What are the limit states for steel plate design in tension?

Three tension limit states per AISC 360, checked at every plate cross-section: (1) GROSS SECTION YIELDING (J4.1a): phi Pn = 0.90 x Fy x Ag. The entire gross cross-section yields in tension. Controls for plates with few or no bolt holes (hole loss < 15% of gross area). For a 6 x 3/8 A36 plate: Ag = 6 x 0.375 = 2.25 in^2. phi Pn = 0.90 x 36 x 2.25 = 72.9 kips. This is the UPPER BOUND — the tension capacity can NEVER exceed gross yielding, regardless of how many bolts are added. (2) NET SECTION RUPTURE (J4.1b): phi Pn = 0.75 x Fu x Ae, where Ae = An x U (effective net area). An = Ag - n_holes x (dh + 1/16) x t per line of holes. U = shear lag factor = 1 - x_bar / L 20-25% of gross area. (3) BLOCK SHEAR RUPTURE (J4.3): a block tears out with shear on one plane + tension on perpendicular plane. Rn = 0.60Fu x Anv + Ubs x Fu x Ant <= 0.60Fy x Agv + Ubs x Fu x Ant. This is the most complex check — it requires identifying potential block shear paths and checking each. Block shear is the #1 governing limit state for gusset plates, coped beam webs, and angle connections.

How do you identify and check block shear paths?

Block shear analysis requires identifying potential tearout blocks and checking each one. The methodology: (1) IDENTIFY POTENTIAL BLOCK SHEAR PATHS: Look for groupings of bolts where a block of plate material could separate. The block is bounded by: one or more shear planes (parallel to load, through bolt lines), and one tension plane (perpendicular to load, across the end of the bolt group). For a bolted plate: shear along a line of bolts, tension across the plate width at the last bolt row. (2) CALCULATE SHEAR AREAS: Agv = gross shear area = L_shear_plane x t. Anv = net shear area = Agv - n_shear_holes x (dh + 1/16) x t. For a 4-bolt line at 3 in spacing + 1.5 in edge: L_shear = 4 x 3 + 1.5 = 13.5 in. Agv = 13.5 x 0.375 = 5.06 in^2. Anv = 5.06 - 4 x 0.875 x 0.375 = 5.06 - 1.31 = 3.75 in^2. (3) CALCULATE TENSION AREAS: Ant = net tension area = (W_tension - n_tension_holes x (dh + 1/16)) x t. For plate width 6 in with 2 bolts across: Ant = (6 - 2 x 0.875) x 0.375 = (6 - 1.75) x 0.375 = 1.59 in^2. (4) APPLY BLOCK SHEAR EQUATION: Rn = 0.60Fu x Anv + Ubs x Fu x Ant. Ubs = 1.0 when tension stress is uniform (most bolted plate cases). Ubs = 0.5 when tension stress is non-uniform (some coped beam cases). For A36 plate (Fu=58): Rn = 0.60 x 58 x 3.75 + 1.0 x 58 x 1.59 = 130.5 + 92.2 = 222.7 kips. Upper bound: 0.60Fy x Agv + Ubs x Fu x Ant = 0.60 x 36 x 5.06 + 1.0 x 58 x 1.59 = 109.3 + 92.2 = 201.5 kips. Controls. phi Rn = 0.75 x 201.5 = 151.1 kips. (5) CHECK ALL POSSIBLE BLOCKS: Multiple shear plane options exist — check the one with: (a) different tension plane locations (at different bolt rows), (b) different shear plane lengths, (c) different numbers of bolts in the shear path. The block with the LOWEST phi Rn governs.

What is the Whitmore effective width for gusset plate tension?

