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Connections

6% of exam

Bolted and welded steel connections, bearing and slip-critical bolts, weld strength, block shear, and embedded, anchored, and post-installed anchors in concrete.

5 concepts

D. Connections

AISC Bolted Connections (Shear, Bearing, Slip)

Bolt shear strength, bearing and tearout at the holes, bearing vs slip-critical behavior, and how the bolt-group capacity is the sum of per-bolt limit states.

A bolted connection fails at whatever limit state reaches its strength first, and the exam's favorite trick is to make you check the wrong one. A row of high-strength bolts can be governed by the bolts shearing off, by the plate crushing against the bolt shanks (bearing), by the material tearing out behind a bolt to the edge (tearout), or — for a slip-critical joint — by the faying surfaces sliding before any of that. The NCEES PE Civil Handbook gives you bolt-group GEOMETRY (§4.2.1) but no member or bolt strength tables, so every number in this concept comes from AISC 360-16 Chapter J and the 15th-edition Steel Construction Manual. Treat the connection as a small system: compute each per-bolt limit state, take the smallest, and only then multiply by the number of bolts.

Bolt shear strength

The nominal shear strength of one bolt is the nominal shear stress

F_{nv}

times the bolt's nominal (gross shank) area

A_b

, multiplied by the number of shear planes. For Group A bolts (A325/F3125 Gr A325),

F_{nv}=54\,\text{ksi}

when threads are in the shear plane (the N condition) and

68\,\text{ksi}

when they are excluded (the X condition); Group B (A490) runs

68

and

84\,\text{ksi}

. The strength-reduction factor for bolts in shear is

\phi=0.75

. A bolt in DOUBLE shear has two planes, so its strength doubles — this is the most common factor-of-two slip on the exam.

\phi R_n = \phi\,F_{nv}\,A_b\,m, \qquad A_b=\frac{\pi d^2}{4}, \quad \phi=0.75

Bearing and tearout at the bolt holes

Where the bolt presses on the connected material, the plate can crush (bearing) or the chunk of plate behind the bolt can shear out to the next hole or the free edge (tearout). AISC 360-16 §J3.10 gives both as a function of the plate thickness

t

and tensile strength

F_u

, with bearing capped at

2.4\,d\,t\,F_u

(deformation at service load a consideration) and tearout at

1.2\,l_c\,t\,F_u

, where

l_c

is the CLEAR distance — edge distance or bolt spacing MINUS the hole dimensions — in the direction of force. Use the standard hole diameter for tearout —

d_h=d+\tfrac{1}{16}\,\text{in}

for bolts up to

\tfrac78

in, and

d+\tfrac18\,\text{in}

for

d\ge1

in (AISC Table J3.3) — and remember the factor is again

\phi=0.75

. For each bolt the controlling value is the smaller of bearing and tearout, summed over the group.

\phi R_n = \phi\min\!\big(2.4\,d\,t\,F_u,\ \ 1.2\,l_c\,t\,F_u\big), \qquad l_c = l_e \ \text{or}\ s - d_h

Bearing-type vs slip-critical connections

In a bearing-type connection (the default, designation -N or -X) the bolts are pretensioned only enough to be snug or fully tightened, but the design relies on the bolts going into shear and the plates into bearing — a little slip into bearing is acceptable. In a slip-critical (SC) connection the design instead resists load through friction at the faying surfaces and is checked so that the joint does NOT slip under service (or factored, depending on the limit state chosen) load. SC is required only where slip would be detrimental: oversized/slotted holes loaded toward the slot, fatigue from load reversal, or connections that cannot tolerate movement. Most simple shear connections are bearing-type — do not default everything to slip-critical.

Slip resistance of a slip-critical bolt

The design slip resistance of one SC bolt follows AISC 360-16 §J3.8:

R_n=\mu\,D_u\,h_f\,T_b\,n_s

, where

\mu

is the mean slip coefficient (

0.30

Class A,

0.50

Class B faying surface),

D_u=1.13

relates mean to specified pretension,

h_f

is a filler factor (

1.0

with no fillers),

T_b

is the minimum bolt pretension from Table J3.1, and

n_s

is the number of slip planes. The resistance factor is

\phi=1.00

for standard holes and short-slotted holes loaded perpendicular to the slot,

0.85

for oversized holes and short-slotted holes loaded parallel to the slot, and

0.70

for long-slotted holes. Note that slip resistance does NOT depend on bolt area or

F_{nv}

— it is a clamping-force-times-friction calculation, a different physical mechanism from shear.

\phi R_n = \phi\,\mu\,D_u\,h_f\,T_b\,n_s, \qquad \phi=1.00\ (\text{std holes})

Bolt-group capacity and block shear

A concentrically loaded bolt group (load through the centroid, no eccentricity) shares the force equally, so its capacity is simply the number of bolts times the smallest per-bolt limit state — but you must check the SAME limit state on every bolt because edge bolts have smaller tearout

l_c

than interior bolts. Above the per-bolt checks sits block shear (§J4.3), in which an entire block of material tears out along a shear plane plus a tension plane:

R_n=0.6F_u A_{nv}+U_{bs}F_u A_{nt}\le 0.6F_y A_{gv}+U_{bs}F_u A_{nt}

, with

\phi=0.75

. The connection's strength is the minimum across bolt shear, bearing/tearout (summed), and block shear.

