Question 1. A single equal angle section of Grade 350 steel is to be used in a truss similar to the one shown in Figure 1. It is to be connected at the ends by welding one leg of the angle to a plate. Select a suitable section from the Liberty Steel Booklet for a factored design axial tension force N* of 105 kN. Figure 1
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- Calculate the nominal flexural and shear strength of an HSS 8×4×1/8 section. ? y =50 ksi . Please use hand calculations to solveA built-up beam made from fastening two channels to a wide flange section carriesloads` as shown. The fasteners are 12mm∅ bolts with allowable shear stress τ= 110 MPa. Forbearing, σb= 220MPa for single shear, and σb= 275 MPa for double shear. E=200 GPA. Fy=420MPa. Do not round off between solutions Determine the flexural stress at fiber 30 mm above Neutral Axis at distance 5 metersfrom the support at A.The cantilever beam is subjected to the load shown Below. Determine 1. the actual stresses acting at point A. (Hint: Tmas TIN EA). 2. IfG =30 GPa find the angle of twist at the free end of beam. 10 KN
- A wooden beam is fabricated from one 2 × 10 and two 2 × 10 dimension lumber boards to form the double-tee cross section shown. The beam flange is fastened to the stem with nails. Each nail can safely transmit a force of 212 lb in direct shear. The allowable shear stress of the wood is 72 psi. Assume b1 = 10 in., b2 = 2 in., d1 = 2 in., d2 = 10 in.(a) If the nails are uniformly spaced at an interval of s = 3.4 in. along the span, what is the maximum internal shear force V that can be supported by the double-tee cross section?(b) What nail spacing s would be necessary to develop the full strength of the double-tee shape in shear? (Full strength means that the maximum horizontal shear stress in the double-tee shape equals the allowable shear stress of the wood.)In the truss loaded and supported as shown, computethe following using the Method of sections.(a) JK = ____________(b) KD = ____________(c) DE = ____________Topic : Uncracked and cracked section 1. A beam has a width of 300mm and effective depth of 540mm, d’ = 60mm. It is reinforced with 4 – 25mm diam tension bars and 2 – 25mm compression bars. f'c = 28mpa and fy= 420. What is the ultimate moment capacity of the beam?
- The cantilever beam shown below has the cross-section on the right portion. If the allowable stresses are 62 MPa shear and 103 MPa tension at point A (just below the flange), determine the maximum allowable load P.Analyze the beam section shown below reinforced with 4 - 16mm bars. Use f'c=21 MPa and fy=415 MPa. a) Determine the ultimate moment capacity of the beam in kN-m b) Determine the required area of reinforcing steel at balanced condition, in mm^2 c) If the given section is reinforced with 10-16mm bars, determine its ultimate moment capacity in kN-mA built-up beam made from fastening two channels to a wide flange section carriesloads` as shown. The fasteners are 12mm∅ bolts with allowable shear stress τ= 110 MPa. Forbearing, σb= 220MPa for single shear, and σb= 275 MPa for double shear. E=200 GPA. Fy=420MPa. Determine the maximum bending stress on the section.Determine the maximum shearing stress at Neutral Axis.
- Problem 2A beam with a rectangular section has a width of 250 mm and a height of 575 mm. It is reinforced with 4∅32distributed in two layers. Use fc’ = 30 MPa and fy = 400 MPa1. Check if this section will have a ductile or brittle behavior. Explain why2. Calculate the depth of the neutral axis.3. Find the strain in steel bars and strength reduction factor.4. Find the nominal moment capacity Mn.5. Is this section appropriate from the design point of view? And why?Design a laterally unsupported beam for the following data. Effective span: 4.5 m Maximum bending moment: 650 kN-m Maximum shear force: 300 kN Steel of grade: Fe410 Find/do the following: a) Section classification b) Check for design bending strength c) Check for shear capacity of section d) Web buckling check e) Web-bearing checkOOO OOOO OOOO SITUATION 2: A W 16 x 58 beam is connected to W 18 x 31 column. A36 steel has an Fy = 248 MPa. Allowable bearing stress is 1.35Fy. Using a 502 grade hot driven rivets, with allowable shearing stress of 120 MPa. The support is to be designed using the full strength 2290x90x10 of W16 x 58 beam based on the gross section. Properties of section: W 16 x 58 W 18 x 31 fv = 0.40 Fy te = = 11 mm d = 403 mm tw = 10 mm W18x31 W16x58 4. Which of the following most nearly gives the minimum diameter of rivets so that allowable shear stress will not be exceeded? 23 mm o b. 25 mm 0 28 mm d. 30 mm 0 5. Which of the following most nearly gives the minimum diameter of rivets not to exceed allowable bearing stress? 23 mm 0 b. 25 mm o 28 mm • d. 30 mm 0 6. Which of the following most nearly gives the minimum diameter of rivets? 23 mm o b. 25 mm o 28 mm • d. 30 mm 0 a. C. a. C. a. C.