Yielding

Structural Design · FE Reference Handbook section

Structural Design

3 formulas

10 exam-style examples

~51 min

All Structural Design lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Yielding within Structural Design. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what yielding describes physically and when it applies.
State every one of the 3 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: f'c and Fy in ksi with areas in in² give kips.

Lecture

Why this section exists. Yielding is the part of Structural Design that lets you connect a reinforced concrete or steel member being checked to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a factored demand compared against φ times a nominal capacity. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. f'c and Fy in ksi with areas in in² give kips. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Bolted connection at the bottom chord of a steel truss bridge. — Photo 1. Where this shows up in practice: yielding.
HAER / Library of Congress, public domain

Structural Design — Yielding: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a reinforced concrete or steel member being checked. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 3 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

Mn	Quantity produced by "Mn = Mp = FyZx" — read its definition and unit from the handbook line directly above the equation.
Fy	Quantity produced by "Fy = specified minimum yield stress" — read its definition and unit from the handbook line directly above the equation.
Zx	Quantity produced by "Zx = plastic section modulus about the x-axis" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

where

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Tension member capacity: yielding versus rupture

An A36 plate 6 in. × ½ in. has two ⅞ in. bolt holes in the critical section. With Fy = 36 ksi, Fu = 58 ksi and U = 1.0, what governs the design tensile strength?

Given

Ag = 6(0.5) = 3.00 in²
Two holes, d_h = 0.875 + 0.125 = 1.00 in.
Fy = 36 ksi, Fu = 58 ksi

Find

φPn (governing limit state)

Start with the thinking

Two limit states: gross-section yielding (φ = 0.90) and net-section rupture (φ = 0.75).
Hole diameter includes the 1/16 in. damage allowance.

Figure for Tension member capacity: yielding versus rupture

Step-by-step solution

Gross yielding — φPn = 0.90 Fy Ag = 0.90(36)(3.00) = 97.2 kips
Net area
$A n = 3.00 - 2 (1.00) (0.5) = 2.00 i n^{2}$
Effective net
$A e = U A n = 1.0 (2.00) = 2.00 i n^{2}$
Rupture — φPn = 0.75 Fu Ae = 0.75(58)(2.00) = 87.0 kips
Governing — the smaller value, 87.0 kips (net-section rupture)

Answer: φPn = 87.0 kips, governed by rupture

Why the other options are there

97.2 kips (yielding taken as governing)
104 kips (holes ignored)

Reference: FE Reference Handbook — Structural Design — Steel tension members

Example 2

Tension member capacity — yielding versus rupture — Yielding

An A992 plate has Ag = 3.50 in² and 3 bolt holes across the critical section (⅞″ bolts, t = 0.438 in.). Find the design tensile strength (Fy = 50 ksi, Fu = 65 ksi).

Given

Ag = 3.50 in²
3 hole(s)
t = 0.438 in.
Fy = 50 ksi, Fu = 65 ksi

Find

φPn governing

Start with the thinking

Two limit states: yielding on Ag (φ = 0.90) and rupture on An (φ = 0.75).
Hole diameter = bolt + 1/8 in.

Figure for Tension member capacity — yielding versus rupture — Yielding

Step-by-step solution

Net area
$A n = A g - n (d_{h}) t = 3.50 - 3 (1.00) (0.438) = 2.188 i n^{2}$
Yielding — φPn = 0.90FyAg = 0.90(50)(3.50) = 157.5 kip
Rupture — φPn = 0.75FuAn = 0.75(65)(2.188) = 106.6 kip
Governing — the smaller controls: 106.6 kip

Answer: φPn ≈ 106.6 kip (rupture governs)

Why the other options are there

157.5 kip (larger value taken)
204.8 kip (Fu with Ag)

Reference: FE Reference Handbook — Structural Design → Yielding

Example 3

Tension member capacity — yielding versus rupture — Yielding (2)

An A992 plate has Ag = 8.25 in² and 2 bolt holes across the critical section (⅞″ bolts, t = 0.688 in.). Find the design tensile strength (Fy = 50 ksi, Fu = 65 ksi).

Given

Ag = 8.25 in²
2 hole(s)
t = 0.688 in.
Fy = 50 ksi, Fu = 65 ksi

Find

φPn governing

Start with the thinking

Two limit states: yielding on Ag (φ = 0.90) and rupture on An (φ = 0.75).
Hole diameter = bolt + 1/8 in.

Figure for Tension member capacity — yielding versus rupture — Yielding (2)

Step-by-step solution

Net area
$A n = A g - n (d_{h}) t = 8.25 - 2 (1.00) (0.688) = 6.875 i n^{2}$
Yielding — φPn = 0.90FyAg = 0.90(50)(8.25) = 371.3 kip
Rupture — φPn = 0.75FuAn = 0.75(65)(6.875) = 335.2 kip
Governing — the smaller controls: 335.2 kip

Answer: φPn ≈ 335.2 kip (rupture governs)

Why the other options are there

371.3 kip (larger value taken)
482.6 kip (Fu with Ag)

Reference: FE Reference Handbook — Structural Design → Yielding

Example 4

Tension member capacity — yielding versus rupture — Yielding (3)

An A992 plate has Ag = 8.25 in² and 3 bolt holes across the critical section (⅞″ bolts, t = 0.625 in.). Find the design tensile strength (Fy = 50 ksi, Fu = 65 ksi).

