Beam To HSS Column Connection: SHEAR PLATE SUBJECT TO SHEAR & AXIAL LOAD

CONNECTION LAYOUT:

MEMBER SIZES:
Beam Size
Column Size
Connection Interface

SHEAR PLATE DATA:
Plate Thickness
Plate Placement (Is CENTERED On Beam?)
Steel Plate Grade

DISTANCES and SPACINGS:
First Bolt Distance from Top of Beam
Beam-Column Gap
Beam Edge Distance
Plate Vertical Edge Distance
Plate Horizontal Edge Distance
Bolt Vertical Spacing
Bolt Horizontal Spacing

BOLT INPUT DATA:
Bolt Diameter
Bolt ASTM Grade
Number of Bolt Columns
Number of Bolt Rows

WELD INPUT DATA:
Fillet Weld Size
Weld Class/Electrode

BEAM END LOADS:
Factored Vertical Load [KIP]	Vu
Factored Axial Load (Compression/Tension) [KIP]	Pu

SOLUTION/DESIGN CALCULATIONS:

A.) GENERAL:
•	Reference: AISC - American Institute of Steel Construction Manual, 14^th Edition
•	Design Method: LRFD - Load and Resistance Factor Design

B.) DESIGN NOTES:

The loading for this connection is a combination of SHEAR and AXIAL LOAD. As a result, the Tension Capacity for any "LIMIT STATE" is reduced by the presence of Shear, or vice versa.

For design bolt tension in the presence of shear, the "INTERACTION" (combined stresses) is handled directly by the AISC Code equations.

For other "LIMIT STATES" in combined stresses such as Bolt Bearing, Gross and Net Shear and Tension, and Block Shear and Tension Tearout, the effect of "INTERACTION" is handled by the use of the formula, P/R_a + (R/R_v)² = 1, as suggested from the following reference:

"Combined Shear and Tension Stress" - by Subhash C. Goel, AISC Journal, 3^rd Qtr.-1986.

Alternately, the effect of "INTERACTION" can also be evaluated by the use of VON MISES' YIELD CRITERION formula, (P/R_a)² + 3(R/R_v)² = 1

Where:
P = Connection Axial Load
R_a = Design Axial Strength for the particular "Limit State" considered
R = Connection Shear Load
R_v = Design Shear Strength for the particular "Limit State" considered

C.) LOAD:
V_u	=	Factored Shear Load
	=	0.00 Kips
P_u	=	Factored Axial Load
	=	0.00 Kips
R_u	=	Factored Resultant Load
	=	$sqrt(V_u^2 + P_u^2)$
	=	0.00 Kips

D.) SHEAR PLATE & DISTANCES:
t_p	=	Thickness of Plate
	=	0.00 inches
h_p	=	Depth of Plate
	=	0.00 inches
w_p	=	Width of Plate
	=	0.00 inches
L_ev	=	Plate Vertical Edge Distance
	=	0.00 inches
L_eh	=	Plate Horizontal Edge Distance
	=	0.00 inches
L_eb	=	Beam Edge Distance
	=	0.00 inches

E.) MATERIAL PROPERTIES:
F_yp	=	Plate Yield Strength
	=	0.00 Ksi
F_up	=	Plate Tensile Strength
	=	0.00 Ksi
F_yb	=	Beam Yield Strength
	=	0.00 Ksi
F_ub	=	Beam Tensile Strength
	=	0.00 Ksi
E	=	Modulus of Elasticity
	=	0.00 Ksi

F.) BEAM PROPERTIES:
Size	=	XX
d	=	Beam Depth
	=	0.00 inches
t_w	=	Beam Web Thickness
	=	0.00 inches
t_f	=	Beam Flange Thickness
	=	0.00 inches
k_det	=	Beam k-Detailing Dimension
	=	0.00 inches

G.) COLUMN PROPERTIES:
Size	=	XX
H_t	=	Depth of HSS
	=	0.00 inches
B_t	=	Width of HSS
	=	0.00 inches
t_nom	=	HSS Nominal Wall Thickness
	=	0.00 inches
t_des	=	HSS Design Wall Thickness
	=	0.00 inches

