Module 9: Steel Connections and Base Plates - Examples & Applications

Bolted Connections

Basic: Number of Bolts Required (Bearing-Type)

A tension member must transfer a factored load (PuP_u) of 450 kN450 \text{ kN} to a gusset plate using a bearing-type connection. The LRFD design shear strength (ϕRn\phi R_n) of one bolt is 80 kN80 \text{ kN} and the design bearing strength (ϕRn\phi R_n) of the plate per bolt is 105 kN105 \text{ kN}. Determine the required number of bolts.

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Intermediate: Slip-Critical Connection Capacity

Calculate the nominal slip resistance (RnR_n) of a single A325 high-strength bolt used in a slip-critical connection. The bolt is fully tensioned to Tb=125 kNT_b = 125 \text{ kN}. The connection consists of two painted steel plates (Class A surface, μ=0.30\mu = 0.30) and has one slip plane (Ns=1N_s = 1). Assume standard holes (Du=1.13D_u = 1.13) and no fillers (hf=1.0h_f = 1.0).

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Welded Connections

Basic: Calculating Nominal Strength of a Fillet Weld

Determine the nominal shear strength (RnR_n) per millimeter length of an 8 mm8 \text{ mm} fillet weld made with an E70XX electrode (FEXX=480 MPaF_{EXX} = 480 \text{ MPa}).

Given Parameters:

  • Weld leg size (ww): 8 mm8 \text{ mm}
  • Electrode ultimate tensile strength (FEXXF_{EXX}): 480 MPa480 \text{ MPa}

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Intermediate: Checking Maximum Weld Size

A structural detailer plans to use a 10 mm10 \text{ mm} fillet weld to attach a thin 8 mm8 \text{ mm} steel plate to a massive, thick column flange. Evaluate if this weld size is permissible according to standard code limitations.

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Design of Axially Loaded Column Base Plates

Basic: Required Base Plate Area Calculation

Determine the required area (A1A_1) for a steel column base plate supporting a factored axial load (PuP_u) of 1,500 kN1,500 \text{ kN}. The base plate rests on a large concrete footing where the full bearing area is much larger than the plate (A24A1A_2 \ge 4A_1). The concrete compressive strength is fc=21 MPaf_c' = 21 \text{ MPa}.

LRFD Factor for bearing on concrete: ϕc=0.65\phi_c = 0.65.

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