Design Procedure and Ultimate Capacity of New Turnbuckle for Measuring Tensile Force

Abstract

A turnbuckle is a device that is installed between tension wires or rods for the purpose of adjusting the tensile force and length of tension members. Tensile force can be adjusted by rotating the body of turnbuckle but the conventional turnbuckle is incapable of measuring tensile force without specific measurement equipment such as a load cell, accelerometer and data logger, etc. New turnbuckles have therefore been developed for adjustment and easy measurement of tensile force. In previous works, the newly developed turnbuckles have been tested and analyzed in order to establish their size, shape and measurable load capacity. The present study addresses the design process and ultimate capacities of new turnbuckles having measurable load capacities of 100, 200 and 300 kN. The ultimate load capacities are obtained via experimental investigation. For each turnbuckle the ultimate loads and the first yield loads are about 5 times and 1.5 times the service load, respectively.

Appendix A Design Examples

The following examples illustrate design procedures for the new turnbuckle installed between tension members. Calculations for both the tension member and the new turnbuckle are presented. The examples are provided for two cases; for a cantilever beam supported by an inclined steel rod at the free end and for a beam retrofitted with a steel rod.

1.1 Cantilever Beam

A cantilever beam has a length of 6 m and carries a uniform dead load of 14 kN/m and a uniform live load of 7 kN/m (Fig. 11). Design the tension member and new turnbuckle.

Fig. 11

Scheme of cantilever

• 1. Calculate required tensile strength (P_r)

  $$- DL = \frac{{3\omega_{DL} L}}{8\sin \theta } = \frac{3 \times 14 \times 6}{{8\sin 30^\circ }} = 63.0{\text{kN}}$$
  $$- LL = \frac{{3\omega_{LL} L}}{8\sin \theta } = \frac{3 \times 7 \times 6}{{8\sin 30^\circ }} = 31.5{\text{kN}}$$

  – P_r = 1.2DL + 1.6LL = 1.2 × 63.0 + 1.6 × 31.5

   = 126.0 kN.

• 2. Assume the tension member

  – Try a thread M27 × 3 of steel rod (ISO 1993) and ASTM A529/A529M Gr. 55 (AISC 2010).

  – Material properties: F_y = 380 MPa, F_u = 485 MPa.

  –Geometric properties: A_g = 572.6 mm², A_e = 459.4 mm².

• 3. Check the lesser of tensile yielding and rupture limit

  –For yielding in the gross section: ϕ_tP_n = ϕ_tF_yA_g = 195.8 kN.

  –For rupture in the net section: ϕ_tP_n = ϕ_tF_uA_e = 167.1 kN.

  –The tensile strength of steel rod is governed by limit state of the rupture in the net section.

  $\therefore$ 167.1 kN > 126.0 kN [o.k.]

• 4. Check load combination for the new turnbuckle

  –Zone A (measurable load state): 1.0DL + 0.5LL = 1.0 × 63.0 + 0.5 × 31.5 = 78.8 kN.

  –Zone B (service load state): 1.0DL + 1.0LL = 1.0 × 63.0 + 1.0 × 31.5 = 94.5 kN.

  –Zone C (limit state): 1.2DL + 1.6LL = 1.2 × 63.0 + 1.6 × 31.5 = 126.0 kN.

  $\therefore$ The new turnbuckle for 100kN is selected.

• 5. Check fracture strength of new turnbuckle using tabulated values (See Table 3)

$\therefore$ 721 kN > F_uA_e = 222.8 kN [o.k.]

1.2 Simple Beam

A beam retrofitted by steel rod has a length of 6 m and carries a uniform dead load of 12 kN/m and a uniform live load of 8 kN/m (Fig. 12). Design the tension member and new turnbuckle.

Fig. 12

Scheme of simple beam

• 1. Calculate required tensile strength (P_r)

  $- DL = \frac{{5\omega_{DL} L}}{16\sin \theta } = \frac{5\times 12\times 6}{{16\sin 10^\circ }} = 129.6$ kN

  $$- LL = \frac{{5\omega_{LL} L}}{16\sin \theta } = \frac{5 \times 8 \times 6}{{16\sin 10^\circ }} = 86.4{\text{kN}}$$

  – P_r = 1.2DL + 1.6LL = 1.2 × 129.6 + 1.6 × 86.4

   = 293.7 kN.

• 2. Assume the tension member

  – Try a thread M39 × 4 of steel rod (ISO 1993) and ASTM A529/A529M Gr. 55 (AISC 2010).

  – Material properties: F_y = 380 MPa, F_u = 485 MPa.

  – Geometric properties: A_g = 1,194.6 mm², A_e = 975.8 mm².

• 3. Check the lesser of tensile yielding and rupture limit

  – For yielding in the gross section: ϕ_tP_n = ϕ_tF_yA_g = 408.5 kN.

  – For rupture in the net section: ϕ_tP_n = ϕ_tF_uA_e = 354.9 kN.

  – The tensile strength of steel rod is governed by limit state of the rupture in the net section.

  $\therefore$ 354.9 kN > 293.7 kN [o.k.]

• 4. Check load combination for the new turnbuckle

  –Zone A (measurable load state): 1.0DL + 0.5LL = 1.0 × 129.6 + 0.5 × 86.4 = 172.8 kN.

  –Zone B (service load state): 1.0DL + 1.0LL = 1.0 × 129.6 + 1.0 × 86.4 = 216.0 kN.

  – Zone C (limit state): 1.2DL + 1.6LL = 1.2 × 129.6 + 1.6 × 86.4 = 293.7 kN.

  $\therefore$ The new turnbuckle for 300kN is selected.

• 5. Check fracture strength of new turnbuckle using tabulated values (See Table 3)

$\therefore$ 1,649 kN > F_uA_e = 473.2 kN [o.k.]