The Whitmore section defines the effective width of gusset plate that resists tension. The tension force from the bolt group SPREADS at a 30-degree angle (tan 30 = 0.577) on each side of the bolt group, following the force path through the plate. The effective width is measured at the last bolt row (furthest from the applied load): Ww = 2 x Lw x tan(30) + sg, where Lw = distance from first to last bolt row along the brace/load axis, sg = bolt gage perpendicular to the load axis (center-to-center of outermost bolts). Example: brace connection with 4 bolt rows at 3 in spacing (Lw = 9 in), bolt gage sg = 4 in. Ww = 2 x 9 x 0.577 + 4 = 10.4 + 4 = 14.4 in. Tension capacity of the gusset at the Whitmore section: (a) Yielding: phi Pn = 0.90 x Fy x Ww x tg = 0.90 x 36 x 14.4 x 0.375 = 175 kips. (b) Net rupture at Whitmore section: phi Pn = 0.75 x Fu x (Ww - n_holes_per_row x (dh + 1/16)) x tg = 0.75 x 58 x (14.4 - 2 x 0.875) x 0.375 = 0.75 x 58 x 12.65 x 0.375 = 206 kips. Yield controls. (c) Block shear at the bolt group (check separately). The Whitmore width must fit within the ACTUAL gusset geometry. If the 30-degree lines intersect the gusset edges before reaching the last bolt row, the available width at the intersection controls — the Whitmore section cannot use plate area that doesn't exist. The 30-degree spread angle comes from elastic stress distribution observed in photoelastic tests by Whitmore (1952). It remains the standard method in AISC Manual Part 9.

How is plate buckling under compression checked?

When a steel plate carries compression (gusset plate compression, base plate under column compression, stiffener in compression), it must be checked for buckling as a column element: (1) EFFECTIVE WIDTH: The plate portion in compression is idealized as a column strip. For gusset plates (Thornton method): the effective column width = Whitmore width Ww. For uniformly compressed plates: the effective width = the full plate width unless local buckling precedes global buckling. (2) EFFECTIVE LENGTH: For gusset plates: the unbraced length is the average distance from the Whitmore section to the nearest restrained edge (beam flange, column flange). Keff = average of three distances: at the two Whitmore corners and at the Whitmore midpoint. K factor: 0.65 for gusset restrained on two edges (beam + column), 1.2 for one-edge restraint, 0.5 for three-edge restraint. (3) RADIUS OF GYRATION: For a rectangular plate strip: r = t / sqrt(12) = tg / 3.464. For 3/8 in gusset: r = 0.375 / 3.464 = 0.108 in. (4) SLENDERNESS: KL/r = K x Leff / r. For Leff = 6 in, K = 0.65: KL/r = 0.65 x 6 / 0.108 = 36.1. (5) CRITICAL STRESS per AISC E3: Fe = pi^2 x E / (KL/r)^2 = pi^2 x 29,000 / 36.1^2 = 219 ksi. Since KL/r = 36.1 100) where the buckling reduction becomes significant. For base plates: compression buckling is not a concern because the plate is continuously supported by the concrete/grout — the plate yields in bearing, not buckling.

When does shear lag reduce plate tension capacity?

Shear lag occurs when only part of a cross-section is connected, causing non-uniform tension stress distribution. The connected portion reaches yield while the unconnected portion lags. Per AISC D3: (1) SHEAR LAG FACTOR U: U = 1 - x_bar / L <= 0.9, where x_bar = distance from the shear plane (connection face) to the centroid of the connected element, L = length of the connection (distance from first to last bolt or weld length). The UNCONNECTED portion of the section experiences reduced stress — the effective area Ae = An x U accounts for this inefficiency. (2) SCENARIOS: (a) Plate connected by longitudinal welds along both edges: U = 1.0 for L >= 2w (w = plate width), U = 0.87 for 2w > L >= 1.5w, U = 0.75 for 1.5w > L >= w. This is because the weld length L controls how far into the plate the tension can spread at a ~30-degree angle. (b) Plate connected by transverse welds only (end weld): U = 1.0 if all plate elements are connected — but the weld must be sized to transfer the full plate yield force. (c) Angle connected through one leg (bolted): x_bar is measured from the back of the connected leg to the angle centroid. For an L4x4x3/8 with 3 bolts in line: x_bar = 1.19 in, L = two bolt spacings = 2 x 3 = 6 in. U = 1 - 1.19/6 = 0.80. This means only 80% of the net area is effective in tension — a 20% reduction! This is why single-leg angle connections carry significantly less tension than the gross area suggests. (d) WT connected through flange: x_bar from WT flange face to WT centroid. U typically 0.75-0.90. (3) DESIGN STRATEGY: increase the connection length L to increase U. But there's a practical limit — adding more bolts beyond L = 2x_bar buys diminishing returns.

