Allowable Shear Stress Of Steel

Steel Fv Allowable Shear Stress

fv = shear stress in steel beam

V = shear force at section investigated

Av = shear surface area, defined as web thickness times beam depth

Shear Stress Force The Steel Column

Shear stress in woods I-beam

Since this is non a rectangular beam, shear stress must be computed by the general shear formula. The maximum shear stress at the neutral centrality every bit well equally shear stress at the intersection between flange and web (shear plane Every bit) will exist computed. The latter gives the shear stress in the glued connection. To compare shear- and bending stress the latter is also computed.

1 Beam of L= ten ft length, with compatible load w= 280plf (W = 2800 lbs)

2 Cantankerous-section of wood I-beam

Bending moment Grand=WL/8 = 2800(10)/viii K = 3500 lb'

For the formula v= VQ/(lb) we must find the moment of inertia of the entire cross-department. Nosotros could apply the parallel axis theorem of Appendix A. However, Hue to symmetry, a simplified formula is possible, finding the moment of inertia for 'he overall dimensions as rectangular axle minus that for two rectangles on both sides of ihe web.

I = (BD3- bd3)/12 = [vi(1Ofc ii(2)60« if

Bending stress fb:

Mote V" x/2 = 5 (ha't the beam depth due to symmetry) Static moment Q of flange about the neutral axis:

Shear stress at flange/spider web intersection:

Static moment Q of flange plus upper half of web nearly the neutral axis Q = SAy = half dozen(2)4+ two(3)one.5 Q = 57 in3

Maximum shear stress at neutral axis:

v = VQ/(lb) = 1400(57)/[428(ii)] five = 93 psi <95, ok

Annotation: Maximum shear stress reaches well-nigh the allowable stress limit, merely angle stress is well below allowable bending stress because the beam is very brusque. We tin can try at what span the beam approaches allowable stress, bold Fifty= 30 ft, using the same full load W = 2800 lbs to go along shear stress constant: Grand= WL/viii = 2800(30)/eight Thousand = 10500 Ib-ft fb = Mc/I = 10500(12)5/428 fb = 1472 psi

1472 >1450, not ok

At 30 ft span bending stress is just over the allowable stress of 1450 psi. This shows that in short beams shear governs, but in long beams bending or deflection governs.

Allowable Spans For Steel Beams

Shear stress in steel beam

This beam, supporting a column point load of 96 g over a door, is a composite beam consisting of a wide-flange base axle with 8x1/2 in plates welded to top and lesser flanges. The beam is analyzed with and without plates. As shown before, for steel beams shear stress is assumed to exist resisted past the web merely, computed as fv = V/Av. The base axle is a W10x49 [10 in (254 mm) nominal depth, 49 lbs/ft (vi.77 kg/m) DL] with a moment of inertia 70= 272 in4 (11322 cm4) (run into Appendix). Shear in the welds connecting the plates to the axle is found using the shear menstruation formula q = VQ/(I).

one Axle of L= 6 ft (1.83 m) span with P = 96 thou point load

2 Composite broad-flange beam W10x49 with 8x1/2 inch stiffener plates

Bending moment M = 48(3) M = 144 g' Wide-flange beam

Shear expanse of web Av = spider web thickness x beam depth

AV = 0.34(10) Av = 3.4 in2

Shear stress fv = V/Av = W'3.4 fv = fourteen ksi

Banoirg ttiBsr. ft, - !VWI = 144(12)5.five/272 fb = 35 ksi

Since the axle would fail in bending, a blended beam is used. Blended beam

Moment of inertia 50=Z(l0o+Ay2) (see parallel axis theorem in Appendix A)

l= 272+2(8)0.53/12+2(8)0.v(v.25)2 I = 493 in4

Bending stress fb= Mc/I = 144(12)5/493 ft, = xix ksi

Since the shear force remains unchanged, the web shear stress is still ok. Shear period q in welded plate connection Q= Ay = viii(.five)v.25 = 21 in3

Since there are two welds, each resists half the full shear period q = qtot/ii q = ane thousand/in

Presume % in weld of 3.2 grand/in * strength 1 < iii.2, ok

* see AISC weld forcefulness table

Annotation: in this steel axle, bending is stress is more critical than shear stress; this is typical for steel beams, except very short ones.

Allowable Shear Stress Steel Aluminum Allowable Deflection For Steel Member

Deflection formulas

Based on the moment-area method, the following formulas for slope and deflection are derived for beams of mutual load and back up conditions. Additional formulas are provided in Appendix A. Although downward deflection would theoretically be negative, it is customary to ignore the sign convention and define up- or downward deflection past inspection. The angle 9 is the gradient of the tangent to the elastic bend at the free end for cantilever beams and at supports for simple beams; A is the maximum deflection for all cases. As derived before, nine is the area of the bending moment diagram divided by El, the elastic modulus and moment of inertia, respectively; A equals 9 times the lever-arm from the centroid of the bending moment diagram between nil and maximum deflection to the point where ix is maximum.

1 Cantilever beam with poi = 9 2/3 Fifty

2 Cantilever beam with jni east-= 1/vi WL2'(EI)

3 Simple beam with betoken load; nine= i/16 PL2/(EI)

4 Elementary axle with uniform load; nine= ane/24 WL2/(EI)

9= 1/ii PL2/(EI) A= ane/3 PL3/(EI)

i Cantilever axle with poi = nine 2/3 L