Required and maximum-permitted stirrup spacing s

Structural Design · FE Reference Handbook section

Structural Design

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10 exam-style examples

~45 min

All Structural Design lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

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Core formulas for this FE topic

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

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

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

Example 1

Concrete shear capacity check — Required and maximum-permitted stirrup spacing s

A beam has b_w = 18 in., d = 20.0 in., f′c = 5000 psi and carries Vu = 50.9 kip. Is shear reinforcement required (φ = 0.75)?

Given

$b_w = 18 in$
$d = 20.0 in$
$f'c = 5000 psi$
$Vu = 50.9 kip$

Find

φVc and the stirrup requirement

Start with the thinking

Vc = 2√f′c·b_w·d in psi units — divide by 1,000 for kips.
Stirrups are required once Vu exceeds φVc/2.

Step-by-step solution

Concrete capacity — Vc = 2√f′c·b_w·d
Substituting
$Vc = 2\sqrt5000(18)(20.0) = 50,912 lb = 50.9 kip$
Design capacity — φVc = 0.75(50.9) = 38.2 kip
Compare — Vu = 50.9 kip > φVc = 38.2 kip
Conclusion — designed stirrups are required

Answer:

φVc ≈ 38.2 kip → stirrups required

Why the other options are there

50.9 kip (φ omitted)
50,912 kip (psi/kip conversion missed)

Reference: FE Reference Handbook — Structural Design → Required and maximum-permitted stirrup spacing s

Example 2

Concrete shear capacity check — Required and maximum-permitted stirrup spacing s (2)

A beam has b_w = 14 in., d = 23.0 in., f′c = 5000 psi and carries Vu = 45.5 kip. Is shear reinforcement required (φ = 0.75)?

Given

$b_w = 14 in$
$d = 23.0 in$
$f'c = 5000 psi$
$Vu = 45.5 kip$

Find

φVc and the stirrup requirement

Start with the thinking

Vc = 2√f′c·b_w·d in psi units — divide by 1,000 for kips.
Stirrups are required once Vu exceeds φVc/2.

Step-by-step solution

Concrete capacity — Vc = 2√f′c·b_w·d
Substituting
$Vc = 2\sqrt5000(14)(23.0) = 45,538 lb = 45.5 kip$
Design capacity — φVc = 0.75(45.5) = 34.2 kip
Compare — Vu = 45.5 kip > φVc = 34.2 kip
Conclusion — designed stirrups are required

Answer:

φVc ≈ 34.2 kip → stirrups required

Why the other options are there

45.5 kip (φ omitted)
45,538 kip (psi/kip conversion missed)

Reference: FE Reference Handbook — Structural Design → Required and maximum-permitted stirrup spacing s

Example 3

Concrete shear capacity check — Required and maximum-permitted stirrup spacing s (3)

A beam has b_w = 14 in., d = 25.5 in., f′c = 4000 psi and carries Vu = 49.7 kip. Is shear reinforcement required (φ = 0.75)?

Given

$b_w = 14 in$
$d = 25.5 in$
$f'c = 4000 psi$
$Vu = 49.7 kip$

Find

φVc and the stirrup requirement

Start with the thinking

Vc = 2√f′c·b_w·d in psi units — divide by 1,000 for kips.
Stirrups are required once Vu exceeds φVc/2.

Step-by-step solution

Concrete capacity — Vc = 2√f′c·b_w·d
Substituting
$Vc = 2\sqrt4000(14)(25.5) = 45,157 lb = 45.2 kip$
Design capacity — φVc = 0.75(45.2) = 33.9 kip
Compare — Vu = 49.7 kip > φVc = 33.9 kip
Conclusion — designed stirrups are required

Answer:

φVc ≈ 33.9 kip → stirrups required

Why the other options are there

45.2 kip (φ omitted)
45,157 kip (psi/kip conversion missed)

Reference: FE Reference Handbook — Structural Design → Required and maximum-permitted stirrup spacing s

Example 4

Concrete shear capacity check — Required and maximum-permitted stirrup spacing s (4)

A beam has b_w = 16 in., d = 18.5 in., f′c = 5000 psi and carries Vu = 58.6 kip. Is shear reinforcement required (φ = 0.75)?

