Check whether a welded W-section (d=500mm, bf=250mm, tf=10mm, tw=8 mm) is compact, non-compact or slender using Fy = 248 MPa.
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Check whether a welded W-section (d=500mm, bf=250mm, tf=10mm, tw=8 mm) is compact, non-compact or slender using Fy = 248 MPa.
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- The angle L 8 x 8 x ½ in tension shown in the figure below must resist aservice dead load of 35 kips, live load of 50 kips, and a roof live load of 2kips. The steel used is A50 (Fy = 50 ksi, Fu = 65 ksi) and its welding length is 4”.Determine if the member has sufficient fracture and yield strength.A simply supported beam shown below is HEA 400 of 5355 steel. It is laterally supported at points with an "X" The material properties and the section properties are given below. Use ASD load combination DL+LL and determine whether the beam has adequate flexural strength. Material: $355 Fy= 355 MPal Fu= 510 MPa E= 200000 MPa Section properties: A: 15900 mm² d: 390 mm h: 298 mm b: 300mm tr: 19 mm W 2311x10³ mm³ Iy: 8564x10*mm* A 4m t: 11 mm. İy: 73.4 mm W: 2563x10³ mm³ J: 189x104 mm DL: 150 KN LL: 100 KN 17 B 4m Cw: 2942x10⁹ mm C E: 200000 MPaInstruction: Answer the questions below: (CLO) A 20mm B Take E= 105 GN/m² 20mm C 20 kN A brass composite bar AB and BC in diameter 30mm and 15 mm respectively was installed rigidly as shown in figure below. Find the total elongation of the composite bar.
- TENSION MEMBERS: THE SINGLE 200 X 10 mm STEEL PLATE IS CONNECTED TO A 12 mm THICK STEEL PLATE BY FOUR 16 mm DIAMETER RIVETS AS SHOWN IN THE FIGURE. THE RIVETS USED ARE A502 GRADE 2, HOT DRIVEN RIVETS. THE STEEL IS ASTM A36 WITH Fy = 248 MPa AND Fu = 400 MPa. DETERMINE THE VALUE OF P. a. P BASED ON TENSION OF GROSS AREA b. P BASED ON TENSION OF NET AREA c. P BASED ON BEARING OF PROJECTED AREA d. P BASED ON SHEAR RUPTURE (BLOCK SHEAR)2. A steel plate is 360 mm wide and 20 mm thick with four bolts hole into the place as shown in the figure. Compute the following: a. Critical net area required by the NSCP specs. b. Max. critical net area required by the NSP specs. Note that, Max. net area is 85% of the critical net area c. Capacity of the joint if the allowable tensile stress is 0.75Fy. Use A36 steel Fy=248 MPa 90m²m 90 mm 90mm 45 45mm In my comin P Scanned with CamScanner CStf Sec Ag tw W16x100 29.4 0.585 0.985 Bolt (7/8 in) A structural analysis was done for -1 the steel section whose shape and characteristics are shown below, and the outputs were arranged in the table below. Choose the appropriate amount, knowing that .all numbers are in units (ksi, in) O fy ( 36 fu 0.85 20.149 584.3 635.04 58 D(bolt) 7/8 P Maximum tensile force (P) that can be exerted Effective cross- sectional area (Ae) The value of T1 Modulus M (mo) !
- SITUATION 14: A PL 300 x 20 mm is to be connected to two plates of the same material with half the thickness by 25 mm Ø rivets as shown in the figure. The rivet holes have a diameter 2 mm larger than the rivet diameter. The plate is A36 steel with Fy= 250 MPa, allowable tensile stress of 0.60Fy and allowable bearing stress of 1.35Fy. The rivets are A502, Grade 2, hotdriven rivets with allowable shear stress of 150 MPa. 25 mm Ø rivets + + PL 300x20 P/2 P P/2 PL 300x10 43. Which of the following most nearly gives the max. load in kN that can be applied to the connection without exceeding the allowable shear stress in the rivets? a. 675 KN b. 490 KN c. 598 KN d. 790.5 kN 44. Which of the following most nearly gives the max. load in kN that can be applied to the connection without exceeding the allowable bearing stress between the plates and the rivets? a. 490 KN b. 675 KN c. 598 KN d. 790.5 kN 45. Which of the following most nearly gives the max. load in kN that can be applied to the…A 15" x 3/8" bar of A572 Gr. 50 steel is used as a tension member. It is connected to a gusset plate with 7/8-in diameter bolts as shown in the figure. Use s = 2.0 and g = 3.0. Determine the design tensile strength of the section based on yielding of the gross area. Answer: |0 O O О оDetermine whether the D = 560 kips L = 68 kips compression member shown is adequate to support the given service loads. Take note Pu = 1.4D. 20' W12 × 79 K = 0.80, r = 3.05 in E = 29000 ksi, Fy = 50 ksi %3D A992 steel Input Yes or No for your final answer. Blank 1 Blank 1 Add your answer
- 6 6a 6b 6c Alaterally supported beam was designed for flexure. The beam is safe for shear & deflection. The most economical section is W 6 x 20 however the said section is not readily available at the time of the construction. If you are the engineer in charge of the construction what alternative section will be the best replacement? Why? Use: Fy=248 MPa: E=200,000 MPa Designation W8 x 24 W8 x 21 Weight Ag (mm2) 3787 36 31 width mm kg/m d (depth) 30 157.48 152.91 9.27 W6 x 20 expla'n briefly your cho'ce. (transform your comparative analys's 'nto a narative form to support your cho'ce) W8 x 28 42 204.72 165.99 11.81 5323 4568 201.42 164.97 10.16 3974 210.31 133.86 flange bf thickness tf Web 10.16 thickness tw 6.60 Elastic Properties mm 4 Ix x 106 7.24 6.22 6.35 17 41 34 31 mm 3 Sxx 103 220 398 342 298 mm rx 67.56 87.63 86.87 88.65 mm 4 lyx 106 6 9 8 4 mm 3 Syx 106 72 109 92 61 mm ry 38.10244 41.15 40.89 32 Plastic properties. mm 3 Zx x 103 244 446 380 334 mm 3 Zyx 103 110 166 140 93…4. Calculate the design strength (ocPn) of W24X76 with length of 12 ft. and pinned ends. A572 Grade50 steel is used. E=29x103 ksi. Show your work in detail. ASTM Classification A36 A572 Grade 50 A992 Grade 50 A500 Grade B (HSS rect, sq) A500 Grade B (HSS round) A53 Grade B Yield Strength F, (ksi) 36 50 50 46 42 35 Ultimate Strength F (ksi) 58 65 65 58 58 60Topic: COMBINED STRESS-AXIAL TENSION AND FLEXURE BENDING: STEEL DESIGN Please solve your Solution in a handwritten Note: It should be handwritten pleaseeee Questions A Tension member with no holes is subjected to axial loads of PD=68kN & PL=64kN.It is also subjected with bending moments of MDy=40kN-m & MLy=55kN-m. Is themember adequate? Steel is of A992 Gr 50 Specs. Use LRFD. Neglect the weight of the beam. Section properties:Lp = 1.863 md=459.99mmtw=9.02mmzx= 1835x10^3 mm^4Lr = 5.305 mLb = 4.8 mbf=191.26mmtf=16mmSx=1611x10^3 mm^4 Sy=195x10^3 mm^4Zy=303x 10^3 mm^4 Cb=1.32