7.3 Non-Threaded Fasteners: Keys, Pins, Clevises, Retaining Rings, Rivets, Lugs & Anchors

7.3 Non-Threaded Fasteners: Pins, Rivets, Retaining Rings & Keys

While threaded fasteners provide clamping preload and straightforward disassembly, mechanical assemblies depend extensively on non-threaded fasteners for positive torque transmission, precise spatial alignment, axial shaft retention, and permanent structural joining. These components include parallel and square shaft keys, splines, machine pins, clevis joints, retaining rings (snap rings), and structural rivets. Each fastener presents unique kinematic constraints, localized stress concentrations, and multi-mode structural failure criteria.

---

Parallel and Square Shaft Keys (ANSI B17.1)

Shaft keys are demountable mechanical elements inserted at the interface between a rotating shaft and a mounted component (such as a gear, pulley, sprocket, or flexible coupling) to provide positive torsional connection while preventing relative rotational slip.

                         Square Key Shaft-Hub Interface
                                [ Hub Body ]
                       +------------------------------+
                       |       Hub Keyseat            |
                       |         +----------+         |
                       |         |   Key    |         |
      -- Shaft Centerline ------>| - - - - -| <-------+-- Horizontal Shear Plane
                       |         | (w x h)  |         |   Area A_s = w * L
                       |         +----------+         |
                       |       Shaft Keyseat          |
                       +------------------------------+
                               [ Solid Shaft ]

Standard Key Dimensions & Geometry

According to ANSI B17.1, keys are standardized based on nominal shaft diameter $d$:

• Square Keys: Standard practice for shafts up to $d = 6.5\text{ in}$ ($165\text{ mm}$). The nominal key width $w$ and height $h$ are equal: $$w = h \approx \frac{d}{4}$$
• Rectangular (Flat) Keys: Used for shafts larger than $6.5\text{ in}$ or for thin-walled hubs where deep keyways would weaken the hub. The width remains $w \approx d/4$, while the height is reduced to $h \approx d/6$.

Keyway Types & Stress Concentration Factors

Cutting a keyseat into a cylindrical shaft creates sharp re-entrant corners, introducing severe geometric stress concentrations that reduce the shaft's torsional and bending fatigue endurance:

• Profile Keyway: Cut with an end mill plunging radially and traversing axially. Its sharp ends produce high stress concentration; Shigley's first-iteration estimates for an end-mill keyseat (bottom radius $r/d \approx 0.02$) are $K_t \approx 2.14$ in bending and $K_{ts} \approx 3.0$ in torsion.
• Sled-Runner Keyway: Cut with a disc milling cutter that runs out gradually at the ends. The gradual runout lowers the bending stress concentration to about $K_t \approx 1.7$, which is why sled-runner seats are preferred near highly stressed shaft sections.

---

Failure Modes: Direct Shear vs. Compressive Bearing

When a shaft of diameter $d$ transmits torque $T$, the tangential force $F$ acting at the shaft surface ($r = d/2$) is:

$$F = \frac{T}{d/2} = \frac{2 T}{d}$$

This force must be resisted by the key across two primary independent failure modes: horizontal shear and side compressive bearing.

1. Direct Horizontal Shear Failure

The tangential force $F$ attempts to shear the key along the horizontal interface separating the shaft keyseat from the hub keyseat. The shear area is $A_s = w L$, where $w$ is key width and $L$ is effective key length:

$$\tau = \frac{F}{A_s} = \frac{2 T}{d \cdot w \cdot L} \le \frac{S_{sy}}{n}$$

Using the Distortion Energy (von Mises) yield criterion for ductile steel ($S_{sy} = 0.577 S_y$):

$$L_{\text{shear}} = \frac{2 T n}{d \cdot w (0.577 S_y)}$$

2. Compressive Bearing Failure

The tangential force compresses the vertical sides of the key. Half the key height ($h/2$) is embedded in the shaft keyseat, and half ($h/2$) is embedded in the hub keyway. The projected compressive bearing contact area is $A_b = (h/2) L$:

$$\sigma_b = \frac{F}{A_b} = \frac{2 T / d}{(h/2) L} = \frac{4 T}{d \cdot h \cdot L} \le \frac{S_{yc}}{n} \approx \frac{S_y}{n}$$

