In the construction of steel structures, there is always a need to join steel members for the purpose of continuity. It is not always possible to have the full lengths of members to span the desired length due to production, handling, and transportation issues. As a result, it is very common to bolt or weld two or more steel members, so as to achieve the desired length.

A bolted beam splice connection is a structural joint used in the construction of steel beams. It is designed to connect two or more beams end-to-end to form a longer beam, increasing the span length or overall structural capacity. The design of a bolted beam splice connection involves several considerations to ensure its strength, stability, and reliability.

The design of a bolted beam splice connection involves careful consideration of load transfer, bolted connection details, splice configuration, stiffness requirements, structural analysis, connection detailing, and material properties. By implication, all the anticipated design forces acting at the point of the beam splice must be taken into consideration during the design.

Therefore, the design involves performing structural analysis to determine the forces and moments acting on the connection. This analysis considers the applied loads, beam properties, and connection geometry. Finite element analysis or other advanced analysis methods may be employed to evaluate the connection’s behaviour under different loading conditions.

In this article, we are going to look at a design example of a beam splice connection (beam-to-beam connection using steel plates). This joint should be able to transmit bending, shear, and axial forces.

**Design and functions of Bolted Splice Connections**

The primary function of a beam splice connection is to transfer the loads between the beams effectively. The connection should be designed to transfer axial forces, shear forces, and bending moments from one beam to another without significant deformation or failure.

The connection typically consists of bolts, washers, and nuts. High-strength bolts are commonly used due to their ability to withstand heavy loads. The number, size, and grade of bolts are determined based on the design requirements and the applied loads.

The configuration of the splice connection depends on the structural requirements and the type of loads being transferred. Common types include full-depth end plate splices, extended end plate splices, and flange plate splices. The choice of splice configuration is influenced by factors such as the beam profile, connection stiffness, and fabrication constraints.

The stiffness of the splice connection plays a crucial role in maintaining the overall structural integrity and load distribution. Adequate stiffness helps to limit deflections and minimize differential movement between the spliced beams. The connection should be designed to ensure that it does not introduce excessive additional stiffness or flexibility to the overall beam system.

**Design Example**

Let us design a bolted beam splice connection for a UB 533 x 210 x 101 kg/m section, subjected to the following ultimate limit state loads;

M_{Ed} = 610 kNm

V_{Ed} = 215 kN

N_{Ed} = 55 kN

At serviceability limit state;

M_{Ed,ser} = 445 kNm

V_{Ed,ser} = 157 kN

N_{Ed,ser} = 41 kN

Properties of UB 533 x 210 x 101

h = 536.7 mm

b = 210 mm

d = 476.5 mm

t_{w} = 10.8 mm

t_{f} = 17.4 mm

r = 12.7 mm

A = 129 cm^{2}

I_{y} = 61500 cm^{4}

I_{z} = 2690 cm^{4}

W_{el,y} = 2290 cm^{3}

W_{pl,y} = 2610 cm^{3};

h_{w} = h – 2t_{f} = 476.5 mm

**Cover plates**

Let us assume 20 mm thick cover plates for the flanges and 15 mm thick plates for the web. The thickness and dimension to be confirmed later in the post.

**Bolts***M24 preloaded class 8.8 bolts*

Diameter of bolt shank d = 24mm

Diameter of hole d_{0} = 26mm

Shear area A_{s} = 353 mm^{2}

**Materials Strength****Beam and cover plates;**

f_{y,b} = f_{y,wp} = 275 N/mm^{2}

f_{u,b} = f_{u,wp} = 410 N/mm^{2}

**Bolts**

Nominal yield strength f_{yb} = 640 N/mm^{2}

Nominal ultimate strength f_{ub} = 800 N/mm^{2}

**Partial Factors for Resistance (BS EN 1993-1- NA.2.15)****Structural Steel**

γ_{m0} = 1.0

γ_{m1} = 1.0

γ_{m2} = 1.1

**Parts in connections**

γ_{m2} = 1.25 (bolts, welds, plates in bearing)

γ_{m3} = 1.0 (slip resistance at ULS)

γ_{m3,ser} = 1.1 (slip resistance as SLS)

**Step 1: Internal Forces at Splice**

For a splice in flexural member, the parts subject to shear (the web cover plates) must carry, in addition to the shear force and moment due to the eccentricity of the centroids of the bolt groups on each side, the proportion of moment carried by the web, without any shedding to the flanges.

