4.4) The design shear strength D,V, is A) 81.1 kips B) 482 kips C) 529 kips D) 567 kips
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- For a beam subjected to a maximum bending moment of 100 kip-ft, design a A36 wide flanged section that will result in a factor of safety of 2 against yielding in normal stress. The emphasis is on the lowest weight section. The allowable shear stress of 24ksi is not exceeded. Question 3 of 4 Given your beam design, what is the ACTUAL maximum shear stress in the cross section? Vmax is 25 Kip 1.2 ksi 2.3 ksi 3.3 ksi 4.4 ksiA 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)…Problem 1. A W30 x 116 beams has the top flange restrained against lateral displacement and rotation and has an unstiffened web. A point load is applied to the top flange at 2m. From the left support. The beam spans 12 m. Use A36 steel with yield strength of Fy = 248 Mpa. 1. Compute the maximum value of point load that may be applied without web crippling occurring. Properties of W30 x 116 d = 762 mm; K = 1.28 mm; bf = 266.7 mm; tf = 21.59 mm; tw = 14.33 mm
- For the structure shown in Figure, with a uniformly distributed service dead load wD= 1.5 kips/ft and a uniformly distributed service live load wL= 2 kips/ft applied to beam BC, answer the following questions: 1. Find the minimum area of steel required to serve as a cross section for member AB. Assume welded connections and shear lag factor U = 0.75 2. Find the lightest standard rolled W section to adequately resist the bending moments in beam BC. Assume that the top flange is laterally braced at points B, C, and at midspan. Compute the value of the bending coefficient cb.A WT305 x41 standard steel shape is used to support the loads shown. Assume W= 50 kN/m, LAB = 1.1 m. LEc-2.2 m, and LCD = 1.1 m. Consider the entire 4.4 m length of the beam and determine: (a) the maximum tensile bending stress at any location along the beam (enter a positive answer), and (b) the maximum compressive bending stress at any location along the beam (enter a negative answer)A compression member adopts a section of 530UB82.0 as shown in Fig. 1. Its steel grade is Grade 350. The effective lengths of the member are Lex=7.6 m and Ley=3.8 m. Determine the maximum design load of this compression member.
- A steel beam having a simple span of 8 m is subjected to a moment M at the left end (Clockwise) and 25% of M at the right end (Counter clockwise). It has no lateral support and has the following properties: bf = 210 mm rt = 53 mmd = 533 mm, Sx = 2.06 x 20^6 mm^3tf = 16 mm Fy = 248 MPa a) Determine the allowable bending stress.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.A timber beam having a simple span of 6m carries a total load including its own weight of 15KN/m. It has a width of 250 mm and a depth of 300mm. Used dressed dimension by reducing its dimension by 10mm. The wooden section is made up of 80% Apitong.Fb = 16.5 MPaEw = 7310 MPaFv = 1.75Mpaa. Determine the maximum flexural stress of the beam.b. Determine the maximum shearing stress of the beam.c. Determine the maximum deflection of the beam.
- The beam in the figure carries a uniform load of 20 kN/m including its own weight, a concentrated load of 400 kN at 2.4 m from the left support and another concentrated load at 280 kNat 1.2 m from right end. The beam is laterally unsupported except at the two vertical ends. The beam is to be made of W 16 x 77. Use AISC Specifications for A36 steel Fy= 248.69 MPa. a. Determine the maximum bending moment of the beam. b. Determine the maximum or actual flexural stress of the beam. c. Determine the allowable flexural stress of the beam. d. Is it safe to use the said steel as beam? Why?A WT305 x 41 standard steel shape is used to support the loads shown. Assume w - 56kN/m, LAB - 1.0m, Lee - 2.0m, and Lco - 1.0m.Consider the entire 4.0 m length of the beam and determine:(a) the maximum tensile bending stress at any location along the beam (enter a positive answer), and(b) the maximum compressive bending stress at any location along the beam (enter a negative answer).The section of simply supported beam is HE450B, under distributed uniform dead load PG=20 kN/m and live load PQ=60 kN/m. Beam flange is laterally supported against compression from support and midpoint of the span. a. Determine safety of the section for bending (compare maximum beam moment with bending moment capacity of the section) b. Check section safety against shear Material is S235.