Solving for the minimum required length to prevent compressive crushing:

$$L_{\text{bearing}} = \frac{4 T n}{d \cdot h \cdot S_y}$$

Comparing Shear vs. Bearing for Square Keys ($w = h$)

Taking the ratio of required lengths for a standard square key where $w = h$:

$$\frac{L_{\text{bearing}}}{L_{\text{shear}}} = \frac{\frac{4 T n}{d \cdot w \cdot S_y}}{\frac{2 T n}{d \cdot w (0.577 S_y)}} = \frac{4}{2 / 0.577} = 2 \times 0.577 = 1.155$$

In practice the key length is usually close to the hub length, commonly about $0.75d$ to $1.5d$. Keys much longer than about $1.5d$ do not share load evenly, because shaft twist concentrates the load at one end.

---

Alternative Keys & Involute Splines

      Woodruff Key                   Involute Spline Profile
    +--------------+                        /\      /\      /\
    |              |                       /  \    /  \    /  \
    \              /                      /    \__/    \__/    \
     \____________/                       -- Circular Pitch P --
   (Semicircular Rocking)                (Self-Centering, High Torque)

1. Woodruff Keys

Woodruff keys are semicircular discs that fit into matching curved slots broached or milled into the shaft. Because the key can rock within its semicircular bed, it easily accommodates tapered shafts and self-aligns with matching hub tapers. However, the deep seat cut for a Woodruff key removes more shaft material than a straight keyway, which weakens the shaft; Woodruff keys are therefore favored for smaller shafts, tapered shaft ends and lighter-duty drives.

2. Taper & Gib-Head Keys

Taper keys feature an axial taper of $1/8\text{ in/ft}$ ($1:96$) along their top face, matching an internal taper in the hub keyway. When driven in axially, the key wedges firmly against the shaft and hub, transmitting torque while simultaneously providing axial retention without setscrews. Gib-head keys include an integral hammer head to allow extraction. Limitation: The radial wedging force introduces shaft eccentricity, causing dynamic runout and vibration in high-speed machinery.

3. Involute Splines (ANSI B92.1)

Splines consist of multiple integral longitudinal teeth machined along the external circumference of a shaft, mating with internal broached teeth in a hub. Modern machinery universally favors involute splines over straight-sided (parallel) splines:

• Pressure Angles: Standardized at $30^\circ$, $37.5^\circ$, or $45^\circ$.
• Self-Centering Action: Under torque, the $30^\circ$ or $45^\circ$ involute tooth flanks generate equalizing radial forces that center the hub precisely on the shaft axis.
• Reduced Stress Concentration: Rounded tooth fillets and shallow tooth depth yield far lower stress concentrations than rectangular keyways.
• Torque Capacity Formula: $$T = \frac{1}{2} p D_m h L N_t k_e$$ where $p$ is the allowable bearing pressure (much lower for sliding splines than for fixed splines), $D_m$ is the pitch diameter, $h$ is active radial tooth engagement height, $L$ is engaged spline length, $N_t$ is the number of teeth, and $k_e < 1$ is a load-sharing factor that accounts for spacing errors, because not every tooth carries its full share.

---

Machine Pins: Dowel, Taper, Roll & Cotter

Machine pins serve two distinct functional roles: precision semi-permanent spatial alignment and direct shear load transfer.

