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

Fy = 36 ksi

 Fu = 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:

G3 1

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 kv = 1.2.

Calculate Aw.

Aw = bt

   = 5.00 in.(¼ in.)

   = 1.25 in.2

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

h/tw = b/t

    = 5.0 in./¼ in.

    = 20

G3 2

Calculate Vn.

G3 3

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

G3 4

Example from AISC Design Examples

Material summaryG3 material

Properties Summary

G3 property

G3 property plot

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

G3 deadload

2..Live load 10.5 kips

G3 liveload

Constraint

G3 constraint

G3 plot

Check 1..ANSI / AISC LRFD 360-10

Beam Characteristics

All (LS1, All Entities)

G3 beam

Shear

All (LS1, 17 Property Shape(s))

G3 shear

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

G3 shear2

Calculate Aw.

G3 shear1

AISC Specification Section G1, the available shear strength is:

G3 shear3

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.

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