Given

$b_w = 16 in$
$d = 18.5 in$
$f'c = 5000 psi$
$Vu = 58.6 kip$

Find

φVc and the stirrup requirement

Start with the thinking

Vc = 2√f′c·b_w·d in psi units — divide by 1,000 for kips.
Stirrups are required once Vu exceeds φVc/2.

Step-by-step solution

Concrete capacity — Vc = 2√f′c·b_w·d
Substituting
$Vc = 2\sqrt5000(16)(18.5) = 41,861 lb = 41.9 kip$
Design capacity — φVc = 0.75(41.9) = 31.4 kip
Compare — Vu = 58.6 kip > φVc = 31.4 kip
Conclusion — designed stirrups are required

Answer:

φVc ≈ 31.4 kip → stirrups required

Why the other options are there

41.9 kip (φ omitted)
41,861 kip (psi/kip conversion missed)

Reference: FE Reference Handbook — Structural Design → Required and maximum-permitted stirrup spacing s

Example 5

Concrete shear capacity check — Required and maximum-permitted stirrup spacing s (5)

A beam has b_w = 16 in., d = 16.5 in., f′c = 4000 psi and carries Vu = 50.1 kip. Is shear reinforcement required (φ = 0.75)?

Given

$b_w = 16 in$
$d = 16.5 in$
$f'c = 4000 psi$
$Vu = 50.1 kip$

Find

φVc and the stirrup requirement

Start with the thinking

Vc = 2√f′c·b_w·d in psi units — divide by 1,000 for kips.
Stirrups are required once Vu exceeds φVc/2.

Step-by-step solution

Concrete capacity — Vc = 2√f′c·b_w·d
Substituting
$Vc = 2\sqrt4000(16)(16.5) = 33,394 lb = 33.4 kip$
Design capacity — φVc = 0.75(33.4) = 25.0 kip
Compare — Vu = 50.1 kip > φVc = 25.0 kip
Conclusion — designed stirrups are required

Answer:

φVc ≈ 25.0 kip → stirrups required

Why the other options are there

33.4 kip (φ omitted)
33,394 kip (psi/kip conversion missed)

Reference: FE Reference Handbook — Structural Design → Required and maximum-permitted stirrup spacing s

Example 6

Concrete shear capacity check — Required and maximum-permitted stirrup spacing s (6)

A beam has b_w = 16 in., d = 21.0 in., f′c = 4000 psi and carries Vu = 59.5 kip. Is shear reinforcement required (φ = 0.75)?

Given

$b_w = 16 in$
$d = 21.0 in$
$f'c = 4000 psi$
$Vu = 59.5 kip$

Find

φVc and the stirrup requirement

Start with the thinking

Vc = 2√f′c·b_w·d in psi units — divide by 1,000 for kips.
Stirrups are required once Vu exceeds φVc/2.

Step-by-step solution

Concrete capacity — Vc = 2√f′c·b_w·d
Substituting
$Vc = 2\sqrt4000(16)(21.0) = 42,501 lb = 42.5 kip$
Design capacity — φVc = 0.75(42.5) = 31.9 kip
Compare — Vu = 59.5 kip > φVc = 31.9 kip
Conclusion — designed stirrups are required

Answer:

φVc ≈ 31.9 kip → stirrups required

Why the other options are there

42.5 kip (φ omitted)
42,501 kip (psi/kip conversion missed)

Reference: FE Reference Handbook — Structural Design → Required and maximum-permitted stirrup spacing s

Example 7

Concrete shear capacity check — Required and maximum-permitted stirrup spacing s (7)

A beam has b_w = 15 in., d = 17.0 in., f′c = 5000 psi and carries Vu = 57.7 kip. Is shear reinforcement required (φ = 0.75)?

Given

$b_w = 15 in$
$d = 17.0 in$
$f'c = 5000 psi$
$Vu = 57.7 kip$

Find

φVc and the stirrup requirement

Start with the thinking

Vc = 2√f′c·b_w·d in psi units — divide by 1,000 for kips.
Stirrups are required once Vu exceeds φVc/2.