Steel Plate Design -- Tension, Net Section, and Block Shear per AISC 360-22

Q: How do you identify and check block shear paths?

Block shear analysis requires identifying potential tearout blocks and checking each one. The methodology: (1) IDENTIFY POTENTIAL BLOCK SHEAR PATHS: Look for groupings of bolts where a block of plate material could separate. The block is bounded by: one or more shear planes (parallel to load, through bolt lines), and one tension plane (perpendicular to load, across the end of the bolt group). For a bolted plate: shear along a line of bolts, tension across the plate width at the last bolt row. (2) CALCULATE SHEAR AREAS: Agv = gross shear area = L_shear_plane x t. Anv = net shear area = Agv - n_shear_holes x (dh + 1/16) x t. For a 4-bolt line at 3 in spacing + 1.5 in edge: L_shear = 4 x 3 + 1.5 = 13.5 in. Agv = 13.5 x 0.375 = 5.06 in^2. Anv = 5.06 - 4 x 0.875 x 0.375 = 5.06 - 1.31 = 3.75 in^2. (3) CALCULATE TENSION AREAS: Ant = net tension area = (W_tension - n_tension_holes x (dh + 1/16)) x t. For plate width 6 in with 2 bolts across: Ant = (6 - 2 x 0.875) x 0.375 = (6 - 1.75) x 0.375 = 1.59 in^2. (4) APPLY BLOCK SHEAR EQUATION: Rn = 0.60Fu x Anv + Ubs x Fu x Ant. Ubs = 1.0 when tension stress is uniform (most bolted plate cases). Ubs = 0.5 when tension stress is non-uniform (some coped beam cases). For A36 plate (Fu=58): Rn = 0.60 x 58 x 3.75 + 1.0 x 58 x 1.59 = 130.5 + 92.2 = 222.7 kips. Upper bound: 0.60Fy x Agv + Ubs x Fu x Ant = 0.60 x 36 x 5.06 + 1.0 x 58 x 1.59 = 109.3 + 92.2 = 201.5 kips. Controls. phi Rn = 0.75 x 201.5 = 151.1 kips. (5) CHECK ALL POSSIBLE BLOCKS: Multiple shear plane options exist — check the one with: (a) different tension plane locations (at different bolt rows), (b) different shear plane lengths, (c) different numbers of bolts in the shear path. The block with the LOWEST phi Rn governs.

Q: What is the Whitmore effective width for gusset plate tension?

The Whitmore section defines the effective width of gusset plate that resists tension. The tension force from the bolt group SPREADS at a 30-degree angle (tan 30 = 0.577) on each side of the bolt group, following the force path through the plate. The effective width is measured at the last bolt row (furthest from the applied load): Ww = 2 x Lw x tan(30) + sg, where Lw = distance from first to last bolt row along the brace/load axis, sg = bolt gage perpendicular to the load axis (center-to-center of outermost bolts). Example: brace connection with 4 bolt rows at 3 in spacing (Lw = 9 in), bolt gage sg = 4 in. Ww = 2 x 9 x 0.577 + 4 = 10.4 + 4 = 14.4 in. Tension capacity of the gusset at the Whitmore section: (a) Yielding: phi Pn = 0.90 x Fy x Ww x tg = 0.90 x 36 x 14.4 x 0.375 = 175 kips. (b) Net rupture at Whitmore section: phi Pn = 0.75 x Fu x (Ww - n_holes_per_row x (dh + 1/16)) x tg = 0.75 x 58 x (14.4 - 2 x 0.875) x 0.375 = 0.75 x 58 x 12.65 x 0.375 = 206 kips. Yield controls. (c) Block shear at the bolt group (check separately). The Whitmore width must fit within the ACTUAL gusset geometry. If the 30-degree lines intersect the gusset edges before reaching the last bolt row, the available width at the intersection controls — the Whitmore section cannot use plate area that doesn't exist. The 30-degree spread angle comes from elastic stress distribution observed in photoelastic tests by Whitmore (1952). It remains the standard method in AISC Manual Part 9.

Q: How is plate buckling under compression checked?