\phi R_n = \phi\big(0.6 F_u A_{nv} + U_{bs} F_u A_{nt}\big) \le \phi\big(0.6 F_y A_{gv} + U_{bs} F_u A_{nt}\big)

Exam strategy

Build a small table: one column per limit state (bolt shear, bearing, tearout, block shear), compute

\phi R_n

for each, and circle the minimum — the connection strength is that minimum, not the bolt-shear value people grab by reflex. Always count shear planes (

m

): a beam web bolted between two angles is in DOUBLE shear. Keep

l_c

as the CLEAR distance (subtract the full

d_h=d+\tfrac{1}{16}

), and use

F_u

— not

F_y

— in bearing and tearout. Decide bearing vs slip-critical from the problem statement; if it is not SC, never compute slip. And do not double-count: a single bolt is limited by the smaller of bearing and tearout, you do not add them.

Key equations

Bolt shear strength

\phi R_n = \phi\,F_{nv}\,A_b\,m

Per bolt. $F_{nv}=54$ ksi (A325-N) / $68$ (A325-X or A490-N) / $84$ (A490-X); $A_b=\pi d^2/4$

Bolt nominal area

A_b = \dfrac{\pi d^2}{4}

Gross shank area from nominal diameter $d$

Bearing strength at a hole

\phi R_n = \phi\,2.4\,d\,t\,F_u

Deformation-considered upper limit. $d$

Tearout strength at a hole

\phi R_n = \phi\,1.2\,l_c\,t\,F_u

Controlling bearing/tearout per bolt

\phi R_n = \phi\min(2.4\,d\,t\,F_u,\ 1.2\,l_c\,t\,F_u)

Slip-critical design resistance

\phi R_n = \phi\,\mu\,D_u\,h_f\,T_b\,n_s

Minimum pretension (Table J3.1, sample)

T_b\!: \ \tfrac{3}{4}''\,\text{A325}=28,\ \tfrac{7}{8}''\,\text{A325}=39,\ \tfrac{7}{8}''\,\text{A490}=49\ \text{kips}

Block shear rupture

\phi R_n = \phi\big(0.6 F_u A_{nv} + U_{bs}F_u A_{nt}\big) \le \phi\big(0.6 F_y A_{gv} + U_{bs}F_u A_{nt}\big)

Group capacity (concentric)

\phi R_{n,\text{group}} = N_b \times \min(\phi R_n)_{\text{per bolt}}

Standard hole diameter

d_h = d + \tfrac{1}{16}\ \text{in}\ (d\le\tfrac78''),\quad d+\tfrac18\ \text{in}\ (d\ge1'')

Worked examples

Bearing-type connection — controlling limit state

Problem. A beam web ( $t=0.50\,\text{in}$ , $F_u=58\,\text{ksi}$ ) is connected by six $\tfrac{3}{4}$ -in A325-N bolts in DOUBLE shear (web between two angles), in a single vertical line at $3\,\text{in}$ spacing with a $1.5\,\text{in}$ end distance. Standard holes. Find the LRFD design strength of the web in this connection (web limit states only).

Solution. Bolt shear:

A_b=\pi(0.75)^2/4=0.442\,\text{in}^2

. With

F_{nv}=54

\phi R_n = 6\,[\,0.75(54)(0.442)(2)\,] = 215\ \text{kips}

Slip-critical check on the same family of bolts

Problem. A connection uses four $\tfrac{7}{8}$ -in A325 slip-critical bolts in single shear, Class A faying surfaces ( $\mu=0.30$ ), standard holes, no fillers. Determine the LRFD slip resistance of the connection.

Solution. Pretension (Table J3.1):

T_b=39\,\text{kips}

Common pitfalls

•Citing handbook §4.2 for bolt STRENGTH. The handbook gives only fastener-group geometry (§4.2.1) and weld symbols (§4.2.3); all $F_{nv}$ , bearing, and slip values are AISC 360-16 Chapter J. Cite the code, not a handbook page.
•Forgetting the second shear plane. A web bolted between two angles is in DOUBLE shear ( $m=2$ ) — using $m=1$ halves the bolt-shear strength.
•Using $F_y$ instead of $F_u$ in bearing/tearout. Both $2.4dtF_u$
•Computing $l_c$ as the center-to-center spacing or full edge distance. $l_c$ is the CLEAR distance — subtract the whole hole diameter $d_h=d+\tfrac1{16}$
•Defaulting every connection to slip-critical. SC is required only for slip-sensitive joints; for a bearing-type connection you check bolt shear and bearing, not slip resistance.
•Adding bearing and tearout for one bolt. A single bolt's strength is the SMALLER of the two, not their sum; you sum the per-bolt minimums across the group.
•Mixing the wrong $\phi$ : bolts in shear/bearing/block-shear use $\phi=0.75$ , but slip-critical with standard holes uses $\phi=1.00$ . Keep them straight.