Given

Ag = 8.25 in²
3 hole(s)
t = 0.625 in.
Fy = 50 ksi, Fu = 65 ksi

Find

φPn governing

Start with the thinking

Two limit states: yielding on Ag (φ = 0.90) and rupture on An (φ = 0.75).
Hole diameter = bolt + 1/8 in.

Figure for Tension member capacity — yielding versus rupture — Yielding (3)

Step-by-step solution

Net area
$A n = A g - n (d_{h}) t = 8.25 - 3 (1.00) (0.625) = 6.375 i n^{2}$
Yielding — φPn = 0.90FyAg = 0.90(50)(8.25) = 371.3 kip
Rupture — φPn = 0.75FuAn = 0.75(65)(6.375) = 310.8 kip
Governing — the smaller controls: 310.8 kip

Answer: φPn ≈ 310.8 kip (rupture governs)

Why the other options are there

371.3 kip (larger value taken)
482.6 kip (Fu with Ag)

Reference: FE Reference Handbook — Structural Design → Yielding

Example 5

Tension member capacity — yielding versus rupture — Yielding (4)

An A992 plate has Ag = 7.00 in² and 2 bolt holes across the critical section (⅞″ bolts, t = 0.625 in.). Find the design tensile strength (Fy = 50 ksi, Fu = 65 ksi).

Given

Ag = 7.00 in²
2 hole(s)
t = 0.625 in.
Fy = 50 ksi, Fu = 65 ksi

Find

φPn governing

Start with the thinking

Two limit states: yielding on Ag (φ = 0.90) and rupture on An (φ = 0.75).
Hole diameter = bolt + 1/8 in.

Figure for Tension member capacity — yielding versus rupture — Yielding (4)

Step-by-step solution

Net area
$A n = A g - n (d_{h}) t = 7.00 - 2 (1.00) (0.625) = 5.750 i n^{2}$
Yielding — φPn = 0.90FyAg = 0.90(50)(7.00) = 315.0 kip
Rupture — φPn = 0.75FuAn = 0.75(65)(5.750) = 280.3 kip
Governing — the smaller controls: 280.3 kip

Answer: φPn ≈ 280.3 kip (rupture governs)

Why the other options are there

315.0 kip (larger value taken)
409.5 kip (Fu with Ag)

Reference: FE Reference Handbook — Structural Design → Yielding

Example 6

Tension member capacity — yielding versus rupture — Yielding (5)

An A992 plate has Ag = 7.25 in² and 3 bolt holes across the critical section (⅞″ bolts, t = 0.500 in.). Find the design tensile strength (Fy = 50 ksi, Fu = 65 ksi).

Given

Ag = 7.25 in²
3 hole(s)
t = 0.500 in.
Fy = 50 ksi, Fu = 65 ksi

Find

φPn governing

Start with the thinking

Two limit states: yielding on Ag (φ = 0.90) and rupture on An (φ = 0.75).
Hole diameter = bolt + 1/8 in.

Figure for Tension member capacity — yielding versus rupture — Yielding (5)

Step-by-step solution

Net area
$A n = A g - n (d_{h}) t = 7.25 - 3 (1.00) (0.500) = 5.750 i n^{2}$
Yielding — φPn = 0.90FyAg = 0.90(50)(7.25) = 326.3 kip
Rupture — φPn = 0.75FuAn = 0.75(65)(5.750) = 280.3 kip
Governing — the smaller controls: 280.3 kip

Answer: φPn ≈ 280.3 kip (rupture governs)

Why the other options are there

326.3 kip (larger value taken)
424.1 kip (Fu with Ag)

Reference: FE Reference Handbook — Structural Design → Yielding

Example 7

Tension member capacity — yielding versus rupture — Yielding (6)

An A992 plate has Ag = 11.75 in² and 3 bolt holes across the critical section (⅞″ bolts, t = 0.750 in.). Find the design tensile strength (Fy = 50 ksi, Fu = 65 ksi).

Given

Ag = 11.75 in²
3 hole(s)
t = 0.750 in.
Fy = 50 ksi, Fu = 65 ksi

Find

φPn governing

Start with the thinking

Two limit states: yielding on Ag (φ = 0.90) and rupture on An (φ = 0.75).
Hole diameter = bolt + 1/8 in.

