Q-A circular column has to be designed using lateral ties to carry an axial load of 1500kN. Using M25 mix and Fe 415 grade steel, what is the required diameter of column section? (assume longitudinal steel percentage = 1% of gross area).
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- 2. A column HP 14 x 102 of A572 Gr. 55 steel has a length of 15 ft is fixed at both ends. Compute the design compressive strength for LRFD and the allowable compressive strength for ASD. (Steel section properties are provided in the next page) ASTM Designation A572 Gr. 42 Gr. 50 Gr. 55 Gr. 60⁰ Gr. 65⁰ Yield Stress (ksi) 42 50 55 60 65 Fu Tensile Stressa (ksi) 60 65 70 75 80Q2. The joint connection shown in Figure Q2 is subjected to a medium term tensile load under service class C24. The joint comprises of six 3mm diameter x 150mm long wire nails. If the characteristic density of the timber is 340kg/m³, determine the load carrying capacity P1 and P2 of the connection and slip per nail P2 P 45 mm thick Figure Q2 75 mm thick P1 P1 P 100 mm thickNeed complete and elaborate solution A.Using ASD, check the adequacy of the base plate (PL400x350x19 mm) centered on a W12X152 column as shown. It is resting on a concrete pedestal that is 450 mm x 400 mm. The loads are PDL = 600kN and PLL = 320 kN. Use A-36 steel and concrete fc’=20 MPa. B. Using LRFD, PL400x350 mm base plate, W12x120 column, 500x500 mm pedestal, PDL = 1200 kN and PLL = 800 kN. Use A-50 steel and concrete fc’ = 30 MPa. Determine the required base plate thickness.
- Problem 1. The composite beam shown below carries a cantilevered load of 10 kN. The beam consists of one 30 x 124 mm plate and four 12 x 50 mm plates. They are pinned together at 120 mm intervals with round pins. The pin material has a shear strength of 159 MPa. Compute the minimum acceptable diameter for the pins. O O O O O O O -0 O 0- O 1000 mm Do O -120 mm (typ) O O O P = 10 KN 30 x 124 mm 12 x 50 mm (typ)In order to reinforce a column in an existing structure, two channels are continuously welded to the column as shown below. Fy = 50 ksi for both the column and the channels. The effective length with respect to each axis is 16 feet. a) calculate the Moment of Inertia about each axis of the built-up shape. a) what is the available axial compressive strength and what is the percentage increase in strength? Use both LRFD and ASD.Topic:Welded Connection - Civil Engineering -Steel Design *Use latest NSCP/NSCP 2015 formula to solve this problem *Please use hand written to solve this problem A channel is used as a tension member with the web of the channel welded to a 9.5 mm thick gusset plate as shown in the figure. The tension member is subjected to the following axial loads. Use LRFD Service dead load = 200 kN Wind load = 276 kN Service live load = 260 kN For channel: Fy = 345 MPa For gusset plate: Fy = 248 MPa Fu = 400 MPa Size of E 70 electrodes = 4 mm Ultimate tensile strength of E 70 electrodes = 480; Fw = 0.6(480) Questions : a) Determine the design factored tensile force. b) Determine the length of longitudinal welds "L". c) Determine the block shear strength of the gusset plate.
- Ce tip.instructure.com Incognito (2) A composite section will be used to support a heavy axial load of P-800 kN with a First Semester SY 2021-2. rigid plate at the top of the section. Determine the average compressive stress in the brass section. Use h=16 mm. Answer in MegaPascals. Home Announcements Brass core (E = 105 GPa) Rigid end plate count Assignments Aluminum plates (E = 70 GPa) Discussions aboard Grades People urses Pages endar Files 300 mm Syllabus |auizes box Modules tory BigBlueButton (Conferences) elp Chat 60 mm T.I.P. Manila Library Video 40 mm Presentation Canvas LMS SatisfactorvThe below figure represents a section of a pre- stressed beam. For a no-tension design where a is the permissible stress in concrete, the total moment carrying capacity is (a) (c) d/2 bd² 6 d/2 bd²oc 3 -b- (b) (d) bd² oc 4 bd²a 122) 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 KN
- Answer the following for the section at Point D Only Calculate the distributed load "w" that: Will cause the section crack Will cause the reinforcement to yield. Material Properties: F'c = 5000 psi Fy = 60000 psi Es = 29000000 psi Ln = 27 ft L wl₂² 16 wl,² 14 CD L wl,2 vl₁² 10 11 win² 16 h: 28 in A=4 in² b=14 in n d: 25 inSubject: Steel design- Bolted Steel Connection *Use NSCP 2015 formula/guide to solve this problem *Use Handwritten A plate having a width of 400 and a thickness of 20mm is to be connected to two plates with thickness of 12mm by four rivets. The allowable shear stress for the rivets is 150 MPa and for bearing is 1.35 Fy. If the yield strength and the ultimate strength is 248 MPa and 400 MPa respectively. Determine the Following: a.) Required diameter of rivets w/o exceeding the aloowable stress in the shear. b.) Required diameter of rivets w/o exceeding the allowable stress in bearing. c.) Required diameter of rivets w/o exceeding the allowable stress in tension.Topic: 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