H.) DETERMINE APPLICABILITY OF SPECIFICATION SECTION K1.3:
» Strength Check:
F_yc	=	Yield Strength of HSS
	=	0.00 Ksi	>	0.00 Ksi	[OK]	AISC 360-10 Table K2.1A
» Ductility Check:
F_uc	=	Tensile Strength of HSS
	=	0.00 Ksi
$F_(yc)/F_(uc)$	=	0.00 Kips	>	0.00 Kips	[OK]	AISC 360-10 Table K2.1A
» Check if SHEAR PLATE connection is suitable (verify HSS wall is not slender):
$b/t$	=	Width-Thickness Ratio				AISC 360-10 Table B4.1b, Case 17
	=	XXX
Thus:
$lambda$	=	Limiting Width-Thickness Ratio
	=	$1.4*[E/F_(yc)]^(1/2)$				AISC 360-10 Table B4.1b, Case 17
	=	0.00	>	0.00	[OK]

I.) CHECK SHEAR PLATE THICKNESS:
T_max	=	Limiting Thickness of Shear Plate such that its strength does not exceed the strength of the HSS wall
	=	$[F_(uc)/F_(yp)]*t_(des)$				AISC 360-10 [Eq. K1-10]
	=	XXX inches
Thus:
t_p	=	Assigned 'Shear Plate' Thickness
	=	0.00 inches	>	0.00 inches	[OK]

J.) CHECK HSS WALL AGAINST AXIAL LOAD:
B	=	HSS Dimension Perpendicular to the Plane of Shear Plate
	=	XXX inches
N_L	=	Load Bearing Length
	=	XXX inches
Q_f	=	Chord Stress Interaction Parameter
	=	XXX inches
Thus:
$phi$ R_n	=	HSS Wall Design Strength against Axial Load
	=	$(F_(yc)t_(des)^2)/(1-[t_p/B])[(2N_L)/B+4sqrt(1-t_p/B)*Q_f]$				AISC 360-10 [Eq. K1-12]
	=	0.00 Kips	>	0.00 Kips	[OK]

K.) BOLT DATA/AVAILABLE STRENGTH:
Bolt Type: A325-N
Designation: ¾"Ø - A325-N-STD
d_b	=	Bolt Diameter
	=	0.00 inches
N	=	Number of Bolt Rows
	=	00
M	=	Number of Bolt Columns
	=	00
S_n	=	Bolt Row Spacing
	=	0.00 inches
S_m	=	Bolt Column Spacing
	=	0.00 inches
e_x	=	Load Eccentricity from Bolt C.G.
	=	0.00 inches
Ø	=	Load Angle From Vertical
	=	$ArcTan [(P_u)/(V_u)]$
	=	0.00 degrees
C	=	Bolt Group Coefficient	By IC Method
	=	0.00
$phi$ _vr_n	=	Available Bolt Shear Strength	AISC 14^th Ed. [Table 7-3]
	=	0.00 Kips
$phi$ _br_n	=	Available Bolt Bearing Strength on Shear Plate/Beam Web
	=	$0.75min[(2.4F_(ub)t_wd_b), (2.4F_(up)t_p*d_b)]$
	=	0.00 Kips

L.) BOLT STRENGTH CHECK:
$phi$ R_n	=	Design Bolt Shear Strength			AISC 14^th Ed. [Eq. 7-19]
	=	$C*min[phi_vr_n, phi_br_n]$
	=	0.00 Kips	0.00 Kips	[OK]

M.) BOLT BEARING on SHEAR PLATE under SHEAR and AXIAL TENSILE Load:
» Bolt Bearing On Shear Plate Under Shear:
Exterior Bolt:
L_c	=	Clear Edge Distance
	=	XXX inches
$phi_er_n$	=	$0.75min[1.2L_ct_pF_(up); 2.4d_bt_p*F_(up)]$				AISC 360-10 [Eq. J3-6a]
	=	XXX Kips/Bolt
Interior Bolt:
L_c	=	Clear Distance Between Interior Bolts
	=	XXX inches
$phi_ir_n$	=	$0.75min[1.2L_ct_pF_(up); 2.4d_bt_p*F_(up)]$				AISC 360-10 [Eq. J3-6a]
	=	XXX Kips/Bolt
Thus:
$phiR_n$	=	Design Bearing Strength of Bolts
	=	$M[phi_er_n + (N-1)phi_ir_n]$
	=	0.00 Kips	>	0.00 Kips	[OK]
» Bolt Bearing On Shear Plate Under Axial Tensile Load:
Exterior Bolt:
L_c	=	Clear Edge Distance
	=	XXX inches
$phi_er_n$	=	$0.75min[1.2L_ct_pF_(up); 2.4d_bt_p*F_(up)]$				AISC 360-10 [Eq. J3-6a]
	=	XXX Kips/Bolt
Interior Bolt:
L_c	=	Clear Distance Between Interior Bolts
	=	XXX inches
$phi_ir_n$	=	$0.75min[1.2L_ct_pF_(up); 2.4d_bt_p*F_(up)]$				AISC 360-10 [Eq. J3-6a]
	=	XXX Kips/Bolt
Thus:
$phiR_n$	=	Design Bearing Strength of Bolts
	=	$N[phi_er_n + (M-1)phi_ir_n]$
	=	0.00 Kips	>	0.00 Kips	[OK]
» Check Bolt Bearing On Shear Plate Under Combined Load:
P	=	x.xx Kips			[Axial Load]
R_a	=	x.xx Kips			[Design Axial Strength]
R	=	x.xx Kips			[Shear Load]
R_v	=	x.xx Kips			[Design Shear Strength]
Substitute to 'Subhash C. Goel' Interaction Equation:
$[P/R_a] + [R/R_v]^2 <= 1.0$
1.00	=	1.00	[OK]
Substitute to 'Von Mises Criterion' Interaction Equation:
$[P/R_a]^2 + 3*[R/R_v]^2 <= 1.0$
1.00	=	1.00	[OK]

