Beam to beam double angle bolted  bearing type connection

LRFD Design Example:

Situation: A structural steel beam to beam double angle bearing type connection is shown below.

Design code: AISC 13th edition. LRFD

Design data:

Connection beam: W12x40, ASTM A992

Supporting beam: W18x35, ASTM A992

Angle: L3x3x1/4, ASTM A36

Bolt: 2/4" diameter, ASTM A325-N (thread not excluded)

Live load reaction: 25 kips

Dead load reaction: 25 kips

Requirement: Check if the connection design is adequate.

• Bookmark:
• Check number of bolts in W12x40
• Check number of bolts in W18x35
• Check bolt bearing in beam web:
• Check shear strength of beam web
• Check block shear of beam web
• Check bolt bearing on angle
• Check shear strength of angle
• Check block shear of angle

Calculated factored reaction

Factored reaction: R_u = 1.2 (25) + 1.6 (25) = 70 kips

Check number of bolts in W12x40:

Resistance factor: f = 0.75

Nominal shear strength of A325-N bolt, F_nv = 48 ksi

Gross section of 3/4" diameter bolts: 0.442 in².

Bolt strength in single shear: fR_n = (0.75)(48)(0.442) = 15.9 kip

This value is shown in AISC manual, 13th ed. Table 7-1.

Number of bolts in W12x40 in double shear: N = R_u / 2 fR_n = 2.2  < 3 O.K.

Check number of bolts in W18x35

Distance from center of bolts in W12x40 to face of support = 1.25+0.5 = 1.75 in

Moment due to eccentricity: M_u = (70)(1.75) = 122.5 in-kip

Assume neutral axis of bolt group is at middle, moment of inertia of bolt group

I = 4 (3 in)² = 36 in².

Tension in bolt: T_u = (122.5)(3)/36 = 10.2 kip

Shear stress in bolt, f_v = 70/[(6)(0.442)] = 26.4 ksi

Nominal tensile stress of A325 bolt is 90 ksi. Available tensile stress in bolt:

F'_nt = (1.3)(90)-[(90)/(0.75)(48)](26.4) = 51 ksi

Available tensile strength of bolt = (0.75)(51)(0.442) = 16.9 kip > 10.2 kip O.K.

Check bolt bearing in beam web:

Web thickness of W12x40: 5/16 inches

Resistance factor: f = 0.75

Bolt hole for 3/4" bolt = 7/8"

Clear distance between bolts: 3" - 7/8" = 2.125"

Clear distance between bolt and edge of angle = 1.5"-7/16" = 1.063"

Minimum tensile strength of A992 material: F_u = 65 ksi

Nominal bearing strength: R_n = 1.2 [2(2.125)+1.063)](5/16)(65) = 129.5 kips

Maximum bearing strength: 2.4d t_w F_u = 2.4 (3/4)(5/16)(65) (3) = 109.7 kips

Bearing strength: (0.75)(109.7) = 82.3 kip > 70 kips O.K.

Check shear strength of beam web

Check shear yield of beam web

Length of copped beam: 12-1.5 = 10.5"

Gross shear area: (10.5)(5/16) = 3.28 in².

Resistance factor: f = 1.0

Shear strength of web, fR_n = f (0.6 F_y A_g) = (1)(0.6)(50)(3.28) = 98.4 kip > 70 kips O.K.

Check shear rupture of beam web:

Net length of copped beam: 12-1.5-(3)(7/8) = 7.875"

Net area A_nv = (7.875)(5/16) = 2.46 in².

Resistance factor: f = 0.75

Shear strength of web: fR_n = f (0.6 F_u A_nv) = (0.75)(0.6)(65)(2.46) = 72 kip > 70 kips O.K.

Check block shear of beam web

Resistance factor: f = 0.75

Net shear area of block shear: A_nv = (7.5-7/16)(5/16) = 2.21 in². 

Assume the gap between end of W12x40 and web of W18x35 is 1/2".

The edge distance of bolt to end of beam is 1.25".

Net tension area of block shear: Ant = (1.25-7/16)(5/16) = 0.254 in².

Gross section of shear area: A_gv = (10.5)(5/16) = 3.28 in².

Factored U_bs = 1 for single column of bolt

Nominal block shear strength: f R_n = 0.75(0.6F_uA_nv+U_bsF_uA_nt) = 77 kip > 70 kips O.K.

Maximum block shear strength: f (0.6F_yA_gv+U_bsF_uA_nt) = 84.9 kip

Check L3x3x1/4" angle

Factored load on each angle = 70/2 = 35 kips

Check bolt bearing on angle

Resistance factor: f = 0.75

Clear distance between bolts: 3 -7/8 = 2.125"

Clear distance between bolt and edge of angle: 1.5-7/16 = 1.063"

Tensile strength of A36 steel, F_u = 58 ksi

Bearing strength: fR_n = (0.75)(1.5)[(2(2.125)+1.063)](1/4) (58) = 69.3 kips 

Maximum bearing strength:  f(2.4d t_w F_u ) = (0.75)[2.4 (3/4)(1/4)(58)](3) = 58.7 kips > 35 ksi O.K.

Check shear strength of angle

Check shear yield of angle

Length of angle: 9"

Gross shear area: (9)(1/4) = 2.25 in².

Resistance factor: f = 1.0

Shear strength of web, fR_n = f (0.6 F_y A_g) = (1)(0.6)(36)(2.25) = 48.6 kip > 35 kips O.K.

Check shear rupture of angle:

Net length of angle: 9-(3)(7/8) = 6.375"

Net area A_nv = (6.375)(1/4) = 1.594 in².

Resistance factor: f = 0.75

Shear strength of web: fR_n = f (0.6 F_u A_nv) = (0.75)(0.6)(58)(1.594) = 41.6 kip > 35 kips O.K.

Check block shear of angle

Resistance factor: f = 0.75

Net shear area of block shear: A_nv = (7.5-7/16)(1/4) = 1.77 in². 

The edge distance of bolt is 1.25".

Net tension area of block shear: A_nt = (1.25-7/16)(1/4) = 0.203 in².

Gross section of shear area: A_gv = (9)(1/4) = 2.25 in².

Factored U_bs = 1 for single column of bolt

Nominal block shear strength: f R_n = 0.75(0.6F_uA_nv+U_bsF_uA_nt) = 54.9 kip 

Maximum block shear strength: f (0.6F_yA_gv+U_bsF_uA_nt) = 45.3 kip > 35 kips O.K.