Example 01: Safe Uniform Load for a Beam that was Notched at the Tension Fibers at Supports

Problem
A 75 mm × 150 mm beam carries a uniform load wo over the entire span of 1.2 m. Square notches 25 mm deep are provided at the bottom of the beam at the supports. Calculate the safe value of wo based on shear alone.

Allowable shear parallel to grain = 1.40 MPa
Allowable shear normal to grain = 1.85 MPa
notched-beam-001-uniform-load.gif

 

Example 02: Maximum Concentrated Load a Box Beam Can Carry

Problem
A beam is built up by nailing together 25 mm thick planks to form a 200 mm × 250 mm box section as shown. The nails are spaced 125 mm apart and each can carry a shearing force of up to 1.3 kN. The beam is simply supported for a span of 3.6 m and to carry a concentrated load P at the third point of the span. The allowable shearing stress of the section is 0.827 MPa.
 

spacing-of-bolts-002-box-beam-cross-section.gif

 

  1. Determine the largest value of P that will not exceed the allowable shearing stress of the beam or the allowable shearing force of the nails.
  2. What is the maximum flexural stress of the beam for the load P computed in Part (1)?

 

Example 01: Spacing of Screws in Box Beam made from Rectangular Wood

Problem
A concentrated load P is carried at midspan by a simply supported 4-m span beam. The beam is made of 40-mm by 150-mm timber screwed together, as shown. The maximum flexural stress developed is 8.3 MPa and each screw can resist 890 N of shear force.
 

spacing-of-bolts-001-box-beam-cross-section.gif

 

  1.   Determine the spacing of screws at A.
  2.   Determine the spacing of screws at B.

 

Example 04: Required Depth of Rectangular Timber Beam Based on Allowable Bending, Shear, and Deflection

Problem
A beam 100 mm wide is to be loaded with 3 kN concentrated loads spaced uniformly at 0.40 m on centers throughout the 5 m span. The following data are given:

Allowable bending stress = 24 MPa
Allowable shear stress = 1.24 MPa
Allowable deflection = 1/240 of span
Modulus of elasticity = 18,600 MPa
Weight of wood = 8 kN/m3
  1. Find the depth d considering bending stress only.
  2. Determine the depth d considering shear stress only.
  3. Calculate the depth d considering deflection only.

 

beam-003-required-depth.gif

 

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.
 

beam-002-wood-reinforced-steel-aluminum.gif

 

Given the following material properties:

Allowable Bending Stress, Fb
Steel = 120 MPa
Aluminum = 80 MPa
Wood = 10 MPa
Modulus of Elasticity, E
Steel = 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.
 

beam-001-circular-log-shear-stress.gif

 

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