A Beginner's Guide to the Steel Construction Manual, 14^th ed.

Chapter 3 - Tension Members

© 2006, 2007, 2008, 2011 T. Bartlett Quimby

Section 3.7

Block Shear Rupture

Last Revised: 01/14/2013

Block shear is, in some ways, similar to tensile rupture in that the main part of the member tears away from the connection. The provisions for block shear are found in SCM J4.3 (SCM page 16.1-129). The difference is that there is now a combination of tension and shear on the failure path.  Like tensile rupture, there frequently is more than one failure path.  Figure 3.7.1 shows three possible block shear failure paths for a WT section.

The Limit State:

The basic limit state follows the standard form.  The statement of the limit states and the associated reduction factor and factor of safety are given here:

The values of P_u and P_a are the LRFD and ASD factored loads, respectively, applied to the member.

Nominal Block Shear Rupture Strength, R_n:

The limit state value computed is the force that STARTS the rupture.  Historically, the SCM equations have assumed that rupture initiates either on the tensile area(s) or on the shear area(s), not on both surfaces simultaneously.  With the AISC specification found in 13th edition of the SCM, revised equations were introduced. The new equations continue in the 14^th edition (Equations J4-5, in SCM J4.3). You should also read the commentary on the section (SCM pg 16.1-411).

The basic concept for computing the block shear rupture strength is that we compute the strength of the tension region and the strength of the shear region and add them together.  If you look at SCM equation J4-5 you will see the two terms.  SCM equation J4-5 computes the nominal resistance, R_n, or strength associated with block shear.

You should also notice that there are actually two equations in equation J4-5.  The term to the left is the situation for shear rupture/tensile rupture and the equation to the right is for shear yielding/tensile rupture.  The way the equations are written, you compute both values and take the smaller of the two.

One thing that you might notice right off is the presence of a "0.6" in the shear terms.  It turns out that the F_u and F_y values are determined by tension tests.  The comparable terms in shear are approximately 0.6 times the values for tension.  Consequently, wherever shear strength is computed, we use the tension values multiplied by 0.6.  Chances are, whenever you see 0.6F_y and 0.6F_u in an equation then that equation is computing shear strength.

The other quantity that will be new to you is the U_bs quantity.  U_bs was new with the 13^th edition and continues in the 14^th edition.  This quantity is used to account for non-uniform tensile stress distribution when the block shear region is non-symmetric or in any other case where non-uniform tensile stress is likely to occur on the tensile fracture area.  This condition generally occurs whenever there is only one shear area in the failure path.  The SCM commentary gives examples of such situations (SCM page 16.1-412).  We have this situation in our example problem 3.1.

Equation J4-5 also contains three area quantities: A_gv, A_nt, and A_nv.  These terms are defined in SCM J4.2 and SCM J4.3 and represent the gross and net areas in tension and shear along the failure path.  The same concepts dealing with holes that we discussed under tensile rupture apply to the computation of these quantities as well.

See example problem 3.1 for implementation of this limit state.

Another Example:

Example 3.1 is not sufficient to show what would happen with a flanged member, so let's assume that a W10x30 is used to replace the tension member used in example problem 3.1.  The same bolt pattern and size is to be used.  The same steel is also used.  The same two failure paths are potentially a problem.  Compute R_n for each path.

For failure path #1, the only difference to the prior calculation is that the plate thickness is now the thickness of the web, t_w, for the given section.  The values for the areas previously computed for path #1 can be multiplied by (t_w/0.75") to get the new values (the areas are linear with plate thickness in this case).

A_gv = (15.75 in²)(0.30"/0.75") = 6.300 in²
A_nv = (11.16 in²)(0.30"/0.75") = 4.46 in²
A_nt = (3.84 in²)(0.30"/0.75") = 1.54 in²

R_n = (563.1 k)(0.30"/0.75")
R_n = 225.3 k

The big difference comes in path #2.  In this case, the added tension area picks up the area of a flange when the path exits to the side of the member.   This makes A_nt very large for failure path #2.  Figure 3.7.2 shows the cross section for A_nt.

Figure 3.7.2
Tension Area

A_gv = (7.875 in²)(0.30"/0.75") = 3.150 in²
A_nv = (5.58 in²)(0.30"/0.75") = 2.231 in²
A_nt = (half A_g of section) + 3"(tw) - (1.5 holes in the web) = 4.926 in²

R_n = 1027 k not to exceed 665 k = minimum(1027 k, 665 k)
R_n = 665 k

In this case, failure path #1 controls since the tension area in failure path #2 more than compensates for the loss of shear failure path from path #1.  The added flange area makes a big difference!  Most experienced engineers would probably have skipped failure path #2 because of this.

Sample Spreadsheet P_n Computation

There are probably many ways to arrange this computation, but here is one.  The values in the shaded cells are either entered manually or computed elsewhere and linked to these locations.  Note that this spreadsheet only ANALYZES a given member.  For design problems where the member is unknown, you will need to solve the limit state equation for the required bolt hole spacing and end distance then make decisions from there.  In the end, however, you need to show that your final selection satisfies the limit state using some calculation similar to the one shown here.

Block Shear (SCM J4.3)				Rn = min[0.6FuAnv + UbsFuAnt, 0.6FyAgv + UbsFuAnt]
	Fy =	36	ksi
	Fu =	58	ksi
	tpl =	1	in
	db =	0.75	in
	Failure Path #1		gross path	number	net path
			length	holes/path	length	# paths	Area
			(in)		(in)		(in^2)
		Agv	10.500	0.000	10.500	2.000	21.000
		Anv1	10.500	3.500	7.438	1.000	7.438
		Anv2	10.500	3.500	7.438	1.000	7.438
		Ant	6.000	1.000	5.125	1.000	5.125
		Ubs	1.0
			Shear	Shear	Use
			Fracture	Yield
			(k)	(k)	(k)
		Rn	814.9	750.9	750.9
	Failure Path #2		gross path	number	net path
			length	holes/path	length	# paths	Area
			(in)		(in)		(in^2)
		Agv	10.500	0.000	10.500	1.000	10.500
		Anv	10.500	3.500	7.438	1.000	7.438
		Ant	7.500	1.500	6.188	1.000	6.188
		Ubs	0.5
			Shear	Shear	Use
			Fracture	Yield
			(k)	(k)	(k)
		Rn	452.8	420.7	420.7
	Controlling Rn =		420.7	k
	IF you need to determine capacity:
	LRFD				ASD
	ft =	0.75			Wt =	2
	ft Pn =	315.5	kips		Pn / Wt =	210.4	kips
	CLF	1.40			CLF	0.90
	Ps,eq =	225.4	kips		Ps,eq =	233.7	kips
	If you need to check capacity:
	LRFD				ASD
	ft =	0.75			Wt =	2
	ft Pn =	315.5	kips		Pn / Wt =	210.4	kips
	Pu	50.00	kips		Pa	50.00	kips
	Pu/ftPn =	15.8%	kips	Pa / (Pn / Wt ) =		23.8%	kips
		Okay				Okay

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