Worked Examples

Interactive Development-Length Tool

Use the simulation below to reproduce the straight-tension examples. It implements the general metric equation and shows the factor and confinement calculations explicitly.

Straight Tension Development Length

Concept and model scope

NSCP 2015 / adopted ACI 318-14 metric teaching subset for straight deformed-bar tension development.

The model applies lightweight, top-bar, epoxy, bar-size, cover/spacing, and transverse-reinforcement modifiers, caps (cb+Ktr)/db(c_b+K_{tr})/d_b at 2.5, and enforces a 300 mm minimum result.

It is not a complete anchorage, lap-splice, hooked-bar, headed-bar, seismic, or confinement-design engine; project use requires the governing code provisions and detailing conditions.

Controls

fc′f'_c28 MPa
fyf_y420 MPa
Bar diameter dbd_b25 mm
Clear cover50 mm
Clear bar spacing100 mm
AtrA_{tr}0 mm²
Tie spacing ss150 mm
Bars along splitting plane nn2

Auditable intermediates

ψt\psi_t
1.00
ψe\psi_e
1.00
ψtψe\psi_t\psi_e used
1.00
ψs\psi_s
1.00
λ\lambda
1.00
cbc_b
62.5 mm
KtrK_{tr}
0.0 mm
Raw (cb+Ktr)/db(c_b+K_{tr})/d_b
2.500
Ratio used
2.500

Result

ld=fyψtψeψs1.1λfc′[(cb+Ktr)/db]dbl_d=\frac{f_y\psi_t\psi_e\psi_s}{1.1\lambda\sqrt{f'_c}[(c_b+K_{tr})/d_b]}d_b

Calculated before minimum: 722 mm

Required ldl_d = 722 mm

Example 1: General Tension Development with the Confinement Cap

Problem: A 25 mm25\ \text{mm} uncoated bottom bar has fy=420 MPaf_y=420\ \text{MPa} and is embedded in normal-weight concrete with fc′=28 MPaf'_c=28\ \text{MPa}. Clear cover is 75 mm75\ \text{mm}, clear spacing to the adjacent developed bar is 100 mm100\ \text{mm}, and Ktr=0K_{tr}=0. Determine the required straight tension development length.

Step-by-Step Solution

0 of 4 Steps Completed
1

Example 2: Explicit K_tr and an Epoxy-Coated Bar

Problem: Four 25 mm25\ \text{mm} epoxy-coated bottom bars are developed in normal-weight concrete with fy=420 MPaf_y=420\ \text{MPa} and fc′=21 MPaf'_c=21\ \text{MPa}. For the bar being checked, cb=40 mmc_b=40\ \text{mm}. Two 10 mm10\ \text{mm} transverse-reinforcement legs cross the potential splitting plane within each s=200 mms=200\ \text{mm} spacing. There are n=4n=4 developed bars along that splitting plane. The coating condition requires ψe=1.5\psi_e=1.5. Determine KtrK_{tr} and ldl_d.

Step-by-Step Solution

0 of 4 Steps Completed
1

Example 3: Top-Bar and Epoxy Product Cap

Problem: A 20 mm20\ \text{mm} top bar has more than 300 mm300\ \text{mm} of fresh concrete cast below it. It is epoxy coated under a condition that gives ψe=1.5\psi_e=1.5. Use fy=420 MPaf_y=420\ \text{MPa}, fc′=28 MPaf'_c=28\ \text{MPa}, normal-weight concrete, and a confinement ratio capped at 2.52.5. Determine ldl_d.

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 4: Simplified NSCP Table Check for a 25 mm Bar

Problem: A 25 mm25\ \text{mm} bottom, uncoated bar in normal-weight concrete has fy=420 MPaf_y=420\ \text{MPa} and fc′=28 MPaf'_c=28\ \text{MPa}. Its clear spacing and cover satisfy the favorable simplified-table conditions. Determine the simplified development length.

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 5: Compression Development Length

Problem: Determine the compression development length of a 25 mm25\ \text{mm} deformed bar with fy=420 MPaf_y=420\ \text{MPa} in normal-weight concrete with fc′=28 MPaf'_c=28\ \text{MPa}. No qualifying confinement reduction is taken.

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 6: Standard 90-Degree Hook in Tension

Problem: A 20 mm20\ \text{mm} uncoated bar in normal-weight concrete has fy=420 MPaf_y=420\ \text{MPa} and fc′=28 MPaf'_c=28\ \text{MPa}. It terminates in a standard 90∘90^\circ hook. The applicable cover conditions permit ψc=0.7\psi_c=0.7, but no hook-confinement reduction is taken, so ψr=1.0\psi_r=1.0. Determine ldhl_{dh}.

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 7: A Hook Does Not Repair Compression Development

Problem: A 25 mm25\ \text{mm} footing dowel carries compression. From Example 5, ldc=476.2 mml_{dc}=476.2\ \text{mm}, but the detail provides only 375 mm375\ \text{mm} of straight embedment before the bar bends into a 90∘90^\circ hook. Is the dowel adequately developed in compression?

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 8: Class B Tension Lap Splice

Problem: A permitted tension lap splice has a development length ld=800 mml_d=800\ \text{mm}. The provided reinforcement is less than twice that required by analysis and all bars are spliced at the same location. Determine the required splice length.

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 9: Compression Lap Splice

Problem: Two 20 mm20\ \text{mm} bars with fy=420 MPaf_y=420\ \text{MPa} are lap-spliced in compression. Use the basic NSCP expression for bars not requiring a higher-strength-steel provision and assume no special low-strength-concrete or confinement modifier. Determine lscl_{sc}.

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 10: Bundled-Bar Development Increase

Problem: A straight 25 mm25\ \text{mm} bar has an individual required development length of 722 mm722\ \text{mm} before the bundled-bar modifier. Determine the required length when that bar is part of (a) a three-bar bundle and (b) a four-bar bundle.

Step-by-Step Solution

0 of 3 Steps Completed
1

Example 11: Available Footing-Bar Anchorage from the Critical Section

Problem: A 3.0 m3.0\ \text{m} square footing supports a 400 mm400\ \text{mm} square column. Bottom 20 mm20\ \text{mm} bars have 75 mm75\ \text{mm} edge cover and a calculated tension development length of 462 mm462\ \text{mm}. Check straight anchorage from the column face to the bar end.

Step-by-Step Solution

0 of 4 Steps Completed
1

Keep Force State and Anchorage Type Matched

Straight tension development, hooked tension development, straight compression development, and lap-splice length are different checks. A numerical length from one provision is not interchangeable with another merely because the same bar diameter is involved.