The second moment of area of the web is;

I_{w} = [(h – 2t_{f})^{3}t_{w}/12]

I_{w} = [(476.5^{3} × 10.8)/12] × 10^{-4} = 9737 cm^{4}

Therefore the web will carry 9737/61500 = 15.8% of the moment in the beam (assuming an elastic stress distribution), while the flange will carry the remaining 84.2%.

The area of the web is;

A_{w} = 476.5 × 10.8 × 10^{-2} = 51.462 cm^{2}

Therefore, the web will carry 51.462/129 = 39.8% of the axial force in the beam, while the flange will carry the remaining 60.1%.

**Forces at ULS**

The force in each flange due to bending is therefore given by;

F_{f,M,Ed} = 0.842 × [(M_{Ed})/(h – t_{f})]

F_{f,M,Ed} = 0.842 × [(610 × 10^{6})/(536.7 – 17.4)] × 10^{-3} = 989 kN

And the force in each flange due to axial force is given by;

F_{f,N,Ed} = 0.601 × (55/2) = 16.53 kN

Therefore

F_{tf,Ed} = 989 – 16.53 = 972.47 kN

F_{bf,Ed} = 989 + 16.53 = 1005.53 kN

The moment in the web = 0.158 × 610 = 96.38 kNm

The shear force in the web = 215 kN

The axial force in the web = 0.398 × 55 = 21.89 kN

**Forces at SLS**

The force in each flange due to bending is therefore given by;

F_{f,M,Ed} = 0.842 × [(M_{Ed,ser})/(h – t_{f})]

F_{f,M,Ed} = 0.842 × [(445 × 10^{6})/(536.7 – 17.4)] × 10^{-3} = 721.53 kN

And the force in each flange due to axial force is given by;

F_{f,N,Ed} = 0.601 × (41/2) = 12.3205 kN

Therefore

F_{tf,Ed} = 721.53 – 12.32 = 709.21 kN

F_{bf,Ed} = 721.53 + 12.32 = 733.85 kN

The moment in the web = 0.158 × 445 = 70.31 kNm

The shear force in the web = 157 kN

The axial force in the web = 0.398 × 41 = 16.318 kN

**Choice of Bolt Number and Configuration****Resistances of Bolts**

The shear resistance of bolts at ultimate limit state

For M24 bolts in single shear = 132 kN

For M24 bolts in double shear = 265 kN

**Flange Splice**

For the flanges, the force of 1005.53 kN at ULS can be provided by 8 M24 bolts in single shear.

The full bearing resistance of an M24 bolt in a 20 mm cover plate (without reduction for spacing and edge distance) is;

F_{b,max,Rd} = (2.5f_{u}dt)/γ_{m2}

F_{b,max,Rd} = [(2.5 × 410 × 24 × 20)/1.25] × 10^{-3} = 393.6 kN

This is much greater than the resistance of the bolt in single shear, and thus the spacings do not need to be such as to maximise the bearing distance. Four lines of 2 bolts at a convenient spacing may be used.

**Web Splice**

For the web splice, consider one or two lines of 4 bolts on either side of the centreline. The full bearing resistance on the 10.8 mm web is;

F_{b,max,Rd} = (2.5f_{u}dt)/γ_{m2}

F_{b,max,Rd} = [(2.5 × 410 × 24 × 10.8)/1.25] × 10^{-3} = 212.54 kN

This is less than the resistance in double shear, and will therefore determine the resistance at ULS. To achieve this value, the spacings will need to be;

e_{1} ≥ 3d_{0} = 3 × 26 = 78mm

p_{1} ≥ 15d_{0}/4 = (15 × 26)/4 = 97.5 mm

e_{2} ≥ 1.5d_{0} = 1.5 × 26 = 39 mm

p_{2} ≥ 3d_{0} = 3 × 26 = 78mm

Initially, try 4 bolts at a vertical spacing of 120mm at a distance of 80mm from the centreline of the splice.

The additional moment due to the eccentricity of the bolt group is;

M_{add} = 215 × 0.08 = 17.2 kNm

**Bolt forces at ULS**

Force/bolt due to vertical shear = 215/4 = 53.75 kN

Force/bolt due to axial compression = 21.89/4 = 5.47 kN

Force/bolt due to moment (considering top and bottom bolts only) = (96.38 + 17.2)/0.36 = 315.5 kN

Thus, a single row is adequate.

