Reinforced Concrete Design

Shear

Use this for “A beam with bw = 12 in and d = 20 in (f′c = 4000 psi) carries Vᵤ = 50 kips. Find φVc and the spacing of two-leg #3 stirrups, fᵧₜ = 60 ksi.”

Diagram

Supports: a pin at A, a roller at B.

s = 7.27 in ≤ sₘₐₓ = 10 in: close enough.

Stirrups

start

s

2

from

each

support;

depth

drawn

as

d

0.9

.

Two stirrup legs cross each crack, so Aᵥ counts both; h is d plus the cover, the stirrup and half a bar.

This is an example. Type your own values to start a new problem.

bw

Web width

example

d

Effective depth

example

f′c

Concrete strength

example · psi

Vc

Concrete’s shear strength

calculated

φVc

Design concrete shear

worked out (not typed)

Vᵤ

Factored shear

example

Vₛ

Stirrups’ shear

calculated

Aᵥ

Stirrup legs’ area

example

fᵧₜ

Stirrup yield strength

example · ksi

s

Stirrup spacing

calculated

sₘₐₓ

Largest spacing

worked out (not typed)

N

Stirrups needed

worked out (not typed)

Formulas

Vc

=

2

×

√f′c

×

bw

×

d

1,000

30.36

=

2

×

√

4

,000

×

12

×

20

1,000

φVc = 0.75 × Vc

22.77 = 0.75 × 30.36

Vᵤ = 0.75 × (Vc + Vₛ)

50 = 0.75 × (30.36 + 36.31)

s

=

Aᵥ

×

fᵧₜ

×

d

Vₛ

7.271

=

0.22

×

60

×

20

36.31

sₘₐₓ

=

min(

d

2,

24)

10

=

min(

20

2,

24)

N

=

none

if

Vᵤ

≤

φVc

2,

the

minimum

if

Vᵤ

≤

φVc,

else

designed

for

Vₛ

50 > 22.77, so designed for Vₛ

Formulas are worked in in, kip, in². The step-by-step shows the conversions.

Assumptions

•

ACI 318-19, normal-weight concrete (λ = 1), φ = 0.75 for shear; f′c in psi under the root.

•

Vertical stirrups, Aᵥ counting every leg across the crack (two #3 legs give 0.22 in²), and the simple Vc = 2λ√f′c bwd.

•

No stirrups up to Vᵤ = φVc ÷ 2, the minimum up to φVc, and stirrups designed for Vₛ above it.

•

sₘₐₓ is the smaller of d ÷ 2 and 24 in, halved once Vₛ passes 4√f′c bwd; round s down. The picture draws a 20 ft span.

Step-by-step

WE KNOW

bw = 30.48 cm, d = 50.8 cm, f′c = 4,000 psi, Vᵤ = 222.4111 kN, Aᵥ = 1.4194 cm², fᵧₜ = 60 ksi

FIND

concrete’s shear strength (Vc), design concrete shear (φVc), stirrups’ shear (Vₛ), stirrup spacing (s), largest spacing (sₘₐₓ), stirrups needed (N)

Step 1 · Convert to the formula’s units

The formulas don’t work directly in the units you chose, so convert first (to in, kip, in²).

bw = 30.48 cm = 12 in (1 in = 2.54 cm)

d = 50.8 cm = 20 in (1 in = 2.54 cm)

Vᵤ = 222.4111 kN = 50 kip (1 kip = 4.44822 kN)

Aᵥ = 1.4194 cm² = 0.22 in² (1 in² = 6.4516 cm²)

Step 2 · Find concrete’s shear strength (Vc)

Use Vc = 2 × √f′c × bw × d ÷ 1,000

ACI’

s

simple

concrete

strength:

2√f′c

(psi)

over

the

web’

s

bw

×

d

gives

pounds;

divide

by

1000

for

kips.

Vc

=

2

×

√

4

,000

×

12

×

20

1,000

Vc

=

2

×

63.25

×

12

×

20

1,000

Vc

=

126.5

×

12

×

20

1,000

Vc

=

1,518

×

20

1,000

Vc

=

30,360

1,000

Vc = 30.3579 kip

Step 3 · Find design concrete shear (φVc)

Use φVc = 0.75 × Vc

The

concrete’

s

design

strength

is

its

nominal

strength

times

φ

=

0.75.

φVc = 0.75 × 30.3579

φVc = 22.7684 kip

Step 4 · Find stirrups’ shear (Vₛ)

Use Vᵤ = 0.75 × (Vc + Vₛ)

The

beam

needs

Vᵤ

φ

in

all;

the

stirrups

carry

what

the

concrete

doesn’t.

Vₛ

=

Vᵤ

0.75

−

Vc

Vₛ

=

50

0.75

−

30.3579

Vₛ = 66.67 − 30.3579

Vₛ = 36.3088 kip

Step 5 · Find stirrup spacing (s)

Use s = Aᵥ × fᵧₜ × d ÷ Vₛ

Each

stirrup

crossing

the

45°

crack

carries

Aᵥfᵧₜ,

and

d

s

of

them

cross

it.

s

=

0.22

×

60

×

20

36.31

s

=

13.2

×

20

36.31

s

=

264

36.31

s

=

7.271

in

Step 6 · Find largest spacing (sₘₐₓ)

Use sₘₐₓ = min(d ÷ 2, 24)

Stirrups

at

most

d

2

(and

24

in)

apart,

so

every

45°

crack

crosses

one;

use

the

smaller

spacing.

36.31

≤

4

×

√

4

,000

×

12

×

20

1,000

:

d

2

sₘₐₓ

=

min(

20

2,

24)

sₘₐₓ = 10 in

Step 7 · Find stirrups needed (N)

Use N = none if Vᵤ ≤ φVc ÷ 2, the minimum if Vᵤ ≤ φVc, else designed for Vₛ

Vᵤ

is

more

than

the

concrete

carries:

stirrups

designed

for

Vₛ

=

Vᵤ

φ

−

Vc.

50 > 22.77: designed for Vₛ

N

=

compare

Vᵤ

with

φVc

N

=

compare

50

with

22.77

N

=

designed

for

Vₛ

Step 8 · Convert the answers

Change each answer back to the unit you chose.

Vc = 30.3579 kip = 135.039 kN (1 kip = 4.44822 kN)

φVc = 22.7684 kip = 101.3 kN (1 kip = 4.44822 kN)

Vₛ = 36.3088 kip = 161.5 kN (1 kip = 4.44822 kN)

s = 7.271 in = 18.4682 cm (1 in = 2.54 cm)

sₘₐₓ = 10 in = 25.4 cm (1 in = 2.54 cm)

Check

Put the numbers back in. Both sides should match.

30.3579

=

2

×

√

4

,000

×

12

×

20

1,000

✓

22.77 = 0.75 × 30.3579 ✓

50 = 0.75 × (30.3579 + 36.31) ✓

7.271

=

0.22

×

60

×

20

36.31

✓

10

=

min(

20

2,

24)

✓

50 > 22.77, so designed for Vₛ ✓