e A fixed-end beam is subjected to a concentrated load (P) as shown in the figure. The beam has two different segments having different plastic moment capacities (M., 2M) as shown. P 21/3 Mp ← L -- The minimum value of load (P) at which the beam would collapse (ultimate load is) 7.5Mp (a) (b) 5.0Mp L L 4.5Mp 2.5Mp (d) L L (c) 2Mp *
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- A flooring system consist of I-beam sections spaced at 3 m(center to center) and with simple spans of 6 m. The beams support a 200 mm thick slab. The flooring system is designed for a live load of 2400 N/m² as well as ceiling load of 750 N/m². The properties of the I-beam sections are : d = 352 mm, weight of beam = 440 N/m, moment of inertia , Ix = 0.0012 m⁴. Consider Fy = 248 MPa, E = 200000 MPa and wt. of concrete of 24 kN/m³. 1. Which of the following most nearly gives the uniform pressure acting on the slab? a. 6.2 kPa b. 7.95 kPa c. 5.2 kPa d. 6.10 kPa 2. Which of the following most nearly gives the total uniform load on the beam in kN/m? a. 19.71 kN/m b. 17.82 kN/m c. 24.29 kN/m d. 28.14 kN/m 3. Which of the following most nearly gives the initial stress on the beam due to deadload only assuming no shoring during construction? a. 12.1 MPa b. 8.91 MPa c. 10.4 MPa…The cantilever beam consists of a rectangular structural steel tube shape [E = 200 GPa; I = 450 × 106 mm4]. Assume P = 50 kN, w = 10 kN/m, LAB = 1.00 m, and LBC = 2.00 m. For the loading shown, determine:(a) the beam deflection at point A.(b) the beam deflection at point B.Determine the component of the beam deflection at point A due to only the uniformly distributed load of w = 10 kN/m.Answer: vA,w = mm. Part 2 Determine the component of the beam deflection at point B due to only the concentrated load of P = 50 kN.Answer: vB,P = mm. Part 3 Determine the component of the rotation angle of the beam at point B due to only the concentrated load of P = 50 kN.Answer: θB,P= rad. Part 4 Determine the component of the beam deflection at point A due to only the concentrated load of P = 50 kN.Answer: vA,P = mm. Part 5 Determine the total beam deflection at point A due to the combined effects of the concentrated…The following tubular A36 steel shaft has an inner diameter of 25mm and an outer diameter of 35mm under the following loading condition. Determine the angle of twist of the end V. Consider: GA-36 = 75 GPa 1) What would happen if a solid shaft was used? 2) As a designer, do you recommend a solid or tubular shaft for the loading above? Why?
- Design the beam to carry the load as shown in Figure. Use rho max with a strain in the tension bars equal to 0.004 and in the design process use 60% of this rho max. f’c= 21MPa and fy=276MPa What is the Mu max that is to be used in the design? Determine the required base and effective depth of the beam. Round up your answers to the nearest 10mm and use 2b=d Determine the most economical number of 20mm diameter bars required. Using the b and d from letter (b) determine the balanced As for the beam.A concrete circular column with a radius of 350mm has f'c=25 MPa and is braced against sidesway. The column is assumed to be pinned on both ends with an unsupported length is 6.1meters. The column is subjected to the following service loads: Axial Dead Load =350 kN Axial Live Load=480 kN Moment due to Dead Load on the top of the column = 125 kN-m Moment due to Dead Load on the bottom of the column = 60 kN-m Moment due to Live Load on the top of the column = 122 kN-m Moment due to Live Load on the bottom of the column = 75 kN-m. If the moments bent the column in a singular curve, determine the following: 1. Critical Load on the column, kN 2. Moment Magnifying factor 3. Design Moment of the column, kN-mA W16X31 A992 steel is used as a beam. The beam has a continuous lateral support. The uniformly distributed load shown in the figure is a service load that consists of 50% dead load (included the beam weight) and 50% live load. Using LRFD, determine the following? Note: A992 STEEL Fy = 50 ksi Ag = 9.13 in^2 Zx = 54.0 in^3 What is the Ultimate (Governing Load Combo) Maximum Positive and Negative Moment (at midspan) in kips-ft
- 1. A column 9 meters long is to carry a total service load (unfactored) of 890 kN. The member will behinged at the top and bottom and will be braced about its minor axis at mid-height. Using A572 Gr. 50steel., design the most economical section based on the following available section. Provide an explanationwhy your chosen section is the most economical. Neglect local buckling. Hint: Calculate all thecapacity and compare the D/C ratio.W10x49 - Area = 9290 mm2,rx = 110.49 mm, ry = 64.52 mmW12 x 50 - Area = 9484 mm2,rx = 131.57 mm, ry = 49.78 mmW8 x 40 - Area = 7612 mm2,rx = 89.66 mm, ry = 51.82 mmW8 x 48 - Area = 9097 mm2,rx = 91.69 mm, ry = 52.83 mmCompute the value of EIy at midspan for the beam loaded as shown.A single-span beam having unsupported length of 8m. has a cross section of 200mm x 350mm. (use nominal dimension). It carries a uniformly distributed load “W” kN/m throughout its span. Allowable bending stress is Fb=9.6 MPa and a modulus of elasticity of 13800 MPa. From the table, the effective length Le=1.92 Lu where Lu=unsupported length of beam. a. Compute the allowable bending stress with the size factor adjustment in MPA Round your answer to 3 decimal places. b. Compute the allowable bending stress with lateral stability adjustment in MPa Round your answer to 3 decimal places. c. Compute the safe uniform load “W” that the beam could carry in KN/m. (choose the smallest of prob. a and b.) use M=wl^2 / 8 Round your answer to 3 decimal places.
- QI/ A beam carries the loads shown in figure, if the tensile stress must not exceed 20 MPa and the compression stress 70MPa. Find the maximum value of load P Som, r W=2-5p W/m - P Lflj Ko ; - 3,‘, ) o Jo 1 er -1 /i e,A steel beam having a span “L” of 7 m is simply supported at its ends and is laterallyunrestrained over its whole length. It supports a uniformly distributed permanent action(dead load) (G) of 8 kN/m over its whole length, and a point load of 20 kN acting as an imposedaction (live load) (Q) at midspan.Design information: • Load combinations: Strength limit state: WL = 1.2G + 1.5QServiceability limit state: WS = G + Q• Maximum deflection under serviceability conditionsis not to exceed L/400• Effective length factors: kt = 1.0; kℓ = 1.4; kr = 1.0• Elastic modulus for steel Esteel = 200 GPa• Self-weight of the beam can be disregarded in all calculations(a) Sketch the load diagram showing the loads on the beam for the strength limit state, andcalculate the design moment (maximum bending moment) and the design shear force(maximum shear force)(b) Calculate the effective length (Le) of the beam from Le = kt×kℓ×kr×L and select from theTables provided the lightest suitable Universal Beam (UB)…The rotating shaft running at n = 900 rpm, shown in the figure below, is machined from AISI 1045 CD steel. It is subjected to a force of F = 10.1 kN. The shaft is experiencing an operating temperature of 35 oC. With the specified loading and for the reliability of 99.9 %, Determine the minimum factor of safety for fatigue based on infinite life*. Determine the maximum safe load (Fmax) that can be applied for a factor of safety of 1.5 and a design life of 5x105 cycles, all other input values remaining the same. Determine the factor of safety against yielding.