Considering the second line of bolts at a horizontal pitch of 100mm

Force/bolt due to vertical shear = 215/4 = 53.75 kN

Force/bolt due to axial compression = 21.89/4 = 5.47 kN

Force/bolt due to moment (considering top and bottom bolts only) = (96.38 + 17.2)/0.36 = 315.5 kN

The additional moment due to eccentricity of this bolt group is;

M_{add} = 215 × (0.08 + 0.1/2) = 27.95 kN.m

The polar moment of inertia of the blot group is given by;

I_{bolts} = 6 × 120^{2} + 8 × (100/2)^{2} = 106400 mm^{2}

The horizontal component of the force on each top and bottom bolt is;

F_{M,horiz} = {[(96.38 + 27.95) × 120]/106400} × 10^{3} = 140.22 kN

The vertical component of the force on each top and bottom bolt is;

F_{M,vert} = {[(96.38 + 27.95) × 50]/106400} × 10^{3} = 58.425 kN

Therefore, the resultant force on the most highly loaded bolt is;

F_{v,Ed} = sqrt[(53.75 + 58.425)^{2} + (140.22 + 5.47)^{2}] = 183.87 kN

This is less than the full bearing resistance, and is therefore satisfactory for such a bolt spacing.

**Chosen Joint Configuration**

**Summary of cover plate dimensions and bolt spacing**

*Flange cover plates;*

Thickness t_{fp} = 20 mm

Length h_{fp} = 660 mm

Width b_{fp} = 200 mm

End distance e_{1,fp} = 70 mm

Edge distance e_{2,fp} = 30 mm

*Spacing*

In the direction of the force p_{1,f } = 100 mm

Transverse to direction of force p_{2,f } = 120 mm

Across the joint in direction of force p_{1,f,j } = 120 mm*Web cover plates;*

Thickness t_{wp} = 12 mm

Length h_{wp} = 460 mm

Width b_{wp} = 460 mm

End distance e_{1,wp} = 50 mm

Edge distance e_{2,wp} = 50 mm

*Spacing*

Vertically p_{1,w } = 120 mm

Horizontally p_{2,w} = 100 mm

Horizontally across the joint p_{1,w,j } = 160 mm**Resistance of flange splices***Resistance of net section*

The resistance of a flange cover plate in tension (N_{t,fp,Rd}) is the lesser of N_{pl,Rd} and N_{u,Rd}.

Here;

N_{u,Rd} = (0.9A_{net,tp}f_{u,fp})/γ_{M2}

Where;

A_{net,fp} = (b_{fp} – 2d_{0})t_{fp} = [200 – (2 × 26)] × 20 = 2960 mm

Therefore;

N_{u,Rd} = [(0.9 × 2960 × 410 )/1.1] × 10^{-3} = 992.945 kN

N_{pl,Rd} = (A_{fp}f_{y,fp})/γ_{M0} = [(200 × 20 × 275)/1.0] × 10^{-3} = 1100 kN

Since 1100 kN > 992.945 KN

N_{t,fp,Rd} = 992.945 kN

For the flange in tension;

F_{tf,Ed} = 989 – 16.53 = 972.47 kN

Therefore, the tension resistance of a flange cover plate is adequate.

**Block Tearing Resistance;**

n_{1,fp} = 3 and n_{2,fp} = 2;

The resistance to block tearing (N_{t,Rd,fp}) is given by:

N_{t,Rd,fp} = {(f_{u,fp}A_{nt,fp})/γ_{m2}} + {[A_{nv,fp}(f_{y,fp}/√3)]/γ_{m0}}

Where:

A_{nt,fp} = t_{fp}(2e_{2,fp} – d_{0})

A_{nt,fp} = 20 × [(2 × 40) – 26] = 1080 mm^{2}

A_{nv,fp} = 2t_{fp}[(n_{1,fp} – 1)p_{1,fp} + e_{1,fp} – (n_{1,fp} – 0.5)d_{0}]

A_{nv,fp} = 2 × 20[(3 – 1) × 100 + 70 – (3 – 0.5) × 26] = 8200 mm^{2}

Therefore;

N_{t,fp,Rd} = {[(410 × 1080)/1.1] + [(8200 × 265/√3)/1.0]} × 10^{-3} = 1657.127 kN

**Resistance of cover plate to compression flange**

Local buckling resistance between the bolts need not be considered if:

p_{1}/t ≤ 9ε

ε = sqrt(235/f_{y,fp}) = sqrt(235/265) = 0.94

9ε = 9 × 0.94 = 8.46

Here, the maximum spacing of bolt across the centreline of the splice p_{1,f,j } = 120 mm

p_{1,f,j}/t_{fp} = 120/20 = 6

Since 8.46 > 6, no local buckling verification required.

