A built-up section was made using PL414x12mm thk plates as shown in the figure below. It is pinned at both ends with additional support against weak axis at middle point. Assume A50 steel. PL414x12 08 Section W16x67 L x-axis Elevation www ANANAN y-axis
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- 2- A 2m COPPER BAR SUBJECTEO TO A TENSILE IS SUSPENDED FROM A LOAD IBO KN SUPPORTED BY TWO IDENTICAL PIN THAT S pOSIS AS SHON OF THE LOWER AFPLIED TENSILE POSTS SUPPORT THE STEEL DETERMINE THE TOTAL DEFORMATION END OF THE LOAD HINT: NOTE THAT THE STEEL PIN WHERE THE COFPER BAR IS ATTACHED. COPPER BAR DUE TO STEEL POST L=0.5m A= 4500mm E- 200GPA COPFER BAR L- am A= 4800 Mn E- l20 GPa 180KN2) 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 KNA 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
- 4 A flanged bolt coupling has ten 12-mm di ameter steel bolts on 500 mm diameter b olt circle and six 16 mm diameter aluminu m bolts on 300 mm diameter bolt circle. T he maximum shear stresses of the materi als are 60 MPa in steel and 40 MPa in al uminum. Use G = 80 GPa for steel and 3 0 GPa for aluminum. What is the maximu m required shear strength of each alumin um bolt to determine the maximum torqu e that can be applied to the system? Dra wing not included in this problem.A flexural member is fabricated from two flange plates 1/2x16 and a web plate 1/4x20, thusforming a built-up ‘I’ shape member. The yield stress of the steel is 50 ksi.a. Compute the plastic section modulus Z and the plastic moment Mp with respect to the majoraxis.b. Compute the section modulus S and the plastic moment My with respect to the major axisA single unequal angle 100 x 75 x 6 is connected to a 10 mm thick gusset plate at the ends with six 16 mm diameter bolts to the 100 mm of leg to transfer tension as shown in figure. What will be the block shear strength (in kN) of the angle section assuming that the yield and the ultimate stress of steel used are 250 MPa and 410 MPa respectively? ISA (100 × 75 x 6) -T 16 mm o bolt 10 mm 40 mm 40 mm| 40 mm 40 mm 40 mm 40 mm 10 mm
- A PL40 mm X 250 mm (smaller member) is connected to a gusset plate (bigger member) as shown. The diameter of the holes are 25 mm. The pitch and gage of the holes are 50 mm and 75 mm, respectively. The yield strength of the steel is 260 MPa while the ultimate tensile strength of the steel is 400 MPa. Determine the design (LRFD) tensile strength of the tension member in kN. Neglect block shear. H G P P D B F C A EThe steel tie bar shown is to be designed to carry a tension force of magnitude P = 120KN when bolted between double brackets at A and B. The bar will be fabricated from 20-mm-thick plate stock. For the grade of steel to be used, the maximum allowable stresses are: o = 180 MPa, t = 120 MPa, op = 380 MPa. Design the tie bar by determining the required values of: a. the diameter d of the bolt, b. the dimension b at each end of the bar, c. the dimension h of the bar.Design 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' = mm2
- Situation 5. The angular section shown below is welded to a 12 mm gusset plate. Both materials are A36 steel with Fy = 250 MPa. The allowable tensile stress is 0.6Fy. The weld is E80 Electrode and 12 mm thickness. INNOVATIONS Properties of L 150x90x12: y = 50 shear stress of weld = 0.3Fu A = 2750 Allowable REVIEW INNOVATIE a K ➜ www A. 234 KN B. 349 KN b 13. What is the value of P without exceeding the allowable tensile the angle? C. 382 kN p. 413 kN 14. Find required length of the weld based on shear? A. 280 mm C. 300 mm D. 380 mm B. 320 mm 15. Find the required value of a? A. 108 mm B. 97.9 mm D. 185 mm NEW INNOVATIONS REVIEW INNOVATIOf REVIEW NEW INNOVATIONSA flexural member is fabricated from two flange plates 1/2in x 16in and a web plate 1/4in x 20in. The yield stress of the steel is 50 ksi. a. Compute the elastic section modulus S and the yield moment My with respect to the major principal axis. b. Compute the plastic section modulus Z and the plastic moment Mp with respect to the major principal axis.1. Two plates each with thickness 7-160mm are bolted together with 6-22 mm dia. bolts forming Bolt 52= 80mm spacing are as follows S₁ = 40mm Bolt hole dia. 25mm S3 = 100mm L fu- 483 тра ту- 345 тра Solve the allowable istrength and the P a Ultimate strength in: 1- Yielding 2. Rupture 3. Shear 4. Block Shear oo lap connection atat