F13 CE470Ch5 Beam Shear Design

8/11/2019 F13 CE470Ch5 Beam Shear Design

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Shear Design of Beams

CE 470 - Steel Design Class

By:Amit H. Varma


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Shear Strength

Beam shear strength is covered in Chapter G of the AISC

specifications. Both rolled shapes and welded built-up shapesare covered.Rolled shapes is the focus here. Built-up shapes, commonlyreferred to as plate-girders are beyond the scope of our course.Consider the behavior of beams in shear


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Shear Strength

The web will completely yield long before the flanges begin

to yield. Because of this, yielding of the web represents oneof the shear limit states.Taking the shear yield stress as 60% of the tensile yieldstress.

This will be the nominal strength in shear provided there isno shear buckling of the web.

Shear buckling of the web depends on its h/t w ratio.If the ratio is too large, then then web can buckle in shearelastically or inelastically.

f v V n A

w

0.6 F y

V n 0.6 F y Aw


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Shear Buckling

Shearyielding

Elastic shearbuckling


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Shear Buckling

in.stiffners,ansverse between tr distanceclearawhere

260a/hor3.0a/hwhen5

)/(5

5

5

260/

2

2

w

v

v

w

t h

hak

webs stiffened for

k

t hwithwebsd unstiffene for


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Shear Design Equations


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= area of web, the overall depth times the web thickness, ,

. h= for rolled shapes, the clear distance between flanges less thefillet or corner radii, in.

= for built-up welded sections, the clear distance between theflanges, in.

= for built-up bolted sections, the distance between fastenerlines, in.

= for tees, the overall depth, in.= thickness of web, in.


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The web plate shear buckling co-efficient, , is determined as

follows:For webs without transverse stiffeners and with < 260:

= 5 except for the stem of tee shapes where = 1.2.

For webs with transverse stiffeners : = 5 + (G2-6)

=5 when a/h >3.0 or a/h >

wherea= clear distance between transverse stiffeners, in.


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Shear Design Equations2. Transverse Stiffeners:

Transverse stiffeners are not required where 2.46 , orwhere the available shear strength provided in accordance withSection G2.1 for = 5 is greater than the required shear strength.The moment of inertia, , of the transverse stiffeners used to develop

the available web shear strength, as provided in Section G2.1, about anaxis in the web center for stiffener pairs or about the face in contactwith the web plate for single stiffeners, shall meet the followingrequirement

3 (G2-7)where

= . 2 0 . 5 (G2-8)

and b is the smaller of the dimensions a and h.


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Block Shear Failure of Beam

Block shear failure was considered earlier in tension member

connections.To facilitate the connections of beams to other beams so thatthe top flanges are at the same elevation, a short length ofthe top flange of one of the beams may be cut away, orcoped .If a coped beam is connected with bolts as shown, segmentABC will tend to tear out.


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Block shear failure of beam ends

The applied load in the case of the beam will be the vertical

reaction, so shear will occur along line AB and there will betension along BC. Thus, the block shear strength will be alimiting value of the reaction.Failure is assumed to occurred by rupture (fracture) on thetension area and rupture or yielding on the shear area.Rn = 0.6 F u Anv + U bs F u Ant 0.6 F y Agv + U bs F u Ant Where, phi =0.75


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Example

Determine the maximum reaction, based on block shear, that

can be resisted by the beam shown below. Treat the bolt enddistance of 1.25 in. as standard.

The effective hole diameter is 3/4 + 1/8 = 7/8 in.The shear areas are:Agv = t w (2+3+3+3) = 0.30 (11) = 3.3 in 2 A

nv = 0.300 [11 -3.5 (7/8)] = 2.381 in 2


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Block shear example

The net tension area is:

Ant = 0.300 [1.25 - 1/2 (7/8) ] = 0.2438 in 2 Since the block shear will occur in a coped beam withstandard bolt end distance U bs = 1.0.Rn = 0.6 F u Anv + F u Ant = 108.7 kips

With an upper limit ofRn = 0.6 F y Agv + F u Ant = 114.85 kips

Therefore, nominal block shear strength = 108.7 kipsFactored block shear strength for design = 0.75 x 108.7 =81.5 kips.

F13 CE470Ch5 Beam Shear Design

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