N.) SHEAR PLATE YIELDING under SHEAR and AXIAL TENSILE Load:
» Plate Shear Yielding:
$phiR_n$	=	Design Shear Yield Strength				AISC 360-10 [Eq. J4-3]
	=	$0.6F_(yp)[t_p*h_p]$
	=	0.00 Kips	>	0.00 Kips	[OK]
» Plate Tensile Yielding:
$phiR_n$	=	Design Tensile Yield Strength				AISC 360-10 [Eq. D2-1]
	=	$0.9F_(yp)[t_p*h_p]$
	=	0.00 Kips	>	0.00 Kips	[OK]
» Check Plate Yielding Under Combined Load:
P	=	x.xx Kips			[Axial Load]
R_a	=	x.xx Kips			[Design Axial Strength]
R	=	x.xx Kips			[Shear Load]
R_v	=	x.xx Kips			[Design Shear Strength]
Substitute to 'Subhash C. Goel' Interaction Equation:
$[P/R_a] + [R/R_v]^2 <= 1.0$
1.00	=	1.00	[OK]
Substitute to 'Von Mises Criterion' Interaction Equation:
$[P/R_a]^2 + 3*[R/R_v]^2 <= 1.0$
1.00	=	1.00	[OK]

O.) SHEAR PLATE RUPTURE under SHEAR and AXIAL TENSILE Load:
» Plate Shear Rupture:
A_nv	=	Net Cross-sectional Area subject to Shear
	=	$t_p[h_p-N(d_b+0.125)]$
	=	XXX in²
Thus:
$phiR_n$	=	Design Shear Rupture Strength				AISC 360-10 [Eq. J4-4]
	=	$0.75[0.6F_(up)*A_(nv)]$
	=	0.00 Kips	>	0.00 Kips	[OK]
» Plate Tensile Rupture:
A_g	=	Gross Cross-sectional Area subject to Tension
	=	$t_p*h_p$
	=	XXX in²
A_n	=	Net Cross-sectional Area subject to Tension
	=	$t_p[h_p-N(d_b+0.125)]$
	=	XXX in²
A_e	=	Effective Cross-sectional Area subject to Tension
	=	$A_n<=0.85*A_g$
	=	XXX in²
Thus:
$phiR_n$	=	Design Tensile Rupture Strength				AISC 360-10 [Eq. D2-2]
	=	$0.9[F_(up)A_e]$
	=	0.00 Kips	>	0.00 Kips	[OK]
» Check Plate Rupture Under Combined Load:
P	=	x.xx Kips			[Axial Load]
R_a	=	x.xx Kips			[Design Axial Strength]
R	=	x.xx Kips			[Shear Load]
R_v	=	x.xx Kips			[Design Shear Strength]
Substitute to 'Subhash C. Goel' Interaction Equation:
$[P/R_a] + [R/R_v]^2 <= 1.0$
1.00	=	1.00	[OK]
Substitute to 'Von Mises Criterion' Interaction Equation:
$[P/R_a]^2 + 3*[R/R_v]^2 <= 1.0$
1.00	=	1.00	[OK]