Step-by-step solution

Concrete capacity — Vc = 2√f′c·b_w·d
Substituting
$Vc = 2\sqrt5000(15)(17.0) = 36,062 lb = 36.1 kip$
Design capacity — φVc = 0.75(36.1) = 27.0 kip
Compare — Vu = 57.7 kip > φVc = 27.0 kip
Conclusion — designed stirrups are required

Answer:

φVc ≈ 27.0 kip → stirrups required

Why the other options are there

36.1 kip (φ omitted)
36,062 kip (psi/kip conversion missed)

Reference: FE Reference Handbook — Structural Design → Required and maximum-permitted stirrup spacing s

Example 8

Concrete shear capacity check — Required and maximum-permitted stirrup spacing s (8)

A beam has b_w = 16 in., d = 21.5 in., f′c = 4000 psi and carries Vu = 52.2 kip. Is shear reinforcement required (φ = 0.75)?

Given

$b_w = 16 in$
$d = 21.5 in$
$f'c = 4000 psi$
$Vu = 52.2 kip$

Find

φVc and the stirrup requirement

Start with the thinking

Vc = 2√f′c·b_w·d in psi units — divide by 1,000 for kips.
Stirrups are required once Vu exceeds φVc/2.

Step-by-step solution

Concrete capacity — Vc = 2√f′c·b_w·d
Substituting
$Vc = 2\sqrt4000(16)(21.5) = 43,513 lb = 43.5 kip$
Design capacity — φVc = 0.75(43.5) = 32.6 kip
Compare — Vu = 52.2 kip > φVc = 32.6 kip
Conclusion — designed stirrups are required

Answer:

φVc ≈ 32.6 kip → stirrups required

Why the other options are there

43.5 kip (φ omitted)
43,513 kip (psi/kip conversion missed)

Reference: FE Reference Handbook — Structural Design → Required and maximum-permitted stirrup spacing s

Example 9

Concrete shear capacity check — Required and maximum-permitted stirrup spacing s (9)

A beam has b_w = 16 in., d = 16.0 in., f′c = 4000 psi and carries Vu = 42.1 kip. Is shear reinforcement required (φ = 0.75)?

Given

$b_w = 16 in$
$d = 16.0 in$
$f'c = 4000 psi$
$Vu = 42.1 kip$

Find

φVc and the stirrup requirement

Start with the thinking

Vc = 2√f′c·b_w·d in psi units — divide by 1,000 for kips.
Stirrups are required once Vu exceeds φVc/2.

Step-by-step solution

Concrete capacity — Vc = 2√f′c·b_w·d
Substituting
$Vc = 2\sqrt4000(16)(16.0) = 32,382 lb = 32.4 kip$
Design capacity — φVc = 0.75(32.4) = 24.3 kip
Compare — Vu = 42.1 kip > φVc = 24.3 kip
Conclusion — designed stirrups are required

Answer:

φVc ≈ 24.3 kip → stirrups required

Why the other options are there

32.4 kip (φ omitted)
32,382 kip (psi/kip conversion missed)

Reference: FE Reference Handbook — Structural Design → Required and maximum-permitted stirrup spacing s

Example 10

Concrete shear capacity check — Required and maximum-permitted stirrup spacing s (10)

A beam has b_w = 16 in., d = 19.0 in., f′c = 5000 psi and carries Vu = 60.2 kip. Is shear reinforcement required (φ = 0.75)?

Given

$b_w = 16 in$
$d = 19.0 in$
$f'c = 5000 psi$
$Vu = 60.2 kip$

Find

φVc and the stirrup requirement

Start with the thinking

Vc = 2√f′c·b_w·d in psi units — divide by 1,000 for kips.
Stirrups are required once Vu exceeds φVc/2.

Step-by-step solution

Concrete capacity — Vc = 2√f′c·b_w·d
Substituting
$Vc = 2\sqrt5000(16)(19.0) = 42,992 lb = 43.0 kip$
Design capacity — φVc = 0.75(43.0) = 32.2 kip
Compare — Vu = 60.2 kip > φVc = 32.2 kip
Conclusion — designed stirrups are required

Answer:

φVc ≈ 32.2 kip → stirrups required

Why the other options are there

43.0 kip (φ omitted)
42,992 kip (psi/kip conversion missed)

Reference: FE Reference Handbook — Structural Design → Required and maximum-permitted stirrup spacing s