Preloaded bolted friction joint to SIA 263

Checks a friction joint between flat steel plates in line, with preloaded bolts, to SIA 263:2013: no slip at the chosen limit state, then the ultimate checks of the bolts and plates. Every step shows its formula and the clause it comes from.

Basic SIA 263:2013

Part of the Basic plan

This module is included in the Basic plan and above, with its calculation note.

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About this check

clauses, worked example and questions

What it checks

Preload: P_B = 0.64 · f_ub,k · A_s, for grades 8.8 and 10.9, with f_ub,k = 800 and 1,000 N/mm².
Slip resistance of one bolt: F_s,Cd = k_s · n · μ · P_B/γ_Mi. The hole factor k_s is 1.0 for normal clearance, 0.85 for oversized or short slotted holes and 0.7 for long slotted holes. n is the number of friction surfaces per bolt, equal to the shear planes.
No slip: E_d ≤ n_b · F_s,Cd, with the number of bolts required on each side, n_b,req, rounded up.
The bolts in shear: F_v,Ed = F_Ed/(n_b · n_s) ≤ F_v,Rd = α_v · f_ub,k · A/γ_M2, with α_v = 0.6, and the long-joint factor β_Lf once L_j exceeds 15 · d.
Bearing, per bolt on the main plate and on the cover plates: F_b,Rd = 0.85 · (e/d₀) · f_u,k · d · t/γ_M2, at most 2.4 · f_u,k · d · t/γ_M2. e is e₁, or p₁ − d₀/2 where p₁ < e₁ + d₀/2.
Gross section of each plate, clause 5.1.2, equation (38): N_Rd = f_y,k · A/γ_M1.
Net section in tension, clause 5.1.2.2 and equation (39): N_u,Rd = 0.9 · f_u,k · A_net/γ_M2.
Block tearing in tension: V_eff,Rd = (0.9 · f_u,k · A_t,net + f_y,k · A_v,brut/√3)/γ_M2.
Placing of the holes: the minimum distances of SIA 263 Table 17 for the bolt size, for e₁, e₂, p₁ and p₂; e₂ at least d₀ and p₂ at least 2 · d₀, the conditions of the bearing resistance (76).

What you enter and what you get

You choose the bolt grade and size, the hole type, whether the shear planes cross the shank or the thread, and the type of joint: double cover plates or a lap joint. On each side of the joint you enter the rows n₁ at pitch p₁ and the lines n₂ at spacing p₂. You enter γ_M2 and γ_M1 (clause 4.1.3), whether the force is tension or compression, F_Ed at the ultimate limit state and the tensile force per bolt F_t,Ed. For the slip check: the surface condition of the faying surfaces, the limit state up to which slip is prevented, the force on the joint E_d and whether the bolts also carry tension at that limit state. For the plates: f_y,k and f_u,k from table 1, and the thickness, width and end distance of the main plate and of each cover plate.

You get the preload, the slip resistance, the bolts required, every ultimate resistance, the utilisation of each check with a pass or fail, and the part that limits the joint. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

A friction joint between flat plates in line. The force is shared equally by the bolts.
μ = 0.25 for most primers and building structures with no special preparation, 0.40 for zinc silicate and 0.50 blast cleaned to Sa 2½. The faying surfaces must be prepared with care so that the coefficient assumed is the one obtained.
γ_Mi = 1.10 when slip is prevented up to the serviceability limit state, 1.25 up to the ultimate limit state, and 1.40 for a joint that must not slip at the ultimate limit state with oversized or slotted holes along the force.
A larger hole lowers k_s, and its larger d₀ weakens the net section and the bearing.
Preloading changes the service behaviour, not the ultimate resistance. The ultimate checks are made in any case.
Where the net section in tension is decisive, no plastic rotation, and so no redistribution of internal forces, may be assumed in that zone.

Worked example

A splice with double cover plates, three rows at p₁ = 80 mm and three lines at p₂ = 80 mm of M24 bolts of grade 10.9 on each side, in normal holes and sheared in the shank, with zinc silicate faying surfaces and slip prevented up to the serviceability limit state. The main plate is 400 × 30 mm, the two cover plates 400 × 16 mm, with e₁ = 70 mm, in f_y,k = 235 N/mm² and f_u,k = 360 N/mm², under F_Ed = 2,100 kN in tension and E_d = 1,470 kN. These are the module's own results for its worked example, computed when this page was built.