References

AISC 360-16 Specification for Structural Steel Buildings — Chapter J (J3 bolts, J3.6–J3.8 shear/slip, J3.10 bearing, J4.3 block shear)
AISC Steel Construction Manual, 15th ed. — Part 7 (bolting) and Tables 7-1 to 7-6 (available bolt strengths)
NCEES PE Civil Reference Handbook — §4.2.1 Fastener Groups in Shear (group geometry only; no strength values)

Welded Connections & Concrete Anchorage

Fillet-weld strength from the effective throat, the directional strength increase, reading weld symbols, and the concrete breakout and pullout limits on headed anchors.

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Eccentric Bolt & Weld Groups

The elastic-vector and instantaneous-center methods for eccentric shear on a fastener group, the polar moment of the group, and how to find the critical fastener.

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Concrete Anchorage (ACI 318 Ch17)

The tension and shear limit states for cast-in and post-installed anchors — steel, concrete breakout, pullout, side-face blowout, pryout — and the governing design value.

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Bearing vs Slip-Critical Bolts

The slip-critical design slip resistance, when slip-critical connections are required, the role of pretension, and how the comparison with a bearing connection plays out.

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(in²);

m

= shear planes;

\phi=0.75

.

(in); a 3/4-in bolt has

A_b=0.442\,\text{in}^2

, a 7/8-in bolt

0.601\,\text{in}^2

.

= bolt dia,

t

= ply thickness,

F_u

= ply tensile strength (ksi);

\phi=0.75

.

$l_c$ = clear distance to next hole or edge in line of force = $l_e$ or $s$ minus the std hole diameter $d_h$ ( $d+\tfrac{1}{16}$ in for $d\le\tfrac78$ in, $d+\tfrac18$ in for $d\ge1$ in, AISC Table J3.3).

Take the smaller; tearout usually governs at edge bolts (small $l_c$ ), bearing at interior bolts.

$\mu=0.30$ (A), $0.50$ (B); $D_u=1.13$ ; $h_f=1.0$ (no filler); $T_b$ from Table J3.1; $n_s$ = slip planes; $\phi=1.00$ std holes.

Specified minimum pretension for fully tensioned high-strength bolts (kips); required for SC and for fatigue/seismic joints.

$A_{nv},A_{gv}$ net/gross shear area; $A_{nt}$ net tension area; $U_{bs}=1.0$ uniform tension; $\phi=0.75$ .

Equal sharing only when the load passes through the group centroid; eccentric groups use the elastic-vector or IC method.

Standard round hole per AISC Table J3.3; oversized and slotted holes are larger and reduce SC $\phi$ and bearing/tearout $l_c$ .

,

m=2

,

\phi=0.75

:

\phi R_n = 0.75(54)(0.442)(2)=35.8\,\text{kips/bolt}.

Group:

6\times35.8 = 215\,\text{kips}

. Hole dia

d_h=0.75+0.0625=0.8125\,\text{in}

. Bearing cap (all bolts):

\phi\,2.4dtF_u = 0.75(2.4)(0.75)(0.50)(58)=39.2\,\text{kips/bolt}

. Edge bolt tearout:

l_c=1.5-0.5(0.8125)=1.094\,\text{in}

(clear to edge),

\phi 1.2 l_c t F_u = 0.75(1.2)(1.094)(0.50)(58)=28.5\,\text{kips}

→ tearout governs the edge bolt. Interior bolt tearout:

l_c=3-0.8125=2.188\,\text{in}

, gives

57.1\,\text{kips}

→ bearing (

39.2

) governs interior bolts. Bearing/tearout group total:

1(28.5)+5(39.2)=224.3\,\text{kips}

. Bolt shear (

215

) is the smaller, so

\phi R_n = 215\,\text{kips}

. Sanity: bolt shear and bearing/tearout are close here (

215

vs

224

), with bolt shear governing by about

4\%

— a near-tie, so the controlling state must be confirmed by calculation, not assumed. Answer:

\boxed{215\ \text{kips}}

(bolt shear controls).

for

\tfrac{7}{8}

-in A325. Factors:

\mu=0.30

,

D_u=1.13

,

h_f=1.0

,

n_s=1

,

\phi=1.00

(std holes). Per bolt:

\phi R_n = 1.00(0.30)(1.13)(1.0)(39)(1)=13.2\,\text{kips}

. Group:

4\times13.2 = 52.9\,\text{kips}

. Sanity: this is well below the same bolts' bearing-type shear strength (

\phi F_{nv}A_b = 0.75(54)(0.601)=24.4\,\text{kips/bolt}

, i.e.

97.4\,\text{kips}

for four). Slip-critical resistance is intentionally lower — the joint is designed not to move, not to fail. Answer:

\boxed{52.9\ \text{kips}}

slip resistance.

\phi R_n = 4\,[\,1.00(0.30)(1.13)(1.0)(39)\,] = 52.9\ \text{kips}

and

1.2 l_c t F_u

use the TENSILE strength

F_u

of the connected ply.

in.