        Dowel Pin                   Roll / Spring Pin               Cotter Pin
     +-------------+                 /-------------\                  __
     |  Hardened   |                |   Slotted     |                /  \
     |  & Ground   |                | Spring Steel  |               | || |
     +-------------+                 \-------------/                |_||_|
   (Precision Locate)              (Tolerance Absorbing)          (Nut Retain)

• Dowel Pins: Solid cylindrical pins manufactured from alloy steel, case- or through-hardened ($R_c 50\text{--}60$), and precision ground to tight diametral tolerances ($\pm 0.0001\text{ in}$). Used in reamed holes to accurately fix the relative spatial position of split housings, gearbox casings, and die plates before tightening clamping bolts.
• Taper Pins: Feature a standard uniform taper of $1/4\text{ in/ft}$ ($1:48$). When driven into matched reamed tapered holes, they wedge solidly, resisting loosening while allowing rapid disassembly.
• Roll / Spring Pins: Formed by rolling high-carbon spring steel into a slotted or coiled tubular cylinder. Manufactured intentionally oversize, they compress elastically when driven into standard drilled holes. The continuous radial expansion spring force holds the pin firmly in place without precision reaming, absorbing shock loads and thermal expansion.
• Cotter Pins: Split-shank wire pins inserted through transverse drilled holes in clevis pins or castle nuts, with the split tines flared outward to provide a positive mechanical retention stop.

---

Clevis Joints: Multi-Mode Failure Criteria

A clevis joint connects an actuator rod (male tang/eye) to a mounting bracket (female fork) via a cylindrical clevis pin, accommodating pure angular pivoting in a single plane. When an axial tensile load $P$ is applied, the joint must be analyzed against five distinct structural failure modes:

                         Clevis Joint Failure Planes
                                      P
                                      ^
                                      | (Tang / Eye)
                           +----------+----------+
                           |         (O)         | <-- Mode 4: Net Tension
                           +----+-----------+----+
                                |   (Pin)   |      <-- Mode 5: Tearout
                       +--------+-----------+--------+
       Mode 1:         |  Fork  |   Tang    |  Fork  |
     Double Shear ===> |  Ear   |    Eye    |  Ear   | <=== Mode 1: Double Shear
                       +--------+-----------+--------+
                           |                     |
                           v                     v
                          P/2                   P/2

1. Mode 1: Pin Double Shear

The cylindrical pin is supported on both sides by the fork ears while the center tang pulls in the opposite direction. The pin must shear across two transverse circular cross sections:

$$\tau = \frac{P}{2 A_{\text{pin}}} = \frac{P}{2 \left(\frac{\pi d_p^2}{4}\right)} = \frac{2 P}{\pi d_p^2} \le \frac{S_{sy}}{n} = \frac{0.577 S_y}{n}$$

2. Mode 2: Pin Bending

If manufacturing clearances create a gap $g$ between the tang and each fork ear, or if the plates are thick, the pin acts as a beam subjected to bending. Assuming uniform bearing pressure across the tang and across each ear of thickness $t_{\text{fork}}$:

$$M_{\max} = \frac{P}{2}\left( \frac{t_{\text{tang}}}{4} + g + \frac{t_{\text{fork}}}{2} \right) = \frac{P}{8}\left( t_{\text{tang}} + 4 g + 2 t_{\text{fork}} \right)$$ $$\sigma_{\text{bending}} = \frac{M_{\max} (d_p/2)}{I} = \frac{32 M_{\max}}{\pi d_p^3} \le \frac{S_y}{n}$$

3. Mode 3: Bearing Stress on Pin and Holes

Compressive contact stress develops along the projected diametral contact areas:

• Bearing on Center Tang: $\sigma_{b,\text{tang}} = \frac{P}{d_p t_{\text{tang}}} \le \frac{S_{y,\text{tang}}}{n}$
• Bearing on Fork Ears: $\sigma_{b,\text{fork}} = \frac{P}{2 d_p t_{\text{fork}}} \le \frac{S_{y,\text{fork}}}{n}$

4. Mode 4: Net Section Tension Across Eye Hole

The eye cross-section is reduced by the pin hole diameter $d_p$:

$$\sigma_t = \frac{P}{(w - d_p) t_{\text{tang}}} \le \frac{S_{ut}}{n}$$

5. Mode 5: Edge Shear Tearout

If the pin hole is placed too close to the end of the tang, the pin will tear out a plug along two parallel shear planes. With $a$ measured from the hole center to the end of the tang, each plane has length of roughly $a - d_p/2$:

$$\tau_{\text{tearout}} = \frac{P}{2 (a - d_p/2)\, t_{\text{tang}}} \le \frac{S_{sy}}{n}$$

To prevent tearout, standard design guidelines require $a \ge 1.5 d_p$.

