Calculation Examples: Tension Members

Interactive Simulation

Note

Use the interactive simulation below to explore how different cross-sectional areas, steel grades, and limit states affect the capacity of a tension member.

Tension Member Capacity Analysis

1.00

Limit State Capacities (phiPn\\phi P_n)

Gross Yielding97.2 kips
Net Rupture108.8 kips
Block Shear75.9 kips

Governing Design Strength

75.9 kips

Example 1: Gross Section Yielding

Determine the nominal gross yielding strength and the LRFD design strength of an ASTM A36 (Fy=36 ksiF_y = 36 \text{ ksi}, Fu=58 ksiF_u = 58 \text{ ksi}) steel plate measuring 1/2×6 inches1/2 \times 6 \text{ inches}.

Given:

  • Steel: A36 (Fy=36 ksiF_y = 36 \text{ ksi})
  • Plate dimensions: t=0.5 int = 0.5 \text{ in}, w=6 inw = 6 \text{ in}

Step-by-Step Solution

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Example 2: Net Section Fracture (Single Line of Bolts)

The same 1/2×6 inch1/2 \times 6 \text{ inch} A36 plate from Example 1 is connected at its ends by two 3/4-inch3/4\text{-inch} diameter bolts in a single line across the width. Determine the LRFD design strength based on net section fracture. Assume U=1.0U = 1.0.

Given:

  • Steel: A36 (Fu=58 ksiF_u = 58 \text{ ksi})
  • Ag=3.0 in2A_g = 3.0 \text{ in}^2, t=0.5 int = 0.5 \text{ in}
  • Bolt diameter: dbolt=3/4 ind_{bolt} = 3/4 \text{ in}
  • U=1.0U = 1.0

Step-by-Step Solution

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Example 3: Net Fracture with Staggered Holes

A 3/8-inch3/8\text{-inch} thick steel plate (Fu=65 ksiF_u = 65 \text{ ksi}) is 10 inches wide and has three 7/8-inch7/8\text{-inch} bolts arranged in a staggered pattern. The holes are spaced longitudinally (pitch ss) by 2.0 inches and transversely (gage gg) by 3.0 inches. Determine the controlling net width and net area.

Given:

  • t=0.375 int = 0.375 \text{ in}, wg=10 inw_g = 10 \text{ in}, Fu=65 ksiF_u = 65 \text{ ksi}
  • dbolt=7/8 ind_{bolt} = 7/8 \text{ in}
  • s=2.0 ins = 2.0 \text{ in}, g=3.0 ing = 3.0 \text{ in}

Step-by-Step Solution

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Example 4: Effective Net Area with Shear Lag ($U$)

An L6×4×1/26\times 4\times 1/2 angle (Ag=4.75 in2A_g = 4.75 \text{ in}^2, xˉ=0.981 in\bar{x} = 0.981 \text{ in} from the long leg face) is connected only by its 6-inch6\text{-inch} long leg to a gusset plate using a welded connection. The length of the longitudinal welds is L=5.0 inL = 5.0 \text{ in}. Determine the effective net area.

Given:

  • Ag=4.75 in2A_g = 4.75 \text{ in}^2, xˉ=0.981 in\bar{x} = 0.981 \text{ in}
  • Weld length L=5.0 inL = 5.0 \text{ in}

Step-by-Step Solution

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Example 5: Block Shear on a Gusset Plate Connection

Determine the LRFD block shear design strength for a 1/2-inch1/2\text{-inch} thick A36 plate (Fy=36 ksi,Fu=58 ksiF_y=36\text{ ksi}, F_u=58\text{ ksi}) connected by two 3/4-inch3/4\text{-inch} bolts in a single line. The bolts are centered, with an edge distance of 1.5 in1.5 \text{ in} and a spacing of 3.0 in3.0 \text{ in}.

Given:

  • t=0.5 int = 0.5 \text{ in}, dh=7/8 ind_h = 7/8 \text{ in}
  • Shear path length: Lv=1.5+3.0=4.5 inL_v = 1.5 + 3.0 = 4.5 \text{ in}
  • Tension path width (half plate width): wt=3.0 inw_t = 3.0 \text{ in}

Step-by-Step Solution

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Example 6: Connection Tear-Out Strength

Check the tear-out strength of the end bolt in the previous example. The bolt has a 1.5-inch1.5\text{-inch} edge distance to the center of the hole.

Given:

  • Ledge=1.5 inL_{edge} = 1.5 \text{ in} to hole center.
  • Hole diameter dh=7/8 ind_h = 7/8 \text{ in}. Bolt diameter d=3/4 ind = 3/4 \text{ in}.
  • t=0.5 int = 0.5 \text{ in}, Fu=58 ksiF_u = 58 \text{ ksi}.

Step-by-Step Solution

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Example 7: Member Selection (Design) using ASD/LRFD

Select the lightest single angle (A36 steel, Fy=36 ksi,Fu=58 ksiF_y=36\text{ ksi}, F_u=58\text{ ksi}) to carry a dead load of 20 kips20\text{ kips} and a live load of 60 kips60\text{ kips}. Assume the effective net area Ae=0.75AgA_e = 0.75 A_g for initial sizing.

Given:

  • PD=20 kipsP_D = 20 \text{ kips}, PL=60 kipsP_L = 60 \text{ kips}

Step-by-Step Solution

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Example 8: Slenderness Limit Check

Check if a single angle L4×4×1/44\times 4\times 1/4 (rz=0.795 inr_z = 0.795 \text{ in}) used as a tension bracing member is acceptable if its unbraced length is 18 feet.

Given:

  • rmin=rz=0.795 inr_{min} = r_z = 0.795 \text{ in}
  • L=18 ftL = 18 \text{ ft}

Step-by-Step Solution

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