12.4 Plug and Slot Welds, Stress Gradients & the Fillet Weld Break Test

Plug and Slot Welds

Plug and slot welds transmit shear across overlapping plates by filling a circular hole or elongated slot with weld metal.

Structural Mechanics and Sizing Equations

1. Effective Shearing Area ($A_w$): Equal to the nominal area of the hole or slot in the plane of the faying surface: $A_w = \frac{\pi d^2}{4} \quad (\text{for circular plug welds})$ $A_w = W \cdot L - (1 - \pi/4) W^2 \approx W \cdot L \quad (\text{for slots with rounded ends})$
2. Design Shear Strength:
   • LRFD: $\phi R_n = 0.75 \times (0.60 F_{EXX}) A_w = 0.45 F_{EXX} A_w$
   • ASD: $R_n / \Omega = \frac{0.60 F_{EXX}}{2.00} A_w = 0.30 F_{EXX} A_w$
3. Dimensional Limitations (AWS D1.1 Clause 4.10 & AISC J2.3):
   • Minimum hole diameter: $d_{\min} = t + 8\text{ mm}$ ($t + 5/16\text{ in}$), where $t$ is plate thickness.
   • Maximum hole diameter: $d_{\max} = 2.25 t$ or $d_{\min} + 3\text{ mm}$.
   • Depth of fill: For plates $t \le 16\text{ mm}$ ($5/8\text{ in}$), depth of fill must equal plate thickness. For $t > 16\text{ mm}$, minimum depth of fill is $t/2$, but not less than $16\text{ mm}$.

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Stress Gradients Across Welded Sections

AWS B5.16 Clause 8.3.5 lists stress gradient alongside sectional properties and stress triaxiality. A stress gradient is the rate at which stress changes with position, $d\sigma/dx$, and welding engineers care about three distinct gradients:

1. Structural (bending) gradient. Pure axial load produces a uniform stress field with zero gradient. Bending produces a linear through-thickness gradient from tension at one surface to compression at the other, crossing zero at the neutral axis. A weld placed near the neutral axis of a beam sees almost no bending stress; the same weld moved to the flange sees the peak. This is why flange-to-web welds in plate girders are sized for horizontal shear flow while the flange butt splice is sized for the full bending stress.
2. Geometric (notch) gradient. A weld toe, reinforcement, undercut or backing bar produces a steep local gradient over a distance of a fraction of a millimetre. The peak stress is $K_t$ times nominal, but it decays rapidly into the plate. A steep gradient means only a small highly stressed volume, so a sharp notch is much less damaging in a small part than the same $K_t$ in a thick section — the mechanism behind the thickness-correction factors in fatigue codes.
3. Residual stress gradient. Longitudinal residual stress peaks near yield at the weld centreline and falls, often reversing into compression, within a few plate thicknesses. Superimposed on applied stress, this shifts the effective mean stress at the toe and is why as-welded fatigue curves are treated as mean-stress independent.

The engineering significance of a gradient is that fracture and fatigue respond to the stressed volume, not just the peak value. A shallow gradient exposes a large volume of material to near-peak stress, giving a high probability of finding a critical defect; a steep gradient shelters the bulk. This is the physical basis of the "highly stressed volume" and weakest-link statistical arguments used to explain why thick sections behave worse than thin sections at identical nominal stress.

$\text{Relative stress gradient: } \chi = \frac{1}{\sigma_{\max}}\cdot\frac{d\sigma}{dx}\bigg|_{\text{surface}}$

A high $\chi$ (steep gradient, small stressed volume) correlates with higher apparent fatigue strength; a low $\chi$ approaches the uniform-stress worst case.

The Fillet Weld Break Test

Clause 8.3.5 also names the fillet break test (influence of second phase and porosity). It is the cheapest destructive check on a fillet weld and appears in both AWS D1.1 welder qualification and WPS qualification for fillet welds.

Method. A fillet weld is deposited on a T-joint test coupon. The specimen is loaded so the fillet is forced to fracture through its throat — typically by pressing or hammering the stem of the tee over until the weld breaks or the stem bends flat against the plate. The fracture surface is then examined, because it exposes the root of the fillet, which no surface method and no radiograph of a tee joint can see reliably.

What the fracture face reveals.

