Module 9 Worked Examples: Steel Connections and Base Plates

These examples emphasize complete force transfer, bolt/weld limit states, joint classification, base-plate behavior, and buildable architectural detailing.

Worked-example data provenance

Unless an example explicitly cites a code table, manufacturer report, or material specification, numerical material properties and adjustment factors are problem-supplied inputs. They demonstrate the calculation procedure and must not be reused as universal NSCP design values for another species, grade, steel grade, section, or product.

Example 1 — Nominal bolt shear

A bolt has nominal area Ab=353 mm2A_b=353\text{ mm}^2 and the applicable nominal shear stress is Fnv=330 MPaF_{nv}=330\text{ MPa}. Determine nominal shear strength for one shear plane.

Step-by-Step Solution

0 of 2 Steps Completed
1

Example 2 — Multiple shear planes

Using the bolt from Example 1, what is the idealized nominal bolt shear resistance if the connection provides two effective shear planes and the provision permits direct multiplication?

Step-by-Step Solution

0 of 2 Steps Completed
1

Example 3 — Bearing-type does not mean always snug-tight

A connection is designed for bearing-type strength. Does that statement by itself prove the bolts are installed only snug-tight?

Step-by-Step Solution

0 of 2 Steps Completed
1

Example 4 — Fillet weld effective throat

Find the effective throat of an 8 mm8\text{ mm} equal-leg fillet weld.

Step-by-Step Solution

0 of 2 Steps Completed
1

Example 5 — Fillet weld nominal strength

A fillet weld has FEXX=490 MPaF_{EXX}=490\text{ MPa}, effective throat 5.66 mm5.66\text{ mm}, and effective length 150 mm150\text{ mm}. Determine the nominal weld-metal strength using Rn=0.60FEXXAweR_n=0.60F_{EXX}A_{we}.

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 6 — Why a bolt-count division is incomplete

A 300 kN300\text{ kN} shear is assigned equally to four bolts, giving 75 kN75\text{ kN} per bolt. What additional checks remain?

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 7 — Eccentric bracket connection

A bracket load acts away from the centroid of its bolt group. Why is equal shear per bolt insufficient?

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 8 — Base-plate bearing concept

A concentric column reaction is increased while base-plate area remains unchanged. What happens to average bearing pressure and plate bending demand?

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 9 — Moment base with uplift

A column base develops enough overturning moment that one side of the plate tends to lift from the concrete. Can the concentric bearing-only model be used?

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 10 — Exposed HSS node review

An architect wants a visually seamless HSS-to-HSS node. List the engineering questions that must be resolved before hiding the connection.

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 11 — Numerical eccentric four-bolt group

A four-bolt group has bolt coordinates (±50,±75) mm(\pm50,\pm75)\text{ mm} from its centroid. A vertical shear V=40 kNV=40\text{ kN} acts with horizontal eccentricity e=200 mme=200\text{ mm}. Using the elastic bolt-group method for this worked case, estimate the largest resultant bolt force.

Step-by-Step Solution

0 of 4 Steps Completed
1

Example 12 — Base plate with partial compression-only contact

A rectangular base plate has B=500 mmB=500\text{ mm} in the eccentricity direction and N=400 mmN=400\text{ mm} perpendicular to it. A compressive resultant Pu=900 kNP_u=900\text{ kN} acts at e=120 mme=120\text{ mm} from the plate centerline. Use the simplified rigid-plate compression-only contact model.

Step-by-Step Solution

0 of 4 Steps Completed
1

Example 13 — Concrete bearing under a concentrically loaded base plate

A 400 mm×400 mm400\text{ mm}\times400\text{ mm} base plate bears concentrically on a geometrically similar 800 mm×800 mm800\text{ mm}\times800\text{ mm} concrete support with fc′=28 MPaf'_c=28\text{ MPa}. Determine the nominal concrete bearing strength benchmark.

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 14 — Bolt tension reduced by simultaneous shear

A bearing-type bolt has problem-supplied Fnt=620 MPaF_{nt}=620\text{ MPa} and Fnv=372 MPaF_{nv}=372\text{ MPa}. Under LRFD, the required bolt shear stress is frv=100 MPaf_{rv}=100\text{ MPa}. Use ϕ=0.75\phi=0.75 and nominal bolt area Ab=245 mm2A_b=245\text{ mm}^2.

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 15 — Concentric W-column base-plate thickness

A concentrically loaded W-column base is checked on the LRFD basis. The factored compression is Pu=900 kNP_u=900\text{ kN}, the plate is B=400 mmB=400\text{ mm} by N=500 mmN=500\text{ mm}, the plate yield stress is Fy=250 MPaF_y=250\text{ MPa}, and the column dimensions used by the AISC 14th Edition Manual Part 14 geometry are d=300 mmd=300\text{ mm} and bf=250 mmb_f=250\text{ mm}. For this worked case, the Manual-defined λn′\lambda n' branch has already been evaluated and is no greater than the mm or nn projections.

Step-by-Step Solution

0 of 4 Steps Completed
1