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- The tensile strength of blue gum timber is 50 MPa at a moisture content of 12 percent. If the strength determined in its green state was 42 MPa, and its fibre saturation point occurs at a moisture content of 25 °C, find the strength of this timber at a moisture content of 8 percent. If the density of the wood was 1.4 g/cm3 atmoisture content of 12 percent, find the density at a moisture content of 8 percent.A wood pole with a diameter of 10 in. has a moisture content of 5%. The fibersaturation point (FSP) for this wood is 30%. The wood shrinks or swells 1%(relative to the green dimensions) in the radial direction for every 5% changein moisture content below FSP.a. What would be the percent change in the wood’s diameter if the wood’smoisture is increased to 55%?b. Would the wood swell or shrink?c. What would be the new diameter?(a) Discuss the visual strength grading of timber. In your solution you should refer to the key,parameters in the process. (b) Describe in detail the methods available for drying timber.
- The moisture content of wood test was performed according to ASTM D4442procedure and produced the following data:Weight of specimen in the green condition = 317.5 gWeight of oven-dry specimen = 203.9 gCalculate the moisture content of the given wood.A wood element had 150 mm diameter (radial direction) and a length of 1 m. The moisture content of the wood is 130% when it was prepared. After seasoning, the moisture content was reduced to 5%. Find the diameter and length of the seasoned element if the moisture- shrinkage relation follows the figure below.(Hint: you should use the curve for radial shrinkage for diameter and longitudinal shrinkage for length change in figure, FSP=28% according to the figure)A pine wood specimen was prepared with actual dimensions of 50 mmx50 mm x 250 mm and grain parallel to its length. The deformation was measured over a gauge length of 200 mm. The specimen was subjected to compression parallel to the grain failure. The load- deformation results are shown in table below: Load (kN) Deformation (mm) 8.9 0.457 17.8 0.597 26.7 0.724 34.5 0.838 43 0.965 52.9 1.118 62.01 1.27 71.03 1.422 80.1 1.588 89 1.765 99.6 1.956 108 2.159 111.3 2.311 112 2.456 a. Using a computer spreadsheet program, plot the stress-strain relationship. b. Calculate the modulus of elasticity. С. What is the failure stress?
- A wood with a dimension 53 mm and 131 mm with 131 mm is side positioned vertically is subjected to ASTM D198 Test the length of the sample is 51 cm. If the modulus of rupture is limited to 100 mpa, using the following method determine if the sample passed in the test. Value of P is 134 kN. a. two-point b. third point c. center-point loadingplease answer these two question 1.Explain the reason that flexural modulus of elasticity and compression test modulus of elasticity must be considered separately when both the flexural test and the parallel to grain compressive test caused normal stresses parallel to the grain of the wood. 2. Discuss the difference between isotropic and orthotropic materials for wood.The moisture content of wood test was performed according to ASTM D4442procedure and produced the following data:Mass of specimen in the green condition = 317.5 gMass of oven-dry specimen = 203.9 gCalculate the moisture content of the given wood.
- 10.6 The moisture content of wood test was performed according to ASTM D4442 procedure and produced the following data: Weight of specimen in the green condition = 317.5 g Weight of oven-dry specimen Calculate the moisture content of the given wood. 203.9 g %3DWhen dimension lumber is used where the moisture content exceeds 19%, what is the applicable wet service factor, CM, for tension stress? 0.8 1.0 0.85 0.9The moisture content of wood test was performed according to AST D4442 procedure and produced the following data: Weight of specimen in the green condition = 317.5 g Weight of oven-dry specimen = 203.9 g Calculate the moisture content of the given wood.