Module 5: Introduction to Structural Steel - Examples & Applications

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.

Case Study 1: Material Ductility and Seismic System Performance

Two buildings experience a major earthquake. One is an unreinforced masonry structure; the other is a properly engineered steel seismic-force-resisting system that develops substantial inelastic deformation without collapse. Explain the role of steel ductility without confusing material behavior with system-level seismic performance.

Step-by-Step Solution

0 of 3 Steps Completed
1

Case Study 2: Modulus of Elasticity vs. Yield Strength

An architect requests a floor system with a very long clear span and strict vibration limits (high stiffness). The structural engineer initially specifies an ASTM A36 steel beam. The architect later asks if upgrading the steel to high-strength ASTM A992 will reduce the bouncing/vibration of the floor. How should the engineer respond?

Step-by-Step Solution

0 of 4 Steps Completed
1

Case Study 1: Selecting the Right Shape and Grade

A structural engineer is detailing a commercial building frame. They need to specify the material for the main wide-flange floor beams, the rectangular columns in the glass atrium, and the small connection plates holding everything together. Recommend the appropriate ASTM grades and shapes for these three applications.

Step-by-Step Solution

0 of 3 Steps Completed
1

Case Study 2: Built-Up Plate Girders vs. Rolled Sections

A highway overpass requires a clear span of 45 meters. The heaviest available standard W-shape rolled by mills is a W36 (roughly 36 inches deep). Preliminary calculations show this rolled section is severely inadequate for both strength and deflection. Propose a structural solution using steel.

Step-by-Step Solution

0 of 3 Steps Completed
1

Basic: ASD vs LRFD Load Combinations

A column must support a dead load (DD) of 100 kN100 \text{ kN} and a live load (LL) of 150 kN150 \text{ kN}. Calculate the required design load using both ASD and LRFD basic load combinations.

Step-by-Step Solution

0 of 2 Steps Completed
1

Observation

LRFD and ASD use different load levels and different available-strength formats. The numerical required loads therefore should not be compared by themselves as though the larger number were more conservative. Adequacy is established only by comparing demand and available strength on the same design basis for every applicable limit state.

Intermediate: LRFD Load Combinations with Roof Live Load

A steel roof beam must support a dead load (DD) of 10 kN/m10 \text{ kN/m}, a roof live load (LrL_r) of 15 kN/m15 \text{ kN/m}, and a wind load (WW) of 12 kN/m12 \text{ kN/m}. Determine the required LRFD ultimate load (wuw_u).

Given LRFD Combinations:

  1. 1.4D1.4D
  2. 1.2D+1.6Lr+0.5W1.2D + 1.6L_r + 0.5W
  3. 1.2D+1.0W+0.5Lr1.2D + 1.0W + 0.5L_r

Step-by-Step Solution

0 of 4 Steps Completed
1

Advanced: Converting Nominal Strength to Allowable Strength

A steel tensile member has a nominal yielding strength Tn=850 kNT_n=850\text{ kN}. Determine the ASD allowable strength using Ωt=1.67\Omega_t=1.67.

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 8 — Section selection beyond area

Two candidate steel columns have similar cross-sectional area, but one has a substantially larger weak-axis radius of gyration. Which property makes it potentially more efficient against flexural buckling?

Step-by-Step Solution

0 of 2 Steps Completed
1

Example 9 — Width-to-thickness classification is action-specific

A rolled I-shape flange in uniform axial compression has b/t=10.0b/t=10.0, E=200,000 MPaE=200{,}000\text{ MPa}, and Fy=345 MPaF_y=345\text{ MPa}. For the stated axial-compression element case, compare with λr=0.56E/Fy\lambda_r=0.56\sqrt{E/F_y}.

Step-by-Step Solution

0 of 3 Steps Completed
1