2.1 Dead Loads and Self-Weight

For a July 2026 PE Civil: Structural candidate, dead-load questions are governed by the April 2024 Civil: Structural specification and the references it lists. Use the PE Civil Reference Handbook active on the test date for mechanics and unit conversions, ASCE 7-16 with IBC 2018 for building applications, and AASHTO LRFD Bridge Design Specifications, 8th edition (2017), including the listed May 2018 errata, for bridge applications. Do not import the April 2027 reference set, newer code editions, or PE Structural/SE assumptions.

What Belongs in Dead Load?

Dead load is the weight of permanent construction. Begin with the structural members, then look for permanently attached nonstructural items. A problem may describe some of these directly and require you to calculate others.

The word permanent controls the classification, but the problem statement and controlling reference control the numerical value. Do not rely on a remembered density when the handbook, standard, or problem supplies one.

The Unit-Preserving Workflow

Use a consistent chain from material to member response:

1. Inventory each permanent component. Mark whether its value is already an area, line, or point load.
2. Convert volume weight to area load. For a uniform layer, q = gamma t, where gamma is unit weight and t is thickness.
3. Add compatible area loads. Do not add psf to plf or kPa to kN/m.
4. Use tributary geometry. For a one-way system, w = q b_t, where b_t is tributary width.
5. Add member self-weight once. If a beam weight is already contained in a model reaction or stated total, do not add it again.
6. Continue the load path. Convert line load to beam reactions, reactions to girder or column loads, and column loads to the foundation.
7. Check scale and units. A missing factor of 12 commonly appears when inches are treated as feet.

Worked Example: Slab to Beam

A simply supported steel beam spans 24 ft and supports a 10 ft tributary width of a 6 in normal-weight concrete slab. Use a concrete unit weight of 150 pcf. Permanent flooring, ceiling, and fixed services total 12 psf. The beam weighs 65 plf. Find the service dead line load and each support reaction.

1. Convert slab thickness.

6 in x (1 ft / 12 in) = 0.50 ft

2. Compute slab self-weight.

q_slab = (150 lb/ft^3)(0.50 ft) = 75 psf

3. Add superimposed permanent load.

q_D = 75 + 12 = 87 psf

4. Convert area load to line load.

w_slab+SDL = (87 psf)(10 ft) = 870 plf

5. Add beam self-weight exactly once.

w_D = 870 + 65 = 935 plf = 0.935 klf

6. Find reactions. For a simply supported beam under uniform load, each reaction is half the total load:

R = wL/2 = (0.935 klf)(24 ft)/2 = 11.22 kips

A useful check is total downward load: 0.935 x 24 = 22.44 kips; the two reactions sum to the same value. The 65 plf beam weight was not hidden inside the 87 psf floor load, so adding it was necessary. If the problem had stated that the 87 psf included supporting framing, adding 65 plf would have been double counting.

SI Cross-Check

The same logic works in SI units. A 150 mm slab with gamma = 24 kN/m^3 weighs (0.150 m)(24 kN/m^3) = 3.6 kPa. If other permanent layers total 1.0 kPa and a beam tributary width is 3.0 m, the supported line load is (3.6 + 1.0) kN/m^2 x 3.0 m = 13.8 kN/m, before adding the beam's own kN/m weight.

Tributary Area Is Geometry, Not Guesswork

For regularly spaced one-way members, an interior beam commonly receives half the bay on each side. Edge beams usually have only the tributary width on the interior side unless another component frames into them. Openings, transfer members, cantilevers, two-way behavior, and discontinuities change that simple picture. Sketch boundaries before multiplying. If a wall bears on a slab or beam, trace where its line load actually travels rather than spreading it across an arbitrary floor area.

Bridge questions require the same equilibrium discipline but a different controlling standard. Identify each permanent component, classify it under the AASHTO framework, and preserve its distribution through deck, girder, bearings, substructure, and foundation. Do not move an ASCE building label or load factor into an AASHTO problem.

Final Error Screen

Before accepting a dead-load result, ask:

• Did every permanent component enter once?
• Were inches converted to feet, or millimeters to meters, before multiplying by unit weight?
• Are the final units appropriate for the object: psf/kPa, plf/kN/m, or lb/kN?
• Does the tributary sketch support the multiplier used?
• Do reactions recover the total applied load?
• Did I keep nominal/service load separate from any later factored combination?

This short screen catches the most common dead-load errors: a missing self-weight, duplicated assembly weight, and an area-to-line conversion made with the wrong tributary width.