DESIGN A SINGLE ANGLE MEMBER FOR THE TRUSS WHOSE TOP CHORD IS 36Kn compression, BOTTOM CHORD IS 30 KN tension, VERTICAL MEMBER IS 25 KN tension and DIAGONAL MEMBER IS 20KN compression IF A36 STEEL IS USE WITH Fy 250 MPA AND Fu = 450 MPA. %3D
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- 2) Find the axial stresses of menbers FD, GD, GE State if it is tensile or Compressive. 4M 3M A LE 3m G 20 RN Go KNdetermine the member stresses of the truss shown. indicate if the stress in the member is in tension or compression. present complete and neat solutions. p1-30kN W-25kN/m P2=51kN AH Av Im P₂ Im P₂ Im P₂ Im Av W 1.6mA built-up section shown, 30 feet long, is fixed at both ends and braced in the weak direction at the midheight. A992 steel is used. Determine the following: 11. Ix 12. Ky 13. effective slenderness ratio 14. Fe 15. LRFD design strength 16. ASD allowable strength 18 4 W10 × 49 A 14.4 in² bf 10 in d. 10 in tw 0.340 in tf 0.560 in Ix= 272 in¹ Iy=93 4in¹ W10×54 A 15-3 in² bf 10 in d = 10 lin tw0-370 in tf 0.615 in Ix= 303 in 4 Ty = 103 in 1
- From the Given Truss Shown: P= 31 KN M=128 KN Calculate the Force Carried by member Gl. If the member is compression put negative (-) sign. PKNJ h=8m B A PKN D PKN MkN MKN MKN PKN I H -6 panels @ 5 m 30 m PKNDesign the reinforcements of the given T beam below. bf=800mm bw=450mm tf=120mm d=600mm d'=80mm fc'=35MPa fy=350MPa USE NSCP 2015 A.Mu = 1300kN-m, As = B.Mu = 1600kN-m, As = mm2 _mm2, As' = mm2W3D 25 Y= 155 %31 1oow= 2500 KN 27=310mm X= 475 Z= 60 %3D 102=600mm
- Check the slenderness of the steel section against Local Buckling. 340 mm Use Fy = 800 MPa. O Compact Slender I O Adequate 400 mm None of the choices 30 mm 20 mm 20 mmQuestion # 1: Determine the force in each member of the truss and state if the members are in tension or compression. 300 lb 600 15 B 60015 GOO G 11. 244in. 6 ft load at point B, D, F = 600 lb load at pount A and H=300 lb distance AC = CE =EG = GH = 8FT -80---SA- [Answer: FAB=15001b-C. Fac-12001b-T, Fac0, For-12001b-T, Fan+ 12001b-C, F₁-601b-C. Form 721h-T, Use symmetry to obtain the rest of the members]Consider the truss shown in (Egure 1). Suppose that F-60 kN and F₂-25 kN Figure 40 KN Part C Fau 50 AN Submit -2m- D 15m 1of1 > 15m 15m μA Value -F₂ Part A Request Answer Determine the force in member ED of the truss, and state if the member is in tension or compression Express your answer to three significant figures and include the appropriate units. Enter negative value in the case of compression and positive value in the case of tension. HA FED Value Submit Part B Request A Determine the force in member EH of the truss, and state if the member is in tension or compression Express your answer to three significant figures and include the appropriate units. Enter negative value in the case of compression and positive value in the case of tension. Submit Fon- Value Determine the force in member GH of the truss, and state if the member is in tension or compression. Express your answer to three significant figures and include the appropriate units. Enter negative value in the case of…
- Use 2015 NSCP. A compression member shown below with Fy = 50 ksi 30′ W12 x 87 A992 steel 1. Which of the following most nearly gives the value of critical buckling stress in ksi? 2. Which of the following most nearly gives the nominal strength in kips? 3. Considering ASD. Which of the following most nearly gives the allowable strength in kips? 4. Considering LRFD. Which of the following most nearly gives the design strength in kips?1.20 Calculate the truss's smallest allowable cross-sectional areas of members CE, BE, and EF. In tension, the working stresses are 20 ksi, while in compression, they are 14 ksi. (In compression, the working stress is lower to lessen the risk of buckling.) 15Kips 15Kips 4ft Vc 4ft VD E 9ft G.A column is built up from four (4)- 125 x 125 x 18 angle shapes as shown. The plates are not continuous but are spaced at intervals along the column length and function to maintain the separation of the angles. They do not contribute to the cross-sectional properties. The effective length is 4 m. Compute the allowable design compressive strength based on flexural buckling. E= 250 MPa. Use ASD. k 375 mm 125mm, HPlate 125mm 4 - 4 125 × 125× l8 section 下好业