Structural Steel Design, AISC 360 Composite Floor Beams & Shear Studs
Composite Steel-Concrete Beam Shear Stud Count Formula
Calculate Governing Horizontal Interface Shear V_prime (kN) between Midspan & Support, Concrete Slab Compression Depth a (mm), Studs Required per Half-Span, and Total Full-Span Studs.
Count shear studs between zero-moment support and maximum-moment midspan—then DOUBLE for the full simple span!
In AISC 360 composite steel wide-flange floor beams welded through metal deck to a concrete slab via headed Nelson shear studs (nominal single-stud shear strength Q_n, kN), full composite plastic behavior transfers Horizontal Interface Shear Force V_prime = MIN(A_s * F_y, 0.85 * f_c_prime * b_eff * t_slab) / 1,000 kN across EACH half-span from the point of maximum positive moment (midspan) to the zero-moment simple support! Forgetting to multiply Half-Span Studs N_half = ROUNDUP(V_prime / Q_n, 0) by 2 leaves the beam 50% under-studded across the total beam length!
Verify on a Full-Composite Office Floor W-Shape—Steel_Section_Area A_s = 8,500.0 mm² (B2), Steel_Yield F_y = 345.0 MPa (B3 -> Steel Yield Force = 2,932.50 kN), Concrete_Strength f_c_prime = 28.0 MPa (B4), Effective_Slab_Width b_eff = 2,000.0 mm (B5), Solid_Slab_Thickness t_s = 130.0 mm (B6 -> Concrete Crushing Limit = 6,193.60 kN, so Steel Governs V_prime = 2,932.50 kN in E2), and Single_19mm_Stud_Capacity Q_n = 95.0 kN (B7): Slab Compression Depth a equals (2,932.5 * 1,000) / (0.85 * 28 * 2,000) = 61.61 mm (E3), Studs per Half-Span N_half = ROUNDUP(2,932.5 / 95, 0) = 31 studs (E4), and Total Full-Span Studs equals 62 studs (E5)!
| AISC 360 Composite Stud KPI (As=8,500 mm², Fy=345, fc'=28, beff=2,000, Qn=95 kN) | Formula | Verified Output |
|---|---|---|
| Half-Span Horizontal Shear V' (kN) | MIN(As*Fy, 0.85*fc'*beff*ts) / 1000 | 2,932.50 kN |
| Slab Compression Depth a (mm) | (V_prime_kN * 1000) / (0.85 * fc' * beff) | 61.61 mm (123.21 @ beff=1,000 mm) |
| Studs Required per Half-Span | ROUNDUP(V_prime_kN / Qn_stud_kN, 0) | 31 studs |
| Total Studs Across Full Span | 2 * Half_Span_Studs | 62 studs |
AISC 360 Full-Composite Horizontal Shear & Total Stud Formula
Validate A_s > 0, F_y in [235, 460] MPa, f_c_prime in [20, 60] MPa, b_eff > 200, t_slab >= 50, and Q_n > 10!
=LET(as_m, B2, fy_s, B3, fc_c, B4, be_m, B5, ts_m, B6, qn_k, B7, IF(OR(as_m <= 0, fy_s < 235, fy_s > 460, fc_c < 20, fc_c > 60, be_m <= 200, ts_m < 50, qn_k <= 10), "Invalid Composite Input", LET(vh_k, MIN((as_m * fy_s) / 1000, (0.85 * fc_c * be_m * ts_m) / 1000), 2 * ROUNDUP(vh_k / qn_k, 0))))Spandrel Edge-Beam Narrower Slab Width Check (Halve Effective Width b_eff = 1,000 mm)
Halve Effective_Slab_Width B5 from 2,000.0 mm (interior beam) to 1,000.0 mm (perimeter spandrel beam): Concrete Crushing Capacity (3,096.80 kN) STILL exceeds Steel Yield (2,932.50 kN) so Total Studs E5 remains 62—yet Slab Compression Block Depth E3 doubles to 123.21 mm (nearly filling the 130 mm slab)!
=IF(ISTEXT(E5), E5, LET(vh_k, MIN((B2 * B3) / 1000, (0.85 * B4 * B5 * B6) / 1000), ROUND((vh_k * 1000) / (0.85 * B4 * B5), 2)))Try the example
Use the original values or change them to test your understanding.