Inputs
Input	Value	Ref.
Bolt grade	10.9	§ t.2
Bolt size	M24	§ t.17
Hole type	Normal clearance — k_s = 1,0
The shear planes cross the	Shank — the unthreaded part
Type of joint	Double cover plates — two shear planes
Rows of bolts in the direction of the force n₁	3
Pitch in the direction of the force p₁	80 mm
Lines of bolts across the force n₂	3
Spacing across the force p₂	80 mm
Resistance factor, bolts and net section γ_M2	1.25	§ 4.1.3
The force in the joint is	Tension
Design force carried by the joint F_Ed	2100 kN
Tensile force per bolt, along its axis F_t,Ed	0 kN
Surface condition	Zinc silicate coating — μ = 0,40
Slip is prevented	Serviceability — slip prevented up to the SLS, γ_Mi = 1,10
Force on the joint at the serviceability limit state E_d	1470 kN
The bolts also carry tension at this limit state	No
Characteristic yield strength f_y,k	235 N/mm²	§ t.1
Characteristic tensile strength f_u,k	360 N/mm²	§ t.1
Thickness t	30 mm
Width b	400 mm
End distance, nearest row to the plate end e₁	70 mm
Thickness t_c	16 mm
Width b_c	400 mm
End distance, outer row to the plate end e_1,c	70 mm
Resistance factor, gross section γ_M1	1.05	§ 4.1.3

Calculation: Common values
Step	Value	Ref.
Bolt	– = M24, grade 10.9
Tensile strength of the bolt steelFrom the grade. The yield strength, 900 N/mm², plays no part: a bolt is checked against its ultimate strength.	f_ub,k = 1,000 N/mm²	§ t.2
Nominal diameter	d = 24 mm
Hole diameterNormal clearance, d + 2 mm.	d₀ = 26 mm	§ t.17
Shank areaA = π·d²/4.	A = 452.4 mm²	§ t.16
Stress areaTabulated: the area of the equivalent cylinder with the failure resistance of the threaded part, 22 % below A.	A_s = 353.0 mm²	§ t.16
Bolts on each side of the jointn_b = n₁·n₂ = 3·3.	n_b = 9
Shear planes per boltDouble cover plates: each bolt is sheared on both faces of the main plate.	n_s = 2
Length of the bolt groupL_j = (n₁ − 1)·p_1, centre to centre of the end rows.	L_j = 160 mm
Force on one shear plane of one boltF_v,Ed = F_Ed/(n_b·n_s) = 2,100/(9·2). The force is shared equally, and one shear plane of the most loaded bolt is checked.	F_v,Ed = 116.67 kN	§ 6.2.2.1

Calculation: Check 1 — the bolts in shear
Step	Value	Ref.
Area in the shear planeThe shear planes cross the shank: the full area.	A = 452.4 mm²
Shear coefficient0,6, from a shear strength of about f_ub,k/√3 rounded for design; the same for every grade.	α_v = 0.60	§ (75)
Shear resistance of one planeF_v,Rd = α_v·f_ub,k·A/γ_M2.	F_v,Rd = 217.15 kN	§ (75)
Long-joint factorL_j = 160 mm is within 15·d = 360 mm: no reduction. Beyond it the end bolts break first and pass their load on to the next.	β_Lf = 1.0000	§ 6.2.2.2
The bolts in shearF_v,Ed ≤ F_v,Rd	η = 0.537 · Pass54 % of the capacity used	§ t.16

Calculation: Common values
Step	Value	Ref.
Edge distance of the main platee₂ = (b − (n₂ − 1)·p₂)/2: the bolt group is centred on the width.	e₂ = 120.0 mm
Edge distance of the cover platese_2,c = (b_c − (n₂ − 1)·p₂)/2.	e_2,c = 120.0 mm
Force in each cover plateF_Ed/2: the two cover plates share the force.	F_c,Ed = 1,050.0 kN
Bearing force per bolt on the main plateF_b,Ed = F_Ed/n_b.	F_b,Ed = 233.33 kN
Bearing force per bolt on each cover plateF_Ed/(2·n_b): each cover plate takes half of every bolt force.	F_b,Ed,c = 116.67 kN
Ultimate resistance of the jointThe smallest of the resistances of the bolts and of the plates, each written as a force on the joint.	R_d = 2,337.7 kN
The joint is limited by	block tearing of the main plate