Figure for Tension member capacity — yielding versus rupture — Yielding (6)

Step-by-step solution

Net area
$A n = A g - n (d_{h}) t = 11.75 - 3 (1.00) (0.750) = 9.500 i n^{2}$
Yielding — φPn = 0.90FyAg = 0.90(50)(11.75) = 528.8 kip
Rupture — φPn = 0.75FuAn = 0.75(65)(9.500) = 463.1 kip
Governing — the smaller controls: 463.1 kip

Answer: φPn ≈ 463.1 kip (rupture governs)

Why the other options are there

528.8 kip (larger value taken)
687.4 kip (Fu with Ag)

Reference: FE Reference Handbook — Structural Design → Yielding

Example 8

Tension member capacity — yielding versus rupture — Yielding (7)

An A992 plate has Ag = 9.25 in² and 1 bolt hole across the critical section (⅞″ bolts, t = 0.625 in.). Find the design tensile strength (Fy = 50 ksi, Fu = 65 ksi).

Given

Ag = 9.25 in²
1 hole(s)
t = 0.625 in.
Fy = 50 ksi, Fu = 65 ksi

Find

φPn governing

Start with the thinking

Two limit states: yielding on Ag (φ = 0.90) and rupture on An (φ = 0.75).
Hole diameter = bolt + 1/8 in.

Figure for Tension member capacity — yielding versus rupture — Yielding (7)

Step-by-step solution

Net area
$A n = A g - n (d_{h}) t = 9.25 - 1 (1.00) (0.625) = 8.625 i n^{2}$
Yielding — φPn = 0.90FyAg = 0.90(50)(9.25) = 416.3 kip
Rupture — φPn = 0.75FuAn = 0.75(65)(8.625) = 420.5 kip
Governing — the smaller controls: 416.3 kip

Answer: φPn ≈ 416.3 kip (yielding governs)

Why the other options are there

420.5 kip (larger value taken)
541.1 kip (Fu with Ag)

Reference: FE Reference Handbook — Structural Design → Yielding

Example 9

Tension member capacity — yielding versus rupture — Yielding (8)

An A992 plate has Ag = 11.00 in² and 2 bolt holes across the critical section (⅞″ bolts, t = 0.563 in.). Find the design tensile strength (Fy = 50 ksi, Fu = 65 ksi).

Given

Ag = 11.00 in²
2 hole(s)
t = 0.563 in.
Fy = 50 ksi, Fu = 65 ksi

Find

φPn governing

Start with the thinking

Two limit states: yielding on Ag (φ = 0.90) and rupture on An (φ = 0.75).
Hole diameter = bolt + 1/8 in.

Figure for Tension member capacity — yielding versus rupture — Yielding (8)

Step-by-step solution

Net area
$A n = A g - n (d_{h}) t = 11.00 - 2 (1.00) (0.563) = 9.875 i n^{2}$
Yielding — φPn = 0.90FyAg = 0.90(50)(11.00) = 495.0 kip
Rupture — φPn = 0.75FuAn = 0.75(65)(9.875) = 481.4 kip
Governing — the smaller controls: 481.4 kip

Answer: φPn ≈ 481.4 kip (rupture governs)

Why the other options are there

495.0 kip (larger value taken)
643.5 kip (Fu with Ag)

Reference: FE Reference Handbook — Structural Design → Yielding

Example 10

Tension member capacity — yielding versus rupture — Yielding (9)

An A992 plate has Ag = 10.50 in² and 3 bolt holes across the critical section (⅞″ bolts, t = 0.500 in.). Find the design tensile strength (Fy = 50 ksi, Fu = 65 ksi).

Given

Ag = 10.50 in²
3 hole(s)
t = 0.500 in.
Fy = 50 ksi, Fu = 65 ksi

Find

φPn governing

Start with the thinking

Two limit states: yielding on Ag (φ = 0.90) and rupture on An (φ = 0.75).
Hole diameter = bolt + 1/8 in.

Figure for Tension member capacity — yielding versus rupture — Yielding (9)

Step-by-step solution

Net area
$A n = A g - n (d_{h}) t = 10.50 - 3 (1.00) (0.500) = 9.000 i n^{2}$
Yielding — φPn = 0.90FyAg = 0.90(50)(10.50) = 472.5 kip
Rupture — φPn = 0.75FuAn = 0.75(65)(9.000) = 438.8 kip
Governing — the smaller controls: 438.8 kip

Answer: φPn ≈ 438.8 kip (rupture governs)

Why the other options are there

472.5 kip (larger value taken)
614.3 kip (Fu with Ag)

Reference: FE Reference Handbook — Structural Design → Yielding

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a reinforced concrete or steel member being checked, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Yielding contains 3 relations; you must be able to find this page in under 15 seconds.
Exam style: a factored demand compared against φ times a nominal capacity.
Unit rule: f'c and Fy in ksi with areas in in² give kips.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

f'c and Fy in ksi with areas in in² give kips
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Yielding

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

Mn=Mp=FyZx

Fy=specifiedminimumyieldstress

Zx=plasticsectionmodulusaboutthex−axis

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section