**RESISTANCE OF WEB SPLICE***Resistance of web cover plate in shear;*

The gross shear area is given by:

V_{wp,g,Rd} = (h_{wp}t_{wp}f_{y,wp}/1.27γ_{m0}√3)

For two web cover plates;

V_{wp,g,Rd} = 2{[(460 × 12)/1.27] × [275/√3]} × 10^{-3} = 1380.185 kN

V_{Ed} = 215 kN. Therefore, the shear resistance is adequate.

The net shear area is given by:

V_{wp,net,Rd} = A_{v,wp,net}(f_{u,wp}/√3)/γ_{m2}

A_{v,net} = (h_{wp} – 3d_{0})t_{wp}

= (460 – 3 × 26) × 12 = 4584 mm^{2}

For two web cover plates;

V_{n,Rd} = {2 × [4584 × (410/√3)]/1.1} × 10^{-3} = 1972.9 kN

Therefore, shear resistance is adequate.

**Resistance to block tearing**

V_{b,Rd} = (f_{u,wp}A_{nt})/γ_{m2} + (f_{y,wp}A_{nv}.A_{nv})/γ_{m0}√3

For a single vertical line of bolts;

A_{nt} = t_{wp}(e_{2} – d_{0}/2)

A_{nt} = 12 × (50 – 26/2) = 444 mm^{2}

A_{nv} = t_{wp} – e_{1} – (n_{1} – 0,5)d_{0})

A_{nv} = 12 × [460 – 50 – (3 – 0.5)26] = 4140 mm^{2}

For two web plates

V_{b,Rd} = 2 × {[(410 × 460)/1.1] + [(275 × 4140)/(√3 × 1.0)]} × 10^{-3} = 1657.53 kN

V_{Rd} = min{1657.53, 1972.9, 1380.185 }

V_{Rd} = 1380.185 kN

V_{Ed}/V_{Rd} = 215/1380.185 = 0.1557 < 1.0

This is ok.

**Resistance of beam web**

Resistance of net shear area

V_{n,w,Rd} = [A_{v,net}(f_{u}/√3)]/γ_{m2}

Where;

A_{v,net} = A_{v} – 3d_{0}t_{w}

A_{v} = A – 2bt_{f} + (t_{w} + 2r)t_{f} but not less than ηh_{w}t_{w}

A_{v }= 12900 – (2 × 210 × 17.4) + (10.8 + 2 × 12.7) × 17.4 = 6221.88 mm^{2}

A_{v }= 6221.88 – (3 × 26 × 10.8) = 5379.48 mm^{2}

Thus;

V_{n,w,Rd} = {[5379.48 × (410/√3)]/1.1} × 10^{-3} = 1157.632 kN

This is ok.

**Resistance of web cover plate to combined bending, shear, and axial force**

V_{wp,Rd} = 1380.185 kN

Vsub>Ed = 215 kN < 1380.185 kN

Therefore, the effects of shear can be neglected

A_{wp} = 12 × 460 = 5520 mm^{2}

Modulus of the cover plate = (12 × 460^{2})/6 = 423200 mm^{3}

N_{wp,Rd} = 12 × 460 × 275 × 10^{-3} = 1518 kN

Therefore, for two web cover plates

M_{c,wp,Rd} = [(2 × 423200 × 275)/1.0] × 10^{-6} = 232.76 kNm

For two web cover plates;

N_{pl,Rd} = 2 × 1518 = 3036 kN

M_{wp,Ed} = 96.38 + 27.95 = 124.33 kNm

N_{wp,Ed} = 21.89 kN

M_{wp,Ed}/M_{c,wp,Rd} + N_{wp,Ed}/N_{wp,Rd}

124.33/232.76 + (21.89/3036) = 0.541 < 1.0

Therefore, the bending resistance of the web cover plates is adequate.

Thank you for visiting Structville today, and God bless.

• Although each body is still unique, there is a jean for every size. Whether you are petite, long-legged, large and tall, or any other way, blue jeans are made to suit the body you're in. Comfort and ease of wearing are prominent strengths in any size. BIM Consulting in USA

BIM documentation in USA

BIM Implementation in USA

Instructions unclear, M20 bolt