Page 13 - Mechanics of Structures – Chapter 1
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Mechanics of Civil Engineering Structures – Chapter 5
Τadd = (12500) (1875)(37.5) = 0.42 N/mm
2
6
((84.11 10 )(25))
Then, τ4 = τ3 + additional stress due to web portion
= τ3 + τadd
2
= 2.6 + 0.42 = 3.02 N/mm
7. Draw shear stress distribution diagram
τ1
τ 3
τ2
N.A 22 τ 4 = τ max
τ 5
τ 6
τ 7
7.3.4 Shear Stress for ‘L’ section of beam
Example 7
A simply supported beam of L section as shown
carries a udl of 20 kN/m for a span of 7 m.
Calculate shear stress at Neutral axis and at
connection of web and flange. 240 mm
10 mm
10 mm 190 mm
NHO JKA PSAS Page 13
