CivilSteel DesignmediumCourseworkPE / Advanced

Shear Capacity of a Bolt Group

A lap splice uses $4$ bolts, each with nominal shear area $A_b = 314\ \text{mm}^2$ , in single shear.

The nominal shear stress is $F_{nv} = 372\ \text{MPa}$ and the resistance factor is $\phi = 0.75$ .

Find the design shear capacity of the bolt group.

Given

n	4 -	Number of bolts
Ab	314 mm^2	Area per bolt
Fnv	372 MPa	Nominal shear stress
φ	0.75 -	Resistance factor

Hint 1

Start with the capacity of a single bolt: stress times area.

Hint 2

How many shear planes cut each bolt in a lap splice?

Hint 3

A lap splice is single shear — one plane per bolt. Then multiply by 4 and by φ = 0.75.

Worked solution — try the problem first

Capacity of one bolt in single shear:

$R_n = F_{nv} A_b = 372 \times 314 = 116{,}808\ \text{N} = 116.8\ \text{kN}$

Group of four, single shear — one shear plane per bolt:

$R_{n,group} = 4 \times 116.8 = 467.2\ \text{kN}$

Design capacity:

$\phi R_n = 0.75 \times 467.2 = 350.4\ \text{kN}$

$\boxed{\phi R_n = 350\ \text{kN}}$

Single vs double shear. In single shear each bolt is cut by one plane; in double shear (a bolt through three plies) there are two planes and the capacity per bolt doubles. Misreading this is a 2× error in either direction, so read the joint geometry carefully.

A complete design check would also verify bearing on the connected plies, net-section tension, block shear, and edge distances — shear on the bolts is only one limit state.

Concepts:Bolted connectionsShear capacityLRFDLimit states

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