Example G.3 Angle in shear

The results are generated with SDC Verifier 3.6 and calculated with FEMAP v11.0.0

Task:

Determine the available shear strength and adequacy of L5×3×¼ (LLV) ASTM A36 with end shears of 3.50 kips from dead load and 10.5 kips from live load.

Solution:

From AISC Manual Table 2-4, the material properties are as follows:

ASTM A36

F_y = 36 ksi

 F_u = 58 ksi

From AISC Manual Table 1-7, the geometric properties are as follows:

L5×3×¼

b = 5.00 in.

t = ¼ in.

From Chapter 2 of ASCE/SEI 7, the required shear strength is:

Note: There are no tables for angles in shear, but the available shear strength can be calculated according to AISC Specification Section G4, as follows.

AISC Specification Section G4 stipulates k_v = 1.2.

Calculate A_w.

A_w = bt

   = 5.00 in.(¼ in.)

   = 1.25 in.²

Determine C_v from AISC Specification Section G2.1(b).

h/t_w = b/t

    = 5.0 in./¼ in.

    = 20

Calculate V_n.

From AISC Specification Section G1, the available shear strength is:

Example from AISC Design Examples

Material summary

Properties Summary

Geometry Property	Value
Height	5.00
Width	3.00
h	5.00
b	3.00
d	0.25
t	0.25

FEM Loads and Constraint

1..Dead load 3.5 kips

2..Live load 10.5 kips

Constraint

Check 1..ANSI / AISC LRFD 360-10

Beam Characteristics

All (LS1, All Entities)

Shear

All (LS1, 17 Property Shape(s))

From Chapter 2 ASCE/SEI 7, the required shear strength is:

Calculate A_w.

AISC Specification Section G1, the available shear strength is:

Comparing results of calculation in SDC Verifier and in Example G.3 we can see that values completely match.

The required shear strength is 21 kips.

The available shear strength is 24.3 kips.