Checks a joint between flat steel plates in line, with non-preloaded bolts in shear, to SIA 263:2013: the bolts, bearing on the plates, the gross and net sections, block tearing and the placing of the holes. Every step shows its formula and the clause it comes from.

What it checks

Bolt data from the grade and the size: f_ub,k = 400, 500, 800 or 1,000 N/mm² for grades 4.6, 5.6, 8.8 and 10.9, the diameter d, the hole d₀ with normal clearance, the shank area A = π · d²/4 and the tabulated stress area A_s.
Force per shear plane: F_v,Ed = F_Ed/(n_b · n_s), with n_b = n₁ · n₂ bolts on each side of the joint and n_s shear planes per bolt. One shear plane of the most loaded bolt is checked.
The bolts in shear: F_v,Ed ≤ F_v,Rd = α_v · f_ub,k · A/γ_M2, with α_v = 0.6 for every grade, on A in the shank or A_s in the thread. The long-joint factor β_Lf reduces the resistance once L_j = (n₁ − 1) · p₁ 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 of each plate in tension, clause 5.1.2.2 and equation (39): A_net = (b − n₂ · d₀) · t and N_u,Rd = 0.9 · f_u,k · A_net/γ_M2.
Block tearing of each plate in tension: V_eff,Rd = (0.9 · f_u,k · A_t,net + f_y,k · A_v,brut/√3)/γ_M2. The two shear lines, of length L_v = e₁ + (n₁ − 1) · p₁, are taken gross; A_t,net is the smaller of the inner block and the path out to the edges.
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).
Resistance of the joint R_d: the smallest of the resistances, written as a force on the joint.

What you enter and what you get

You choose the bolt grade and size, whether the shear planes cross the shank or the thread, and the type of joint: double cover plates with two shear planes, or a lap joint with one. On each side of the joint you enter the rows n₁ at pitch p₁ along the force and the lines n₂ at spacing p₂ across it. You enter γ_M2 and γ_M1 (clause 4.1.3), whether the force is tension or compression, F_Ed, and the tensile force per bolt along its axis F_t,Ed, zero for a joint in pure shear. For the plates: f_y,k and f_u,k from table 1, then t, b and e₁ of the main plate and t_c, b_c and e_1,c of each cover plate.

You get every force and 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

Flat plates in line, loaded along their axis. Angles connected by one leg, eccentric bolt groups and moment joints have their own modules.
Normal holes. Oversized and slotted holes are treated with preloaded bolts.
The force is shared equally by the bolts. β_Lf corrects this over a long joint.
The bolt group is centred on the width of every plate: e₂ = (b − (n₂ − 1) · p₂)/2.
A bolt is checked against its ultimate strength; its yield strength plays no part.
In compression the force passes by contact and only the gross section is checked.
Where the net section in tension is decisive, no plastic rotation, and so no redistribution of internal forces, may be assumed in that zone.
Non-preloaded bolts are very sensitive to repeated tension: preloaded bolts are strongly recommended under tension, and required under fatigue.

Worked example

A splice with double cover plates in tension, F_Ed = 650 kN, with three rows at p₁ = 60 mm and two lines at p₂ = 70 mm of M20 bolts of grade 4.6 on each side, sheared in the shank, a 150 × 24 mm main plate and two 150 × 12 mm cover plates with e₁ = 40 mm, in f_y,k = 235 N/mm² and f_u,k = 360 N/mm². These are the module's own results for its worked example, computed when this page was built.

Inputs
Input	Value	Ref.
Bolt grade	4.6	§ t.2
Bolt size	M20	§ t.17
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	60 mm
Lines of bolts across the force n₂	2
Spacing across the force p₂	70 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	650 kN
Tensile force per bolt, along its axis F_t,Ed	0 kN
Characteristic yield strength f_y,k	235 N/mm²	§ t.1
Characteristic tensile strength f_u,k	360 N/mm²	§ t.1
Thickness t	24 mm
Width b	150 mm
End distance, nearest row to the plate end e₁	40 mm
Thickness t_c	12 mm
Width b_c	150 mm
End distance, outer row to the plate end e_1,c	40 mm
Resistance factor, gross section γ_M1	1.05	§ 4.1.3

