Calculate the maximum allowable load that a nominal 350mm x 350 mm sawn Ilumber column, 3.5m long, can carry. The applied loading is for 15-year duration and the wood used is visually graded sawn Yakal (E = 9.78 GPa, Fc = 15.8 MPa). The column is effectively held in position at both ends and is fixed for both ends. %3D
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- Calculate the maximum allowable load (in kN) that a nominal 350mm x 350 mm sawn lumber column, 3.5m long, can carry. The applied loading is for 15-year duration and the wood used is visually graded sawn Yakal (E = 9.78 GPa, Fc = 15.8 MPa). The column is effectively held in position at both ends and is fixed for both ends. Use k = 0.65. Blank 1Calculate the maximum allowable load that a nominal 203.2x 203.2 mm (8”x8”) sawn lumber column, 4.0m long, can carry. The applied loading is for 20-year duration and the wood used is visually graded sawn Apitong (E = 7.31 GPa, Fc = 9.56 MPa). The column is both pin at ends. Consider actual size of 7.25”x 7.25” and CD = 1.0.A beam will support a combination of dead, live, and earthquake loads. The Load Duration Factor for Dead is 0.9, Live is 1.0, and Earthquake is 1.6. The Dead load causes a stress of 300 psi, the live load causes a stress of 200 psi, and the earthquake causes a stress of 100 psi. The tabulated strength of the wood is 1000 psi. What is the critical (largest) Demand/Capacity ratio for the beam? (Demand/Capacity is fb/Fb'.)
- Compute for the Modulus of Elasticity of the Wood if the Clear Span Length is 2300 mm, applied load is 10lbs and the deflection of the wood is about 16.5mm. Moment of Inertia of the wood is 2.75x105 mm4 o 2.59 GPa o 2.49 GPa o 2.74 GPa o 2.63 GPaIf the allowable shear strength of the wood is 80 psi, determine the maximum load P that may be applied at midspan.Compute for the Modulus of Elasticity of the Wood if the Clear Span Length is 2300 mm, applied load is 10lbs and the deflection of the wood is about 16.5mm. Moment of Inertia of the wood is 2.75x105 mm4 a 2.63 GPa b 2.49 GPa c 2.59 GPa d 2.74 GPa
- The wood has an allowable normal stress of Oallow = 15 MPa and an allowable shear stress of Tallow = 1.33 MPa. (Figure 1) Part A Determine the minimum dimension h of the beam's cross section to safely support the load. Express your answer to three significant figures and include the appropriate units. µA ? h = Value Units Submit Request Answer Figure Provide Feedback 25 kN/m B 2 m 100 mmCalculate the maximum allowable load that a nominal 177.8 x 177.8 mm (7”x7”) sawn lumber column, 5.0m long, can carry. The applied loading is for 20-year duration and the wood used is visually graded sawn Apitong (E = 7.31 GPa, Fc = 9.56 MPa). The column is both pin at ends. Consider actual size of 6.5”x 6.5” (165.1 mm x 165.1 mm) and CD = 1.0Q1. Determine flexural strength for a wood specimen with given data below. V Span length = 22 inches , %3D Mc R = I L = span length, mm (in.) distance from neutral axis to edge of sample = ½h I = moment of inertia = bh³/12 b = average width, mm (in.) h V Cross section , b = 2" d= 2" C = where V Maximum load = 420O Ibf %3D modulus of rupture, MPa (psi) bending moment P = maximum load, N (lb) B = M = PL/4 average depth, mm (in.) || Q2. Determine shear strength for a wood specimen with given data below. Load block V a= 1.75 “ , b=0.25" , c = 2" , d= 2" V Maximum load = 510O Ibf Pmax Shear Strength || Area Test fixture Q3. Determine compressive strength for a wood specimen with given data below. V Cross section dimensions of the face on which load is applied = 2" x 2" V Maximum load = 4500 Ibf Load Pmax т Compressive Strength Area a * b
- A third-point bending test was performed on a 100 * 150 mm wood lumber according to ASTM D198 procedure with a span of 1.8 m and the 150 mm side is positioned vertically. A strain gauge is glued at the center point between the supports. The strain gauge is zeroed at the start of the test when the force is zero. A strain of 0.00456 is measured when the combined load is 6.672 kN and the load–strain curve is in the linear range. Determine the flexural modulus of elasticity. If the maximum combined load on both loading bearings was 22.722 kN, calculate the modulus of rupture.A short round wood column with an actual diameter of 250 mm is to be con-structed. If the failure stress is 29.6 MPa, what is the maximum load that can be applied to this column, using a factor of safety of 1.3?(NOS 2 – 13) PROBLEM 1. A timber beam shown in the figure is made up of Sasalit (50% stress grade), Use LRFD. All the given loads shown in the figure below are unfactored service loadings only, it does not include the weight of the beam itself. 2) What is the unit weight of the given wood, in KN/m3 ? 3) What is the unfactored weight of the beam itself in KN/m? 4) What is the actual reaction R1 in KN? 5) What is the actual reaction R2 in KN? 6) What is the distance in m. from the point of zero shear to the left support? 7) What is the maximum actual shear in KN? PDL = 5 KN %3D 500 mm PLL = 3 KN WD =6 KN/m WL=3 KN/m 800 mm R1 R2 2 m 1 m 1 m