---

Retaining Rings (Snap Rings) & Axial Retention

Retaining rings (snap rings) are stamped or wire-formed spring steel rings fitted into shallow circumferential grooves machined into shafts (external rings) or housing bores (internal rings) to axially position bearings, gears, and pins.

                       Retaining Ring Axial Thrust
                                 Retained Component
                                  +--------------+
                                  |              | --> Axial Thrust Load F_r
       Shaft Surface              |              |
       --------------+            +-------+------+  <-- Critical Corner Chamfer
                     | Groove             | Ring |
                     +------+             +------+  <-- Groove Wall Yielding
                            | Depth y     |
                            +-------------+

Failure Modes & Allowable Thrust Capacity

Retaining ring assemblies fail under three distinct mechanisms:

1. Groove Wall Yielding ($F_g$): The compressive contact stress against the soft shaft or housing material exceeds its yield strength, causing the groove lip to shear or deform plastically: $$F_g = \frac{\pi d y S_y}{n}$$ where $d$ is nominal shaft/bore diameter, $y$ is groove depth, and $S_y$ is the yield strength of the shaft material.
2. Ring Shear Capacity ($F_r$): Direct transverse shear across the hardened spring steel ring: $$F_r = \frac{\pi d t \tau_{\text{allow}}}{n}$$ where $t$ is the ring axial thickness and $\tau_{\text{allow}} \approx 0.577 S_y$ of the spring steel ring ($R_c 45\text{--}52$).
3. Dishing / Prying Failure (Corner Chamfer Effect): Retaining rings rely on sharp, square contact against the retained component. If the retained bearing or gear has a large corner radius or chamfer ($R$), the axial thrust load acts eccentrically away from the groove edge. This induces a severe prying bending moment that dishes the ring outward, causing it to pop out of the groove at loads far below its catalog rating. Always install a sharp-cornered square spacer washer between a radiused bearing and a retaining ring.

---

Riveted Joints & Structural Efficiency

Rivets are permanent fasteners comprising a cylindrical shank with a preformed head. After insertion through aligned holes, the plain end is mechanically upset (bucked) to form a second closing head.

Failure Modes of Riveted Lap & Butt Joints

1. Rivet Shear: $\tau = \frac{P}{m \cdot n_r \cdot \left(\frac{\pi}{4} d^2\right)}$, where $m = 1$ for single shear (lap joint) and $m = 2$ for double shear (butt joint with two cover plates), and $n_r$ is the total number of rivets.
2. Plate Bearing: $\sigma_b = \frac{P}{n_r \cdot d \cdot t}$, where $t$ is plate thickness.
3. Plate Net-Section Tension: $\sigma_t = \frac{P}{(w - n_{\text{row}} d) t}$, where $n_{\text{row}}$ is the number of rivet holes in the critical fracture row.
4. Plate Edge Tearout: Avoided by maintaining minimum edge margin $e \ge 1.5 d$ to $2.0 d$.

Joint Efficiency ($\eta$)

Joint efficiency is the ratio of allowable joint load capacity to the tensile strength of the solid, unpunched plate:

$$\eta = \frac{P_{\text{allowable joint}}}{P_{\text{solid plate}}} = \frac{P_{\text{allow}}}{w \cdot t \cdot S_t}$$

Well-designed structural rivet joints achieve efficiencies of $\eta \approx 70\%\text{ to }85\%$.

---

Lifting Lugs, Shackles & Anchors

Pin-Connected Lifting Lugs

A lifting lug (padeye) is a plate with a pin hole, so it is checked exactly like the clevis tang above: net-section tension beside the hole, shear tear-out beyond the hole, bearing of the pin on the plate, pin shear and bending, and the weld or bolts that attach the lug to the load. ASME BTH-1 (Design of Below-the-Hook Lifting Devices) sets the design factor for such hardware: $N_d = 2.00$ for Design Category A and $N_d = 3.00$ for Design Category B (service where loads or conditions are less predictable).