Acceptance. The current edition of AWS D1.1 accepts a fillet weld break specimen that either does not fracture at all, or — if it does fracture — shows complete fusion to the root of the joint with no single inclusion or pore greater than $3/32\text{ in}$ ($2.5\text{ mm}$) in its greatest dimension, and with the sum of the greatest dimensions of all inclusions and porosity not exceeding $3/8\text{ in}$ ($10\text{ mm}$) in the $6\text{ in}$ ($150\text{ mm}$) specimen. Always read the acceptance table in the edition invoked by the contract, since the dimensional limits are edition-specific.

Exam Trap: "A bent-but-unbroken specimen fails the fillet break test." The opposite is true. A specimen that bends flat without fracturing demonstrates that the weld is stronger and more ductile than the base plate stem, and it passes. The fracture surface criteria apply only when fracture actually occurs.

Comprehensive Worked Engineering Example: Sizing a Tension Lap Joint

Problem Statement

A structural tension member consisting of an ASTM A572 Gr 50 flat bar ($F_y = 345\text{ MPa}$, $F_u = 450\text{ MPa}$, thickness $t = 12\text{ mm}$, width $b = 180\text{ mm}$) is spliced to a gusset plate using a lap joint. The joint is subjected to a service dead load $P_D = 140\text{ kN}$ and service live load $P_L = 260\text{ kN}$.

Welding is executed using the GMAW process with matching AWS A5.18 ER70S-6 filler metal ($F_{EXX} = 70\text{ ksi} \approx 485\text{ MPa}$). Fillet weld leg size is specified as $w = 8.0\text{ mm}$.

Calculate the required length of longitudinal fillet welds using: (a) LRFD methodology, and (b) ASD methodology. Then, (c) verify the connection capacity if a transverse fillet weld of length $b = 180\text{ mm}$ is added across the end of the plate under LRFD.

                   +----------------------------------+
                   |                                  |========== Longitudinal Weld (L)
                   |          A572 Gr 50 Plate        | 
             Pu <--|          t = 12 mm, b = 180 mm   |          
                   |                                  |========== Longitudinal Weld (L)
                   +----------------------------------+
                                                      | Transverse Weld (b = 180 mm)

Step-by-Step Engineering Solution

Step 1: Compute Factored and Service Loads

• LRFD Factored Load ($P_u$): $P_u = 1.2 P_D + 1.6 P_L = 1.2(140\text{ kN}) + 1.6(260\text{ kN}) = 168 + 416 = 584\text{ kN} = 584,000\text{ N}$
• ASD Service Load ($P_a$): $P_a = P_D + P_L = 140\text{ kN} + 260\text{ kN} = 400\text{ kN} = 400,000\text{ N}$

Step 2: Effective Throat and Unit Weld Capacities

• Equal leg $w = 8.0\text{ mm} \implies t_e = 0.7071 w = 0.7071(8.0) = 5.657\text{ mm}$.
• Nominal shear stress for longitudinal weld ($\theta = 0^\circ$): $F_{nw} = 0.60 F_{EXX} = 0.60 \times 485\text{ MPa} = 291.0\text{ MPa} = 291.0\text{ N/mm}^2$

Step 3: Required Longitudinal Weld Length under LRFD

• Design shear strength per linear mm of longitudinal weld ($\phi = 0.75$): $\phi q_{n} = \phi F_{nw} t_e = 0.75 \times 291.0\text{ N/mm}^2 \times 5.657\text{ mm} = 1234.6\text{ N/mm} = 1.2346\text{ kN/mm}$
• Total required weld length for two longitudinal welds: $L_{\text{total}} = \frac{P_u}{\phi q_n} = \frac{584.0\text{ kN}}{1.2346\text{ kN/mm}} = 473.0\text{ mm}$
• Length per side: $L_{\text{side}} = 473.0 / 2 = 236.5\text{ mm} \implies$ Specify $240\text{ mm}$ per side.

Step 4: Required Longitudinal Weld Length under ASD

• Allowable shear strength per linear mm of longitudinal weld ($\Omega = 2.00$): $q_{\text{allow}} = \frac{F_{nw} t_e}{\Omega} = \frac{291.0 \times 5.657}{2.00} = 823.1\text{ N/mm} = 0.8231\text{ kN/mm}$
• Total required weld length: $L_{\text{total}} = \frac{P_a}{q_{\text{allow}}} = \frac{400.0\text{ kN}}{0.8231\text{ kN/mm}} = 485.9\text{ mm}$
• Length per side: $L_{\text{side}} = 485.9 / 2 = 243.0\text{ mm} \implies$ Specify $245\text{ mm}$ per side.

