Module 2: Timber Tension and Compression Members - Examples & Applications

Module 2: Timber Tension and Compression Members - Examples & Applications

Timber Tension Members

Basic: Calculating Tension Capacity with Bolt Holes

50 \text{ mm} \times 150 \text{ mm}

Apitong timber member is subjected to axial tension. The member is connected at its ends using two

16 \text{ mm}

diameter bolts placed in a single line across the width.

Given Parameters:

Adjusted allowable tensile stress parallel to grain ( $F_t'$ ): $10.5 \text{ MPa}$
Assume the hole diameter is $2 \text{ mm}$ larger than the bolt diameter to allow for clearance.

Determine the maximum safe axial tensile load ( $P_t$ ) the member can carry.

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Intermediate: Designing a Tension Diagonal with Notches

A wooden diagonal brace in a truss must resist a tensile force of

45 \text{ kN}

. It is made of Tanguile (Group II) with an allowable tensile stress of

F_t' = 8.5 \text{ MPa}

. The member has a

20 \text{ mm}

deep notch cut into one side at the connection to accommodate a steel plate. The member thickness is fixed at

50 \text{ mm}

Determine the required total width (

b

) of the brace to safely carry the load.

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Advanced: Evaluating a Multi-Bolt Tension Member

100 \text{ mm} \times 250 \text{ mm}

solid timber member is in tension. The connection utilizes two rows of

20 \text{ mm}

bolts (hole size

22 \text{ mm}

). Row 1 has two bolts side-by-side, while Row 2 has one bolt staggered. Calculate the capacity based on the critical net section.

Given Parameters:

Adjusted allowable tensile stress ( $F_t'$ ): $10.0 \text{ MPa}$
Thickness ( $t$ ): $100 \text{ mm}$

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Column Classification

Basic: Short Column Capacity Calculation

Determine the maximum safe axial compressive load (

P

) for a solid wood column (Guijo) with dimensions

150 \text{ mm} \times 150 \text{ mm}

and an unbraced length of

1.5 \text{ m}

Assume pinned-pinned end conditions (

K_e = 1.0

Given Parameters:

Adjusted allowable compressive stress ( $F_c'$ ): $15 \text{ MPa}$

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Intermediate: Long Column Capacity using Euler Buckling

Determine the maximum safe axial compressive load (

P

) for a solid timber column with dimensions

100 \text{ mm} \times 100 \text{ mm}

and an unbraced length of

4.0 \text{ m}

. Assume pinned-pinned end conditions (

K_e = 1.0

Given Parameters:

Adjusted allowable compressive stress parallel to grain ( $F_c'$ ): $12 \text{ MPa}$
Modulus of Elasticity ( $E$ ): $10,000 \text{ MPa}$

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Advanced: Column Stability Factor ($C_P$) Calculation

Calculate the adjusted allowable compressive stress using the modern NDS/NSCP Column Stability Factor (

C_P

) approach for a sawn lumber column.

Given Parameters:

Reference compression design value ( $F_c^*$ ): $10.5 \text{ MPa}$ (Assume all other factors like $C_D, C_M = 1.0$ )
Minimum Modulus of Elasticity ( $E_{min}'$ ): $4,500 \text{ MPa}$
Slenderness Ratio ( $L_e/d$ ): $25$
Interaction parameter ( $c$ ): $0.8$ (for sawn lumber)

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Spaced Columns

Conceptual: Advantages of Spaced Columns

A structural design calls for a

6

-meter tall timber column to support a roof truss. A single solid

150 \text{ mm} \times 150 \text{ mm}

timber post is proposed. However, the slenderness ratio check fails. The engineer decides to replace it with a spaced column assembly utilizing two

75 \text{ mm} \times 150 \text{ mm}

members separated by spacer blocks. Explain why this resolves the issue.

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