When a steel plate carries compression (gusset plate compression, base plate under column compression, stiffener in compression), it must be checked for buckling as a column element: (1) EFFECTIVE WIDTH: The plate portion in compression is idealized as a column strip. For gusset plates (Thornton method): the effective column width = Whitmore width Ww. For uniformly compressed plates: the effective width = the full plate width unless local buckling precedes global buckling. (2) EFFECTIVE LENGTH: For gusset plates: the unbraced length is the average distance from the Whitmore section to the nearest restrained edge (beam flange, column flange). Keff = average of three distances: at the two Whitmore corners and at the Whitmore midpoint. K factor: 0.65 for gusset restrained on two edges (beam + column), 1.2 for one-edge restraint, 0.5 for three-edge restraint. (3) RADIUS OF GYRATION: For a rectangular plate strip: r = t / sqrt(12) = tg / 3.464. For 3/8 in gusset: r = 0.375 / 3.464 = 0.108 in. (4) SLENDERNESS: KL/r = K x Leff / r. For Leff = 6 in, K = 0.65: KL/r = 0.65 x 6 / 0.108 = 36.1. (5) CRITICAL STRESS per AISC E3: Fe = pi^2 x E / (KL/r)^2 = pi^2 x 29,000 / 36.1^2 = 219 ksi. Since KL/r = 36.1 100) where the buckling reduction becomes significant. For base plates: compression buckling is not a concern because the plate is continuously supported by the concrete/grout — the plate yields in bearing, not buckling.

Q: When does shear lag reduce plate tension capacity?

Shear lag occurs when only part of a cross-section is connected, causing non-uniform tension stress distribution. The connected portion reaches yield while the unconnected portion lags. Per AISC D3: (1) SHEAR LAG FACTOR U: U = 1 - x_bar / L <= 0.9, where x_bar = distance from the shear plane (connection face) to the centroid of the connected element, L = length of the connection (distance from first to last bolt or weld length). The UNCONNECTED portion of the section experiences reduced stress — the effective area Ae = An x U accounts for this inefficiency. (2) SCENARIOS: (a) Plate connected by longitudinal welds along both edges: U = 1.0 for L >= 2w (w = plate width), U = 0.87 for 2w > L >= 1.5w, U = 0.75 for 1.5w > L >= w. This is because the weld length L controls how far into the plate the tension can spread at a ~30-degree angle. (b) Plate connected by transverse welds only (end weld): U = 1.0 if all plate elements are connected — but the weld must be sized to transfer the full plate yield force. (c) Angle connected through one leg (bolted): x_bar is measured from the back of the connected leg to the angle centroid. For an L4x4x3/8 with 3 bolts in line: x_bar = 1.19 in, L = two bolt spacings = 2 x 3 = 6 in. U = 1 - 1.19/6 = 0.80. This means only 80% of the net area is effective in tension — a 20% reduction! This is why single-leg angle connections carry significantly less tension than the gross area suggests. (d) WT connected through flange: x_bar from WT flange face to WT centroid. U typically 0.75-0.90. (3) DESIGN STRATEGY: increase the connection length L to increase U. But there's a practical limit — adding more bolts beyond L = 2x_bar buys diminishing returns.

Steel plates in tension are fundamental elements in structural connections: gusset plates, splice plates, shear tabs, and flange cover plates. Their design involves three tension limit states (gross yielding, net section fracture, and block shear rupture) plus buckling checks for compression. This reference covers the AISC 360-22 Chapter D and J4 provisions with worked examples.

Limit States for Tension Members (AISC 360 Chapter D)

Tensile Yielding in the Gross Section (D2)

The simplest limit state: the plate yields across its full cross-section.

Rn = Fy * Ag
phi_t = 0.90

where Ag = gross cross-sectional area = plate width x thickness. This limit state controls when the plate has no holes, or when the net section has higher capacity than the gross section.

Tensile Rupture in the Net Section (D2)

At bolt holes, the cross-sectional area is reduced. Fracture initiates at the hole:

Rn = Fu * Ae
phi_t = 0.75

where Ae = effective net area = An x U, An = Ag - sum(dh x t) for a straight chain of holes, U = shear lag factor = 1.0 for plates with uniform stress distribution.