Sling geometry raises lug loads. For a symmetric two-leg sling with each leg at angle $\theta$ from vertical, each leg carries

$$T = \frac{W}{2\cos\theta}$$

so a $30^\circ$ leg angle raises leg tension by about $15\%$ over a vertical pick, and a $60^\circ$ angle doubles it.

Shackles (ASME B30.26)

Shackles are rated by their rated load (working load limit). ASME B30.26 requires a design factor of at least $5$ for shackles rated up to 150 tons ($4$ above that). Loading a shackle off its in-line axis reduces its rating:

Mini-example: A $20\text{ kN}$ load is lifted by two sling legs at $30^\circ$ from vertical, so each leg carries $20/(2 \times 0.866) = 11.5\text{ kN}$. If each leg attaches through a shackle with a rated load of $19.6\text{ kN}$ (2 t) that is side-loaded $30^\circ$, its usable rating is $0.70 \times 19.6 = 13.7\text{ kN} \ge 11.5\text{ kN}$, which is acceptable.

Anchors in Concrete (ACI 318, Chapter 17)

Cast-in anchors (headed studs, hooked bolts) and post-installed anchors (expansion, undercut and adhesive anchors) transfer machine-base loads into concrete. ACI 318 Chapter 17 requires checks of every failure mode and takes the lowest: steel strength in tension and in shear, concrete breakout (a cone-shaped failure whose projected area for a single anchor is $9 h_{ef}^2$, extending $1.5 h_{ef}$ on each side of the anchor, where $h_{ef}$ is the effective embedment depth), pullout, side-face blowout, bond failure for adhesive anchors, and concrete pryout in shear. Close edge distances and tight anchor spacing overlap the breakout areas and reduce capacity, which is why base-plate anchors are kept away from slab edges.

---

Worked Engineering Scenario: Drive Key & Clevis Pin Sizing

     Part 1: Shaft Drive Key                Part 2: Actuator Clevis Joint
     =======================                =============================
     Shaft Diameter: d = 2.00 in            Actuator Tension: P = 6,000 lbf
     Power: 50 hp @ 1,750 rpm               Double Shear Pin Sizing
     Key Material: AISI 1018 CD             Pin Material: AISI 1045 CD

Problem Statement

An electric motor drives an industrial centrifugal blower:

• Part 1 (Drive Key): The motor delivers $50.0\text{ hp}$ at $1,750\text{ rpm}$ through a solid shaft of diameter $d = 2.00\text{ in}$. Size a standard square key made of AISI 1018 cold-drawn steel ($S_y = 54.0\text{ kpsi}$, $S_{sy} = 0.577 S_y = 31.16\text{ kpsi}$) using a design factor of safety $n = 2.0$. Determine the minimum required key length $L$ for shear and bearing.
• Part 2 (Clevis Pin): The blower throttle damper linkage is positioned by a hydraulic actuator exerting an axial tensile force of $P = 6,000\text{ lbf}$ on a clevis joint. The clevis pin is fabricated from AISI 1045 cold-drawn steel ($S_y = 77.0\text{ kpsi}$, $S_{sy} = 0.577 S_y = 44.43\text{ kpsi}$). Using a design factor $n = 2.5$, size the clevis pin diameter $d_p$ for double shear, and verify bearing stress on a tang of thickness $t_{\text{tang}} = 0.625\text{ in}$.