Calculation: Check 1 — bearing on the main plate
Step	Value	Ref.
End distance usedp₁ = 80 mm is below e₁ + d₀/2 = 83.0 mm: the bolt behind limits the tear-out, so p₁ − d₀/2 replaces e₁.	p₁ − d₀/2 = 67.0 mm	§ (76)
Bearing resistance per boltF_b,Rd = 0,85·(e/d₀)·f_u,k·d·t/γ_M2, at most 2,4·f_u,k·d·t/γ_M2.	F_b,Rd = 454.20 kN	§ (76)
Bearing on the main plateF_b,Ed ≤ F_b,Rd	η = 0.514 · Pass51 % of the capacity used	§ (76)

Calculation: Check 2 — bearing on the cover plates
Step	Value	Ref.
End distance usedp₁ = 80 mm is below e_1,c + d₀/2 = 83.0 mm: the bolt behind limits the tear-out, so p₁ − d₀/2 replaces e_1,c.	p₁ − d₀/2 = 67.0 mm	§ (76)
Bearing resistance per boltF_b,Rd = 0,85·(e/d₀)·f_u,k·d·t_c/γ_M2, at most 2,4·f_u,k·d·t_c/γ_M2.	F_b,Rd = 242.24 kN	§ (76)
Bearing on the cover platesF_b,Ed ≤ F_b,Rd	η = 0.482 · Pass48 % of the capacity used	§ (76)

Calculation: Check 3 — gross section of the main plate
Step	Value	Ref.
Gross areaA = b·t.	A = 12,000 mm²
Resistance of the gross sectionN_Rd = f_y,k·A/γ_M1.	N_Rd = 2,685.7 kN	§ (38)
Gross section of the main plateF_Ed ≤ N_Rd	η = 0.782 · Pass78 % of the capacity used	§ 5.1.2

Calculation: Check 4 — gross section of the cover plates
Step	Value	Ref.
Gross areaA = b_c·t_c.	A = 6,400 mm²
Resistance of the gross sectionN_Rd = f_y,k·A/γ_M1.	N_Rd = 1,432.4 kN	§ (38)
Gross section of the cover platesF_c,Ed ≤ N_Rd	η = 0.733 · Pass73 % of the capacity used	§ 5.1.2

Calculation: Check 5 — net section of the main plate
Step	Value	Ref.
Net areaA_net = (b − n₂·d₀)·t = (b − 3·26)·t, across the row that carries the whole force.	A_net = 9,660 mm²	§ 5.1.2.2
Resistance of the net sectionN_u,Rd = 0,9·f_u,k·A_net/γ_M2.	N_u,Rd = 2,503.9 kN	§ (39)
Net section of the main plateF_Ed ≤ N_u,Rd	η = 0.839 · Pass84 % of the capacity used	§ 5.1.2

Calculation: Check 6 — net section of the cover plates
Step	Value	Ref.
Net areaA_net = (b_c − n₂·d₀)·t_c = (b_c − 3·26)·t_c, across the row that carries the whole force.	A_net = 5,152 mm²	§ 5.1.2.2
Resistance of the net sectionN_u,Rd = 0,9·f_u,k·A_net/γ_M2.	N_u,Rd = 1,335.4 kN	§ (39)
Net section of the cover platesF_c,Ed ≤ N_u,Rd	η = 0.786 · Pass79 % of the capacity used	§ 5.1.2

Calculation: Check 7 — block tearing of the main plate
Step	Value	Ref.
Length of each shear lineL_v = e₁ + (n₁ − 1)·p_1, along an outer bolt line from the last row to the plate end.	L_v = 230 mm
Gross area in shearA_v,brut = 2·L_v·t: two shear lines, taken gross because that zone yields before the tension line breaks.	A_v,brut = 13,800 mm²	§ (83)
Net area in tensionAcross the inner block, (n₂ − 1)·(p₂ − d₀)·t, smaller than out to the edges, (2·e₂ − d₀)·t.	A_t,net = 3,240 mm²	§ (83)
Block tearing resistanceV_eff,Rd = (0,9·f_u,k·A_t,net + f_y,k·A_v,brut/√3)/γ_M2: the tension line at f_u,k because it tears, the shear lines at f_y,k because they yield first.	V_eff,Rd = 2,337.7 kN	§ (83)
Block tearing of the main plateF_Ed ≤ V_eff,Rd	η = 0.898 · Pass90 % of the capacity used	§ (83)