P.) PLATE BLOCK SHEAR and TENSION TEAR-OUT due to SHEAR and AXIAL TENSILE Load:
» Plate Block Shear Strength: (consider L-Pattern Failure for combined stress)
$U_(bs)$	=	XXX
A_gt	=	Gross Area with Tensile Resistance
	=	$[(M-1)S_m+L_(eh)]t_p$
	=	XXX in²
A_nt	=	Net Area with Tensile Resistance
	=	$A_("gt") - (M-0.5)(d_b+0.125)t_p$
	=	XXX in²
A_gv	=	Gross Area with Shear Resistance
	=	$[h_p-L_(ev)]*t_p$
	=	XXX in²
A_nv	=	Net Area with Shear Resistance
	=	$A_(gv)-(N-0.5)(d_b+0.125)tp$
	=	XXX in²
Thus:
$phiR_n$	=	Design Block Shear Strength				AISC 360-10 [Eq. J4-5]
	=	$0.75min[(0.6F_(up)A_(nv)+U_(bs)F_(up)A_(nt)); (0.6F_(yp)A_(gv)+U_(bs)F_(up)*A_(nt))]$
	=	0.00 Kips	=	0.00 Kips	[OK]
» Plate Tension Tear_Out Strength: (consider L-failure Pattern for combined stress)
$U_(bs)$	=	XXX
A_gv	=	Gross Area with Shear Resistance
	=	$[(M-1)S_m+L_(eh)]t_p$
	=	XXX in²
A_nv	=	Net Area with Shear Resistance
	=	$A_(gv)-(M-0.5)(d_b+0.125)t_p$
	=	XXX in²
A_gt	=	Gross Area with Tensile Resistance
	=	$[h_p-L_(ev)]*t_p$
	=	XXX in²
A_nt	=	Net Area with Tensile Resistance
	=	$A_("gt")-(N-0.5)(d_b+0.125)t_p$
	=	XXX in²
Thus:
$phiR_n$	=	Design Tensile Tear-Out Strength				AISC 360-10 [Eq. J4-5]
	=	$0.75min[(0.6F_(up)A_(nv)+U_(bs)F_(up)A_(nt)); (0.6F_(yp)A_(gv)+U_(bs)F_(up)*A_(nt))]$
	=	0.00 Kips	>	0.00 Kips	[OK]
» Check Plate Combined Block Shear and Tension Tear-Out:
P	=	x.xx Kips			[Axial Load]
R_a	=	x.xx Kips			[Design Axial Strength]
R	=	x.xx Kips			[Shear Load]
R_v	=	x.xx Kips			[Design Shear Strength]
Substitute to 'Subhash C. Goel' Interaction Equation:
$[P/R_a] + [R/R_v]^2 <= 1.0$
1.00	=	1.00	[OK]
Substitute to 'Von Mises Criterion' Interaction Equation:
$[P/R_a]^2 + 3*[R/R_v]^2 <= 1.0$
1.00	=	1.00	[OK]

Q.) Beam Web BLOCK SHEAR and TENSION TEAR-OUT due to SHEAR and AXIAL TENSILE Load:
» Beam Web Block Shear Strength:
** There is NO vertical block shear for this case. Shear PATH just runs into the flange.
» Beam Web Tension Tear-Out Strength: (C-Pattern Failure)
$U_(bs)$	=	XXX
A_gv	=	Gross Area with Shear Resistance
	=	$2[L_(eb)+(M-1)S_m]*t_w$
	=	XXX in²
A_nv	=	Net Area with Shear Resistance
	=	$A_(gv)-2(M-0.5)(d_b+0.125)*t_w$
	=	XXX in²
A_gt	=	Gross Area with Tensile Resistance
	=	$[(N-1)S_n]t_w$
	=	XXX in²
A_nt	=	Net Area with Tensile Resistance
	=	$A_("gt")-[(N-1)(d_b+0.125)]t_w$
	=	XXX in²
Thus:
$phiR_n$	=	Design Tensile Tear-Out Strength				AISC 360-10 [Eq. J4-5]
	=	$0.75min[(0.6F_(ub)A_(nv)+U_(bs)F_(ub)A_(nt)); (0.6F_(yb)A_(gv)+U_(bs)F_(ub)*A_(nt))]$
	=	0.00 Kips	=	0.00 Kips	[OK]

