For shear stress W at a section y 1 from the neutral axis the incremental

# For shear stress w at a section y 1 from the neutral

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For shear stress W at a section y 1 from the neutral axis, the incremental horizontal force dN over the slice of thickness t is dN = W tdx or W ±µ¶ t ² dN / dx The total horizontal (normal) forces due to the bending moment are given as ࠵? = ∫ ࠵?࠵?࠵? and ࠵? + ࠵?࠵? = ∫ (࠵? + ࠵?࠵?)࠵?࠵? ࠵? 2 ࠵? 1 ࠵? 2 ࠵? 1 Subtracting the two terms, we have: ࠵?࠵? = ∫ ࠵?࠵?࠵?࠵? = ∫ ࠵?࠵? ࠵? ࠵?࠵?࠵? = ࠵?࠵? ࠵? ࠵?࠵?࠵? = ࠵?࠵? ࠵? ࠵? ࠵? 2 ࠵? 1 ࠵? 2 ࠵? 1 ࠵? 2 ࠵? 1 where ࠵?࠵? = ࠵?࠵? ࠵? ࠵? and ࠵? = ∫ ࠵?࠵?࠵? (= ∑ ࠵? ࠵? ̅ ࠵? ࠵? ) ࠵? 2 ࠵? 1 is termed the first moment of area. The shear stress is thus given as ࠵? = 1 ࠵? ࠵?࠵? ࠵?࠵? = 1 ࠵? ࠵?࠵? ࠵?࠵? ࠵? ࠵? = ࠵?࠵? ࠵?࠵? where the shear force (from mechanics of materials, is defined as the change of moment) ࠵? = ࠵?࠵? ࠵?࠵? . It should be noted that the shear stress W is defined over the slice with thickness t . 113 Example : A W 16 × 100 ( I x = 1490 in 4 ) is subjected to a vertical shear force of 80 kips. Determine the maximum shear stress and plot the distribution of shear stress for the cross section. Solution : The maximum shear stress will be at the neutral axis where Q , the statical moment about the neutral axis, is maximum. 4 ࠵? = 10.425(0.985) (7.50 + 0.985 2 ) + 7.50(0.585) ( 7.50 2 ) = 98.5 in 3 Shear stress at NA: ࠵? ࠵? = ࠵?࠵? ࠵?࠵? = 80(98.5) 1490(0.585) = 9.04 ksi The shear stresses over the cross section can be calculated as shown. _______________________ Note that the flanges resist very low shear stresses. It is the web that predominantly resists the shear in wide flange beams. AISC allows the use of average web shear for calculating shear stress: ࠵? ࠵? = ࠵? ࠵?࠵? ࠵? = 80 16.97(0.585) = 8.06 ksi where d is the full depth of beam and t w is the web thickness of the beam. This is less than the shear stress of 9.04 ksi calculated by the general shear shear stress formula! 4 Shear stress between flange and web: ࠵? = 10.425(0.985) (7.50 + 0.985 2 ) = 82.07 in 3 Bottom of flange: ࠵? ࠵? = 80(82.07) 1490(10.425) = 0.42 ksi Top of web: ࠵? ࠵? = 80(82.07) 1490(0.585) = 7.53 ksi b f = 10.425 in t f = 0.985 in 7.50 in t w = 0.585 in d = 16.97 in N.A. 0.42 ksi 7.53 ksi 9.04 ksi Cross section Shear distribution 114 Design Shear Strength : V u ≤ φ v V n , φ v =1.0 (for W-shapes, see AISC Pg. 16.1-67). V n = 0.6 F y A w C v (Eq. G2-1) where C v is termed web shear Coefficient. (Note: Shear yield strength F v = 0.6 F y .) For webs of rolled I-shape members with ℎ/࠵? ࠵? ≤ 2.24 ࠵?/࠵? ࠵? (which applies to all F y = 50 ksi W-shapes and most W-shapes tabulated in the manual, see pg. 16.1-68), C v =1. Generally, web shear strength may need to consider possible shear buckling limit state. For example, for webs of doubly symmetric shapes (except rolled W-shapes), etc. (see AISC Section G2.1(b)), Web yielding (plastic zone): ࠵? ࠵? ≤ 1.10 ࠵?࠵? ࠵? ࠵? ࠵? (≈ 418 ࠵? ࠵? ) C v =1 (Eq. G2-3) • Inelastic buckling of web : 1.10 ࠵?࠵? ࠵? ࠵? ࠵? < ࠵? ࠵? ≤ 1.37 ࠵?࠵? ࠵? ࠵? ࠵? (≈ 522 ࠵? ࠵? ) ࠵? ࠵? = 1.10√࠵?࠵?  #### You've reached the end of your free preview.

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• Spring '14
• GregoryG.Deierlein
• Shear Stress, FY, Shear strength
• • • 