Calculation: Common values
Step	Value	Ref.
Bolt	– = M20, grade 4.6
Tensile strength of the bolt steelFrom the grade. The yield strength, 240 N/mm², plays no part: a bolt is checked against its ultimate strength.	f_ub,k = 400 N/mm²	§ t.2
Nominal diameter	d = 20 mm
Hole diameterNormal clearance, d + 2 mm.	d₀ = 22 mm	§ t.17
Shank areaA = π·d²/4.	A = 314.2 mm²	§ t.16
Stress areaTabulated: the area of the equivalent cylinder with the failure resistance of the threaded part, 22 % below A.	A_s = 245.0 mm²	§ t.16
Bolts on each side of the jointn_b = n₁·n₂ = 3·2.	n_b = 6
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 = 120 mm
Force on one shear plane of one boltF_v,Ed = F_Ed/(n_b·n_s) = 650/(6·2). The force is shared equally, and one shear plane of the most loaded bolt is checked.	F_v,Ed = 54.17 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 = 314.2 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 = 60.32 kN	§ (75)
Long-joint factorL_j = 120 mm is within 15·d = 300 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 = 0.898 · Pass90 % 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 = 40.0 mm
Edge distance of the cover platese_2,c = (b_c − (n₂ − 1)·p₂)/2.	e_2,c = 40.0 mm
Force in each cover plateF_Ed/2: the two cover plates share the force.	F_c,Ed = 325.0 kN
Bearing force per bolt on the main plateF_b,Ed = F_Ed/n_b.	F_b,Ed = 108.33 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 = 54.17 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 = 659.4 kN
The joint is limited by	net section of the main plate

Calculation: Check 1 — bearing on the main plate
Step	Value	Ref.
End distance usedp₁ = 60 mm is at or above e₁ + d₀/2 = 51.0 mm: e₁ is used as it is.	e₁ = 40.0 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 = 213.64 kN	§ (76)
Bearing on the main plateF_b,Ed ≤ F_b,Rd	η = 0.507 · Pass51 % of the capacity used	§ (76)

Calculation: Check 2 — bearing on the cover plates
Step	Value	Ref.
End distance usedp₁ = 60 mm is at or above e_1,c + d₀/2 = 51.0 mm: e_1,c is used as it is.	e_1,c = 40.0 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 = 106.82 kN	§ (76)
Bearing on the cover platesF_b,Ed ≤ F_b,Rd	η = 0.507 · Pass51 % of the capacity used	§ (76)

Calculation: Check 3 — gross section of the main plate
Step	Value	Ref.
Gross areaA = b·t.	A = 3,600 mm²
Resistance of the gross sectionN_Rd = f_y,k·A/γ_M1.	N_Rd = 805.7 kN	§ (38)
Gross section of the main plateF_Ed ≤ N_Rd	η = 0.807 · Pass81 % 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 = 1,800 mm²
Resistance of the gross sectionN_Rd = f_y,k·A/γ_M1.	N_Rd = 402.9 kN	§ (38)
Gross section of the cover platesF_c,Ed ≤ N_Rd	η = 0.807 · Pass81 % 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 − 2·22)·t, across the row that carries the whole force.	A_net = 2,544 mm²	§ 5.1.2.2
Resistance of the net sectionN_u,Rd = 0,9·f_u,k·A_net/γ_M2.	N_u,Rd = 659.4 kN	§ (39)
Net section of the main plateF_Ed ≤ N_u,Rd	η = 0.986 · Pass99 % 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 − 2·22)·t_c, across the row that carries the whole force.	A_net = 1,272 mm²	§ 5.1.2.2
Resistance of the net sectionN_u,Rd = 0,9·f_u,k·A_net/γ_M2.	N_u,Rd = 329.7 kN	§ (39)
Net section of the cover platesF_c,Ed ≤ N_u,Rd	η = 0.986 · Pass99 % 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 = 160 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 = 7,680 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 = 1,152 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 = 1,132.2 kN	§ (83)
Block tearing of the main plateF_Ed ≤ V_eff,Rd	η = 0.574 · Pass57 % 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 = 160 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 = 3,840 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 = 576 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 = 566.1 kN	§ (83)
Block tearing of the cover platesF_c,Ed ≤ V_eff,Rd	η = 0.574 · Pass57 % of the capacity used	§ (83)

Calculation: Check 9 — end distance
Step	Value	Ref.
End distancee₁ ≥ 30 mm	η = 0.750 · Pass75 % of the capacity used	§ t.17

Calculation: Check 10 — edge distance
Step	Value	Ref.
Edge distancee₂ ≥ 25 mm	η = 0.625 · Pass63 % 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 ≥ 45 mm	η = 0.750 · Pass75 % of the capacity used	§ t.17

Calculation: Check 12 — spacing across the force
Step	Value	Ref.
Spacing across the forcep₂ ≥ 45 mm	η = 0.643 · Pass64 % of the capacity used	§ t.17

How this module is verified

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

Questions

What limits the joint in the worked example?

The net section of the main plate: N_u,Rd = 659.4 kN against F_Ed = 650 kN, η = 0.986. The bolts in shear follow at η = 0.898.

When does the pitch change the bearing resistance?

Where p₁ < e₁ + d₀/2, the bolt behind limits the tear-out rather than the free end, and p₁ − d₀/2 replaces e₁. In the worked example p₁ = 60 mm is above 51.0 mm, so e₁ is used as it is.

Can I print the calculation?

The full calculation is shown on screen, free and without an account. A print-ready A4 calculation note is part of the Basic plan.

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

Non-preloaded bolted plate joint to SIA 263

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