Step 5: Incorporating a Transverse End Fillet under LRFD (AISC J2.4)

• Sizing a transverse weld across plate width $L_{\text{trans}} = b = 180\text{ mm}$:
  • For transverse weld ($\theta = 90^\circ$): $F_{nwt} = 1.50 \times (0.60 F_{EXX}) = 1.50 \times 291.0 = 436.5\text{ MPa}$ $\phi R_{nwt} = \phi F_{nwt} t_e L_{\text{trans}} = 0.75 \times 436.5\text{ N/mm}^2 \times 5.657\text{ mm} \times 180\text{ mm} = 333,354\text{ N} = 333.35\text{ kN}$
• Using AISC Equation J2-10b to find required longitudinal length ($L_{\text{long}}$): $\phi R_n = 0.85 \phi R_{nwl} + 1.5 \phi R_{nwt} \implies P_u = 0.85 (\phi q_n \cdot L_{\text{long}}) + \phi R_{nwt}$ $584.0\text{ kN} = 0.85(1.2346 \cdot L_{\text{long}}) + 333.35$ $0.85(1.2346) L_{\text{long}} = 584.0 - 333.35 = 250.65\text{ kN}$ $1.0494 L_{\text{long}} = 250.65 \implies L_{\text{long}} = 238.8\text{ mm}$
• Length per longitudinal side: $238.8 / 2 = 119.4\text{ mm} \implies$ Specify $120\text{ mm}$ per side.
• Result: Adding the $180\text{ mm}$ transverse weld cuts the required longitudinal weld length in half (from $240\text{ mm}$ to $120\text{ mm}$ per side), drastically compacting the joint overlap length.

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Industrial Scenarios & Certified Welding Engineer Exam Pitfalls

Real-World Field Disaster Scenario

A mobile crane outrigger support arm fabricated from ASTM A514 high-yield steel ($F_y = 690\text{ MPa}$, $100\text{ ksi}$) experienced catastrophic shear fracture during a maximum-rated lift. The structural engineer specified an overmatched electrode (AWS A5.5 E11018-M) for all fillet welds, reasoning that higher strength equaled higher safety. During full-load lifting, while the weld metal remained elastic, the concentrated shear strain localized abruptly into the narrow, un-tempered martensitic Heat-Affected Zone (HAZ) of the A514 base plate. Because the overmatched weld provided zero plastic strain relaxation, a cleavage shear crack initiated at the weld root and ripped through the outrigger box section, dropping the crane boom onto an active highway. Subsequent failure analysis demonstrated that had an undermatched E7018 filler metal been used, the weld throat would have yielded plastically by $0.8\text{ mm}$, redistributing stress safely across the entire outrigger assembly without cracking.

Common Exam Traps

Exam Trap 1: Forgetting the 3 mm (1/8 in) Deduction on 45° PJP Welds When an exam problem specifies a Partial Joint Penetration (PJP) groove weld prepared at a $45^\circ$ included angle welded with SMAW or FCAW-S without backing, candidates almost invariably take $E = S$. You must subtract $3\text{ mm}$ ($1/8\text{ in}$) from the depth of preparation: $E = S - 3\text{ mm}$. If the preparation depth is $12\text{ mm}$, the design effective throat is only $9\text{ mm}$.

Exam Trap 2: Direct Addition of Longitudinal and Transverse Capacities When calculating the combined capacity of a weld group containing both longitudinal and transverse welds, you cannot simply compute $R_n = 1.0 R_{nwl} + 1.5 R_{nwt}$. Transverse welds reach their peak strength and fracture at deformations of only $\approx 0.05 w$, well before longitudinal welds develop their full yield strength. You must evaluate both AISC Equation J2-10a ($R_n = R_{nwl} + R_{nwt}$) and Equation J2-10b ($R_n = 0.85 R_{nwl} + 1.5 R_{nwt}$) and select the higher value.

Exam Trap 3: Overlooking Base Metal Shear Rupture Along Fusion Boundary Sizing the weld throat is only half the problem. When thin plates ($t < 1.414 w$) are joined by large fillet welds, the base metal along the fusion boundary will shear rupture before the weld throat yields. Always verify that base metal allowable shear ($0.40 F_y t L$ or $0.30 F_u t L$) exceeds the applied load.