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'.)
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- Since the safe normal stress of the wood used for the beam and loading condition shown in the figure is 12 MPa, determine the optimum cross-section height of the beam (mm).SHOW COMPLETE SOLUTION AND ANSWER IN 3 DECIMAL PLACES The piece of wood ,100mm by 100mm in cross section, contains a glued joint inclined at the angle Ø with the vertical. The working stresses are 20 MPa for wood in tension . 8 MPa for glue in tension and 12 MPa for glue in shear . If Ø =50 , Determine the largest allowable axial force P.A wood specimen was subjected to bending until failure by applying a load inthe middle of its span. The specimen has a cross section of 25 mm * 25 mm(actual dimensions) and a span of 350 mm between the simple supports. Theload and the deflection in the middle of the span were recorded as shown inTable P10.22.a. Using a computer spreadsheet program, plot the load–deflectionrelationship.b. Plot the proportional limit on the graph.c. Calculate the modulus of rupture (flexure strength).
- A simply supported wood beam with a span of L = 15 ft supports a uniformly distributed load of w = 320 lb/ft. The allowable bending stress of the wood is 1,200 psi. If the aspect ratio of the solid rectangular wood beam is specified as h/b = 2.0, calculate the following: 4. Calculate the required section modulus S considering the allowable bending stress of the wood in in^3. 5. Calculate the minimum allowable beam width in inches.Compute the modulus of elasticity of the wood species whose stress–strainrelationship is shown in Figure 10.12, using both the SI and English units.Compare the results with the typical values shown in Table 1.1 in Chapter 1and comment about the results.A wood specimen was subjected to bending until failure by applying a loadin the middle of its span. The specimen has a cross section of 1 in. * 1 in.(actual dimensions) and a span of 14 in. between the simple supports. Theload and the deflection in the middle of the span were recorded as shown inTable P10.22.a. Using a computer spreadsheet program, plot the load–deflectionrelationship.b. Plot the proportional limit on the graph.c. Calculate the modulus of rupture (flexure strength).
- Two steel plates, each 4 in. wide and 0.25 in. thick, reinforce a wood beam that is 3 in. wide and 8 in. deep. The steel plates are attached to the vertical sides of the wood in a position such that the composite shape is symmetric about the z-axis, as shown in the figure. Determine the maximum bending stress produced in both the wood and the steel if a bending moment of Mz = +50 kip-in is applied about the z-axis. Assume Ewood = 2000 ksi and Esteel = 30000 ksi.if the stress at failure and bending is 7200 PSI for small, perfect samples of a certain species of wood, what is the allowable bending stress for a piece of lumber with defects that are judged to reduce the strength by 40%?UPVOTE will be given! Please write the solutions completely and legiby. Box the final answer. Answer in 3 decimal places! Strength of Materials The assembly shown is made up of T-shape steel and two wood as shown. A bending moment M is applied to the composite beam. Given: Es = 200 GPa Ew = 25 GPa M = 90 kN.M h = 200 mm a. Calculate the distance of the centroid from the bottom of the beam in mm. b. Calculate the absolute maximum stress in the wood in MPa.
- A wood specimen was prepared with actual dimensions of 25 mm * 25 mm *150 mm and grain parallel to its length. Displacement was measured over a100 mm gauge length. The specimen was subjected to compression parallel to the grain to failure. The load–deformation results are as shown in Table P10.24. a. Using a computer spreadsheet program, plot the stress–strain relationship.b. Calculate the modulus of elasticity.c. What is the failure stress?A timber column 6.5 m long and is laterally supported at 3 m from the bottom carries an axial load of 250 kN. If the allowable compressive stress parallel to the grain is 13.7 MPa, Design a column section by selecting the correct answer from the following questions. Use Forest Laboratory Formula. Modulus of Elasticity, E of the wood is 13,800 MPa. Considering the top part of the column for L1/D1 the ratio is equal to A. 12.5 B. 15.5 C. 16.5 D. 17.5 Considering the full length of the column for L2/D2, the ratio is equal to A. 25.35 B. 22.22 C. 27.30 D. 32.5 Comparing the larger L/D to K, the column should be considered a A. short colum B. intermediate column C. long column D. none of the aboveCalculate the modulus of elasticity of the wood specimen with the stress-strain relationshipshown in figure below.