For staggered holes, the net width uses the stagger formula:

wn = wg - sum(dh) + sum(s^2 / (4g))

where s = longitudinal stagger (pitch) and g = transverse spacing (gage). The s^2/(4g) term accounts for the diagonal path between staggered holes being longer than the direct transverse path.

Example: PL 1/2 x 8 with two 13/16 in. holes in a single chain: Ag = 0.5 x 8 = 4.0 in^2 An = 4.0 - 2 x 0.8125 x 0.5 = 4.0 - 0.8125 = 3.1875 in^2 Ae = An (U = 1.0 for flat plate, uniform stress)

Tensile yielding: phi Rn = 0.90 x 36 x 4.0 = 129.6 kip Tensile rupture: phi Rn = 0.75 x 58 x 3.1875 = 138.7 kip

Yielding controls (129.6 kip < 138.7 kip). The plate yields before it ruptures -- ductile behavior.

Example with more holes (rupture controls): PL 3/8 x 6 with four 7/8 in. bolts (dh = 15/16 in.): Ag = 0.375 x 6 = 2.25 in^2 An = 2.25 - 4 x 0.9375 x 0.375 = 2.25 - 1.406 = 0.844 in^2

Yielding: phi Rn = 0.90 x 36 x 2.25 = 72.9 kip Rupture: phi Rn = 0.75 x 58 x 0.844 = 36.7 kip

Rupture controls (36.7 kip << 72.9 kip). The plate fractures at the net section before full yielding -- a brittle failure mode. Per AISC 360 Section J4.1, the net section fracture strength should exceed the gross yielding strength to ensure ductility. This plate fails that requirement and should be thickened.

Block Shear Rupture (AISC 360 Section J4.3)

Block shear is the combination of shear rupture on one plane and tension rupture on a perpendicular plane. It governs for gusset plates, beam copes, and shear tabs where the bolt group is near the plate edge.

Rn = 0.60 Fu Anv + Ubs Fu Ant <= 0.60 Fy Agv + Ubs Fu Ant
phi = 0.75

where Anv = net area in shear, Ant = net area in tension, Agv = gross area in shear, Ubs = 1.0 for uniform tension stress distribution.

The first term is shear rupture + tension rupture. The cap is shear yielding + tension rupture. The smaller controls.

Block Shear Worked Example -- Gusset Plate at Brace Connection

A brace gusset plate PL 1/2 x 14 wide at the connection. Six 3/4 in. A325 bolts in two rows of three. Pitch = 3 in., gage = 4 in. Edge distance = 1.5 in. to the plate edge in tension, 1.5 in. to the plate edge in shear. Load direction is parallel to the bolt rows.

Tension plane (perpendicular to load): Ant_gross = (gage + 2 x edge) x t = (4.0 + 3.0) x 0.5 = 3.50 in^2 Holes in tension plane: 3 holes x 13/16 in. = 2.4375 in. total Ant_net = 3.50 - 2.4375 x 0.5 = 3.50 - 1.219 = 2.281 in^2

Shear plane (parallel to load): Agv = (2 x 3 + 1.5) x 0.5 = 7.5 x 0.5 = 3.75 in^2 (the length of the bolt group plus edge distance) Holes in shear plane: 2.5 holes x 13/16 in. (counting half-holes at the edge of the shear plane) Anv = 3.75 - 2.5 x 0.8125 x 0.5 = 3.75 - 1.016 = 2.734 in^2

Block shear strength: Shear rupture + tension rupture: 0.60 x 58 x 2.734 + 1.0 x 58 x 2.281 = 95.1 + 132.3 = 227.4 kip Shear yield + tension rupture: 0.60 x 36 x 3.75 + 1.0 x 58 x 2.281 = 81.0 + 132.3 = 213.3 kip (controls)

phi Rn = 0.75 x 213.3 = 160.0 kip

Check against the factored brace force. This is the block shear capacity of the gusset at the bolted connection.

Plate Buckling in Compression (AISC 360 Section E3)

When a plate is loaded in compression (gusset plate in a compression brace, stiffener in a base plate), buckling must be checked.