R.) SHEAR PLATE under Combined COMPRESSION and BENDING:
» Geometric Properties:
K	=	Effective Length Factor
	=	XXX
L	=	Plate Unsupported Length
	=	XXX inches
r	=	Least Radius of Gyration
	=	$[t_p/3.464]$
	=	XXX inches
A_g	=	Cross-sectional Area
	=	$t_p * h_p$
	=	XXX in²
Z	=	Plastic Section Modulus
	=	$[t_p * h_p^2]/4$
	=	XXX in³
» Evaluate Plate Strength:
F_e	=	Elastic Critical Buckling Stress		AISC 360-10 [Eq. E3-4]
	=	$(pi^2E)/[(KL)/r]^2$
	=	XXX Ksi
Since Fe > 0.44Fy, therefore:
F_cr	=	Elastic Critical Buckling Stress
	=	$[0.658^(F_(yp)/F_e)]*F_(yp)$		AISC 360-10 [Eq. E3-2]
	=	$0.877*F_e$		AISC 360-10 [Eq. E3-3]
	=	XXX Ksi
P_r	=	Required Axial Compressive Strength
	=	P_u
	=	XXX Kips
P_cy	=	Available Axial Compressive Strength
	=	$0.9[F_(cr)A_g]$		AISC 360-10 [Eq. E3-1]
	=	XXX Kips
M_r	=	Required Flexural Strength
	=	$V_u*L$
	=	XXX Kip-in
M_cx	=	Available Flexural Strength
	=	$0.9[F_(yp)Z]$		AISC 360-10 [Eq. E7-1]
	=	XXX Kip-in
C_b	=	Lateral-Torsional Buckling Modification Factor
	=	XXX (assumed)
For Doubly Symmetric Members in Single Axis Bending, the Interaction of Flexure and Compression in Column is controlled by the below equation:
$P_r/P_(cy)[1.5-0.5P_r/P_(cy)]+[M_r/(C_b*M_(cx))]^2 <= 1.0$				AISC 360-10 [Eq. H1-2]
Substituting values to Interaction Equation H1-2, therefore:
1.00	=	1.00	[OK]

S.) Plate to HSS Column Wall WELD:
F_EXX	=	Tensile Strength of Weld
	=	0.00 Ksi
ct_fw	=	Column Flange Thickness
	=	0.00 inch
w_a	=	Assigned Fillet Weld Size
	=	0.00 inch
w_u	=	Useful weld based on Column Flange rupture @ weld zone				AISC 14^th Ed. [Eq. 9-2]
	=	$(F_(ub)*[(ct_(fw))/16])/3.09$
	=	0.00 inch
w_e	=	Effective Fillet Weld Size
	=	$min[w_a, w_u]$
	=	0.00 inch
Ø	=	Load Angle relative to Weld Line
	=	$Arc tan[P_u/V_u]$
	=	0.00 radian
	=	0.00 degrees
V_w	=	Available Shear Strength of Weld per linear inch
	=	$0.75[0.6F_(EXX)(1.0+0.50sin^1.5theta)0.707w_e]$
	=	0.00 Kips/in
L_w	=	Total Length of Fillet Weld (back-to-back)
	=	0.00 inches
Thus:
$phi$ R_n	=	Design Shear Strength of Weld
	=	$V_w*L_w$
	=	0.00 Kips	>	0.00 Kips	[OK]

S.) Through-Plate to HSS Column Wall WELD:
F_EXX	=	Tensile Strength of Weld
	=	0.00 Ksi
ct_des	=	HSS Design Wall Thickness
	=	0.00 inch
w_a	=	Assigned Fillet Weld Size
	=	0.00 inch
w_u	=	Useful weld based on Column Flange rupture @ weld zone				AISC 14^th Ed. [Eq. 9-2]
	=	$(F_(ub)*[(ct_(fw))/16])/3.09$
	=	0.00 inch
w_e	=	Effective Fillet Weld Size
	=	$min[w_a, w_u]$
	=	0.00 inch
H	=	HSS Dimension Parallel to Shear Plate Plane
	=	0.00 inch
R_h	=	Horizontal Reaction on Either Weld
	=	$P_u/2$
	=	0.00 Kips
R_v	=	Vertical Reaction on Either Weld
	=	$max(V_u[(H+e_x)/H]; V_u[e_x/H])$				AISC 14^th Ed. [Fig. 10-47]
	=	0.00 Kips
R_w	=	Resultant Reaction on Either Weld
	=	$sqrt(R_h^2 + R_v^2)$
	=	0.00 Kips
Ø	=	Load Angle relative to Weld Line
	=	$Arc tan[R_h/R_v]$
	=	0.00 radian
	=	0.00 degrees
V_w	=	Available Shear Strength of Weld per linear inch
	=	$0.75[0.6F_(EXX)(1.0+0.50sin^1.5theta)0.707w_e]$
	=	0.00 Kips/in
L_w	=	Total Length of Fillet Weld (back-to-back)
	=	0.00 inches
Thus:
$phi$ R_n	=	Design Shear Strength of Weld
	=	$V_w*L_w$
	=	0.00 Kips	>	0.00 Kips	[OK]

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Saturday, April 6, 2019

Beam To HSS Column Connection: SHEAR PLATE SUBJECT TO SHEAR & AXIAL LOAD

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