---

Step-by-Step Mathematical Solution

Part 1: Shaft Drive Key Sizing

1. Calculate transmitted motor torque $T$:

$$T = \frac{63,025 \times \text{hp}}{N} = \frac{63,025 \times 50.0}{1,750} = 1,800.7\text{ lbf}\cdot\text{in}$$

1. Sizing standard square key dimensions for $d = 2.00\text{ in}$:

$$w = h = \frac{d}{4} = \frac{2.00\text{ in}}{4} = 0.500\text{ in}$$

1. Determine tangential force at shaft radius ($r = 1.00\text{ in}$):

$$F = \frac{T}{d/2} = \frac{1,800.7\text{ lbf}\cdot\text{in}}{1.00\text{ in}} = 1,800.7\text{ lbf}$$

1. Required length for direct horizontal shear ($n = 2.0$, $S_{sy} = 31,160\text{ psi}$):

$$\tau_{\text{allow}} = \frac{S_{sy}}{n} = \frac{31,160\text{ psi}}{2.0} = 15,580\text{ psi}$$ $$L_{\text{shear}} = \frac{F}{w \tau_{\text{allow}}} = \frac{1,800.7\text{ lbf}}{0.500\text{ in} \times 15,580\text{ psi}} = \frac{1,800.7}{7,790} = 0.231\text{ in}$$

1. Required length for compressive bearing ($n = 2.0$, $S_y = 54,000\text{ psi}$):

$$\sigma_{b,\text{allow}} = \frac{S_y}{n} = \frac{54,000\text{ psi}}{2.0} = 27,000\text{ psi}$$ $$L_{\text{bearing}} = \frac{F}{(h/2) \sigma_{b,\text{allow}}} = \frac{1,800.7\text{ lbf}}{0.250\text{ in} \times 27,000\text{ psi}} = \frac{1,800.7}{6,750} = 0.267\text{ in}$$

1. Summary and Specification: Compressive bearing governs, requiring $L \ge 0.267\text{ in}$. To ensure stable hub engagement and prevent key rocking, mechanical practice recommends $L \ge 0.75 d = 1.50\text{ in}$. Specify a $0.500\text{ in} \times 0.500\text{ in} \times 1.50\text{ in}$ square key.

---

Part 2: Actuator Clevis Pin Sizing

1. Allowable shear stress for clevis pin ($n = 2.5$, $S_{sy} = 44,430\text{ psi}$):

$$\tau_{\text{allow}} = \frac{S_{sy}}{n} = \frac{44,430\text{ psi}}{2.5} = 17,772\text{ psi}$$

1. Double shear equation for clevis pin under tensile load $P = 6,000\text{ lbf}$:

$$\tau = \frac{P}{2 A_{\text{pin}}} = \frac{P}{2 \left(\frac{\pi d_p^2}{4}\right)} = \frac{2 P}{\pi d_p^2} \le \tau_{\text{allow}}$$

1. Solve for required pin diameter $d_p$:

$$d_p \ge \sqrt{\frac{2 P}{\pi \tau_{\text{allow}}}} = \sqrt{\frac{2 \times 6,000\text{ lbf}}{\pi \times 17,772\text{ psi}}} = \sqrt{\frac{12,000}{55,833}} = \sqrt{0.2149} = 0.4636\text{ in}$$

Select standard fractional pin diameter: $d_p = 0.500\text{ in}$ ($1/2\text{ in}$).

1. Verify bearing stress on center tang ($t_{\text{tang}} = 0.625\text{ in}$):

$$\sigma_{b,\text{tang}} = \frac{P}{d_p t_{\text{tang}}} = \frac{6,000\text{ lbf}}{0.500\text{ in} \times 0.625\text{ in}} = \frac{6,000\text{ lbf}}{0.3125\text{ in}^2} = 19,200\text{ psi} = 19.2\text{ kpsi}$$

1. Bearing factor of safety on tang (AISI 1018 steel, $S_y = 54.0\text{ kpsi}$):

$$n_{\text{bearing}} = \frac{S_{y,\text{tang}}}{\sigma_{b,\text{tang}}} = \frac{54.0\text{ kpsi}}{19.2\text{ kpsi}} = 2.81$$

Because $n_{\text{bearing}} = 2.81 > 2.5$, the $0.500\text{ in}$ pin and $5/8\text{ in}$ tang assembly are fully verified against double shear and compressive bearing failure.