Calculation: Check 8 — block tearing of the cover plates
Step	Value	Ref.
Length of each shear lineL_v = e_1,c + (n₁ − 1)·p_1, along an outer bolt line from the last row to the plate end.	L_v = 230 mm
Gross area in shearA_v,brut = 2·L_v·t_c: two shear lines, taken gross because that zone yields before the tension line breaks.	A_v,brut = 7,360 mm²	§ (83)
Net area in tensionAcross the inner block, (n₂ − 1)·(p₂ − d₀)·t_c, smaller than out to the edges, (2·e_2,c − d₀)·t_c.	A_t,net = 1,728 mm²	§ (83)
Block tearing resistanceV_eff,Rd = (0,9·f_u,k·A_t,net + f_y,k·A_v,brut/√3)/γ_M2: the tension line at f_u,k because it tears, the shear lines at f_y,k because they yield first.	V_eff,Rd = 1,246.8 kN	§ (83)
Block tearing of the cover platesF_c,Ed ≤ V_eff,Rd	η = 0.842 · Pass84 % of the capacity used	§ (83)

Calculation: Check 9 — end distance
Step	Value	Ref.
End distancee₁ ≥ 35 mm	η = 0.500 · Pass50 % of the capacity used	§ t.17

Calculation: Check 10 — edge distance
Step	Value	Ref.
Edge distancee₂ ≥ 30 mm	η = 0.250 · Pass25 % of the capacity used	§ t.17

Calculation: Check 11 — pitch in the direction of the force
Step	Value	Ref.
Pitch in the direction of the forcep₁ ≥ 55 mm	η = 0.688 · Pass69 % of the capacity used	§ t.17

Calculation: Check 12 — spacing across the force
Step	Value	Ref.
Spacing across the forcep₂ ≥ 55 mm	η = 0.688 · Pass69 % of the capacity used	§ t.17

Calculation: Common values
Step	Value	Ref.
Preload forceF_p,Cd = 0,64·f_ub,k·A_s, for grades 8.8 and 10.9 with controlled preload.	F_p,Cd = 225.9 kN	§ (80)
The preload of EN 1993-1-8 would be9 % more preload, and as much more slip resistance: a difference between the two norms, not a rounding.for comparison	0,7·f_ub,k·A_s = 247.1 kN	EN 1993-1-8 §3.9.1
Preload stress in the bolt0,64·f_ub,k, which is 71 % of the yield strength of the bolt steel.	F_p,Cd/A_s = 640 N/mm²
Hole factor1,0 for normal clearance, 0,85 for oversized or short slotted holes, 0,7 for long slotted.	k_s = 1.00	§ 6.2.3.2
Friction coefficient0,25 for most primers, 0,40 for zinc silicate, 0,50 blast cleaned to Sa 2½.	μ = 0.400	§ 6.2.3.5
Resistance factor for slipSet by the limit state and the holes, not by the material.	γ_Mi = 1.10	§ 6.2.3.2
Friction surfaces per boltThe shear planes of the joint: the same interfaces either slip or shear.	n = 2	§ 6.2.3.2
Slip resistance of one boltF_s,Cd = k_s·n·μ·F_p,Cd/γ_Mi.	F_s,Cd = 164.31 kN	§ (79)
Slip resistance of the jointn_b = 9 bolts on each side of the joint.	n_b·F_s,Cd = 1,478.7 kN

Calculation: Check 1 — no slip at the serviceability limit state
Step	Value	Ref.
No slip at the serviceability limit stateE_d ≤ n_b·F_s,Cd	η = 0.994 · Pass99 % of the capacity used	§ (79)
Bolts required on each sideE_d over the slip resistance of one bolt, rounded up: 1,470/164.3 = 8.95.	n_b,req = 9

How this module is verified

Verified clause by clause against SIA 263:2013 and against independent calculations used on real projects, before release.

Questions

Does the preload raise the ultimate resistance of the joint?

No. The ultimate resistance of a bolted joint does not depend on the preload. In the worked example the joint is limited by block tearing of the main plate, R_d = 2,337.7 kN.

How close is the worked example to slipping?

E_d = 1,470 kN against n_b · F_s,Cd = 1,478.7 kN, η = 0.994. The bolts required are 1,470/164.3 = 8.95, rounded up to 9, the number provided on each side.

Which plan includes this module?

The Basic plan, with its calculation note.

Norm: SIA 263:2013 · Page updated: 8 November 2026