# Working Stresses of Philippine Woods at 50% Stress Grade

**Working Stresses for Visually Stress-Graded Unseasoned Structural Timber of Philippine Woods**

Species |
50% Stress Grade |
||||

Bending & Tension Parallel to Grain | Modulus of Elasticity in Bending | Compression Parallel to grain | Cpmpression Perpendicular to Grain | Shear parallel to Grain | |

MPa | GPa | MPa | MPa | MPa |

# Working Stresses of Philippine Woods at 80% Stress Grade

**Working Stresses for Visually Stress-Graded Unseasoned Structural Timber of Philippine Woods**

Species |
80% Stress Grade |
||||

Bending & Tension Parallel to Grain | Modulus of Elasticity in Bending | Compression Parallel to grain | Compression Perpendicular to Grain | Shear parallel to Grain | |

MPa | GPa | MPa | MPa | MPa |

# Allowable Loads on One Bolt in Timber Connection

## Group Species of Philippine Wood

I. High Strength Group |
II. Moderately High Strength Group |
III. Medium Strength Group |
IV. Moderately Low Strength Group |

Agoho Liusin Malabayabas Manggachapui Molave Narig Sasalit Yakal |
Antipolo Binggas Bokbok Dao Gatasan Guijo Kamagong Kamatog Katmon Kato Lomarau Mahogany, Big-leafed Makaasim Malakauayan Narra Pahutan |
Apitong Bagtikan Dangkalan Gisau Lanutan-bagyo Lauan Malaanonang Malasaging Malugai Miau Nato Palosapis Pine Salakin Vidal lanutan |
Almaciga Bayok Lingo-lingo Mangasinoro Raintree Yemane |

# Example 03: Moment Capacity of a Timber Beam Reinforced with Steel and Aluminum Strips

**Problem**

Steel and aluminum plates are used to reinforced an 80 mm by 150 mm timber beam. The three materials are fastened firmly as shown so that there will be no relative movement between them.

Given the following material properties:

Allowable Bending Stress, F_{b}Steel = 120 MPa Aluminum = 80 MPa Wood = 10 MPa |
Modulus of Elasticity, ESteel = 200 GPa Aluminum = 70 GPa Wood = 10 GPa |

Find the safe resisting moment of the beam in kN·m.

# Example 02: Required Diameter of Circular Log Used for Footbridge Based on Shear Alone

**Problem**

A wooden log is to be used as a footbridge to span 3-m gap. The log is required to support a concentrated load of 30 kN at midspan. If the allowable stress in shear is 0.7 MPa, what is the diameter of the log that would be needed. Assume the log is very nearly circular and the bending stresses are adequately met. Neglect the weight of the log.

# Example 01: Maximum bending stress, shear stress, and deflection

**Problem**

A timber beam 4 m long is simply supported at both ends. It carries a uniform load of 10 kN/m including its own weight. The wooden section has a width of 200 mm and a depth of 260 mm and is made up of 80% grade Apitong. Use dressed dimension by reducing its dimensions by 10 mm.

Bending and tension parallel to grain = 16.5 MPa

Shear parallel to grain = 1.73 MPa

Modulus of elasticity in bending = 7.31 GPa

- What is the maximum flexural stress of the beam?
- What is the maximum shearing stress of the beam?
- What is the maximum deflection of the beam?

# Equation of the Diameter of Parabola Bisecting Parallel Chords of Given Slope

**Problem**

A parabola has an equation of *y*^{2} = 8*x*. Find the equation of the diameter of the parabola, which bisect chords parallel to the line *x* – *y* = 4.

A. y = 2 |
C. y = 4 |

B. y = 3 |
D. y = 1 |

# Safe Dimensions of the Notch at a Joint of a Timber Truss

**Situation**

The truss shown in is made from timber Guijo 100 mm × 150 mm. The load on the truss is 20 kN. Neglect friction.

Compression parallel to grain = 11 MPa

Compression perpendicular to grain = 5 MPa

Shear parallel to grain = 1 MPa

1. Determine the minimum value of *x* in mm.

A. 180 | C. 160 |

B. 150 | D. 140 |

2. Determine the minimum value of *y* in mm.

A. 34.9 | C. 13.2 |

B. 26.8 | D. 19.5 |

3. Calculate the axial stress of member *AC* in MPa.

A. 1.26 | C. 1.57 |

B. 1.62 | D. 1.75 |

# Maximum Stress of Truss Member Due to Moving Loads

**Situation**

The bridge truss shown in the figure is to be subjected by uniform load of 10 kN/m and a point load of 30 kN, both are moving across the bottom chord

Calculate the following:

1. The maximum axial load on member *JK*.

A. 64.59 kN | C. -64.59 kN |

B. -63.51 kN | D. 63.51 kN |

2. The maximum axial load on member *BC*.

A. 47.63 kN | C. -47.63 kN |

B. -74.88 kN | D. 74.88 kN |

3. The maximum compression force and maximum tension force on member *CG*.

A. -48.11 kN and 16.36 kN |

B. Compression = 0; Tension = 16.36 kN |

C. -16.36 kN and 48.11 kN |

D. Compression = 48.11 kN; Tension = 0 |