \text{ mm}$.

$$A_w = 0.707 h L_{\text{total}} = 0.707 \times h \times 200 = 141.4 h \text{ mm}^2$$

Step 2: Set Primary Shear Stress to Allowable Limit

$$\tau = \frac{P}{A_w} \implies 80 \text{ N/mm}^2 = \frac{50,000 \text{ N}}{141.4 h \text{ mm}^2}$$

Step 3: Solve for Leg Size $h$

$$h = \frac{50,000}{141.4 \times 80} = \frac{50,000}{11,312} = 4.42 \text{ mm}$$

Rounding up to standard weld sizes yields a minimum required leg size of $5.0 \text{ mm}$.