The effective length factor K for gusset plate buckling per AISC DG29 uses a modified Whitmore section approach. The effective column length is the average of L1, L2, and L3 (distances from the Whitmore section to the restraint lines at beam, column, and brace). For a gusset plate of thickness t and effective width b_eff:

Whitmore width = 2 x L_w x tan(30 degrees) + brace width

The radius of gyration for out-of-plane buckling is r = t / sqrt(12) -- the plate buckles in the weak direction.

Example: Gusset plate PL 3/8 x 12 effective width, unbraced length = 8 in.: Ag = 0.375 x 12 = 4.5 in^2 r = t / sqrt(12) = 0.375 / 3.464 = 0.108 in. KL/r = 1.2 x 8 / 0.108 = 88.9

Fe = pi^2 x E / (KL/r)^2 = pi^2 x 29,000 / 88.9^2 = 286,219 / 7,903 = 36.2 ksi

Fcr = 0.658^(Fy/Fe) x Fy = 0.658^(36/36.2) x 36 = 0.658^0.994 x 36 = 0.659 x 36 = 23.7 ksi

phi Pn = 0.90 x 23.7 x 4.5 = 96.0 kip

Steel Plate Material Specifications

ASTM Spec	Fy (ksi)	Fu (ksi)	Typical Applications
A36	36	58	General structural plates, base plates, shear tabs
A572 Gr 50	50	65	Higher-strength plates, gussets, heavy splice plates
A572 Gr 55	55	70	Bridge gusset plates
A588	50	70	Weathering steel plates (unpainted, exposed)
A514 Gr 100	100	110-130	Quenched and tempered, crane runway plates

A36 is the default for plates up to 2 in. thick. A572 Gr 50 is specified when the higher strength allows thinner plate or for material consistency with W-shapes.

Minimum Plate Thickness for Practical Applications

Application	Min. t	Rationale
Shear tab	1/4 in.	Practical minimum for bolting and welding without warping
Gusset plate (light brace)	3/8 in.	Minimum for bolt bearing on 3/4 in. bolts
Gusset plate (heavy brace)	1/2 in.	Buckling resistance for compression braces
Base plate	5/8 in.	Constructability -- thinner plates warp during welding
Flange cover plate	3/8 in.	Match flange thickness for uniform stress distribution
Column splice plate	5/8 in.	Must develop required cross-sectional area
Stiffener plate	3/8 in.	Minimum for fillet weld on both sides

Worked Example -- Tension Splice Plate Design

Problem: Design flange splice plates for a W12x65 column splice. Pu_flange = 210 kip (tension from column uplift). Use A572 Gr 50 plate. 7/8 in. A325-N bolts.

Step 1 -- Required plate area: Ag_req = Pu / (phi x Fy) = 210 / (0.90 x 50) = 210 / 45 = 4.67 in^2

Step 2 -- Trial plate: Two plates, one on each face of the flange. Each plate carries 105 kip. Try PL 5/8 x 8 (Ag = 5.0 in^2 each, total = 10.0 in^2).

Step 3 -- Net section check (rupture): Four 15/16 in. holes per plate. An = (8 - 2 x 0.9375) x 0.625 = (8 - 1.875) x 0.625 = 3.828 in^2 per plate.

phi Rn_rupture = 0.75 x 65 x 3.828 = 186.6 kip per plate > 105 kip. OK.

Step 4 -- Bolt bearing on plate: Lc per bolt = 2.0625 in. Tearout: 1.2 x 2.0625 x 0.625 x 65 = 100.5 kip. Bearing: 2.4 x 0.875 x 0.625 x 65 = 85.3 kip. Bearing controls. phi Rn = 0.75 x 85.3 = 64.0 kip per bolt. 4 bolts per plate: 256 kip >> 105 kip.

Step 5 -- Block shear: Ant = 0.996 in^2, Anv = 5.098 in^2, Agv = 6.563 in^2. Rn = 0.60 x 65 x 5.098 + 1.0 x 65 x 0.996 = 263.5 kip. Cap = 0.60 x 50 x 6.563 + 1.0 x 65 x 0.996 = 261.6 kip. phi Rn = 0.75 x 261.6 = 196.2 kip > 105 kip. OK.

Final plate: 2 PL 5/8 x 8 x 1'-4-1/2 (A572 Gr 50).

Related References

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