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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 80A simply supported beam is 25 × 50 cm deep and has 2-20 mm Fe415 grade steel bars going into the support shown in figure below. If the shear force at the centre of support is 110 kN at service loads. Assume M20 mix, design bond stress (ta) = 1.2 MPa. Take clear cover to steel = 25 mm 2-20 mmo with referecne to the information provided, (a) The required development length (b) The anchorage length 00 mmTopic: 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
- 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)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 SatisfactorvDesign Specifications 1st and 2nd floor Member Roof Deck Load (Typical) Live Load Item Superimposed Dead Load (Typical) Live Load Superimposed Dead a. b. C. For the structural plan attached, Hint: Self-weight B1 B2 Notes: 1. For simplicity, neglect self-weight of concrete slab B3 G1 G2 C1 C2 Design Loadings Application Slabs Slabs Slabs Slabs Value 4 3 2 3 4 4 4 Value 4 means pinned means continuous Assume all members have the same stiffness (same El) for simplicity 8 4 6 Unit kN/m kN/m kN/m kN/m kN/m kN/m kN/m Unit kPa kPa kPa kPa Determine the factored bending moment (LRFD) to be used in the design of Beam B1 in kNm. Determine the factored shear force (LRFD) to be used in the design of Girder G2 in kN. Determine the factored axial load (LRFD) to be used in the design of Column C1 in kN. -I T C2 G2 C1 G1 C1 5000 B3 1 B2 B1 B2 C2 C1 ∙H. GROUND FLOOR FRAMING PLAN C1 I G2 H G1 İ 8 C2 C1 G1 C1 5000 B3 1 B2 B1 + B2 1 B3 . C2 C1 I C1 SECOND FLOOR AND ROOF DECK FRAMING PLAN G1 G1
- Problem2. The compression member is shown in figure. Find the following: a. The Euler stress Fe. b. The buckling stress Fcr c. The design strength d. The allowable strength e. Does the member satisfactorily meet the design requirements? Why? HSS 8x 8x4 ASTM AS00, Grade B steel (Fy = 46 ksi) 15'For the column shown what is the nominal compressive strength (Pn) ? Pn=? W12x50 10 12X50 THT X combilevered WI2. 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 CS
- Refer to the figure shown below. 350 Given are the following: fc 55 MPa fou 1,350 MPa Aps - 1.500 mnm2 fpy - 1,100 MPa Use bonded tendons as prestressed reinforcement. Determine the design monment capacity of the section. (kN-m)Fotoğraflar - Ekran Görüntüsü (10)_LI.jpg Tüm fotoğrafları görüntüle + Şuraya ekle: Z Düzenle ve Oluştur v E Paylaş The stress-strain diagram of a reinforcement steel having a cross-sectional diameter of 12 mm diameter and 100 mm gage length is determined after its tensile strength test as follows. Based on the stress- strain diagram determine the followings properties of the material (Poisson's ratio of the material is 0.32) : 600 a) Modulus of elasticity 550 500 b) Yield strength 450 c) Toughness 400 350 d) Resilience 300 e) Shear modulus 250 200 f) Bulk modulus 150 g) Ductility as described by 100 percent change in length 50 0.01 0.02 O.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 Strain 18:12 18.01.2021 10 Stress, MPaIn the connection shown, 1/4-in. side and end fillet welds are used to connect the 3-in.-by-1 in. tension member to the plate. The applied load is 60,000 lb. Find the required dimension L. The steel is ASTM A36 and the electrode used is an E70. - Fillet welds L 3" Activate Windows Go to Settings to activate Wir TABLE 19.2 Allowable stresses in ksi (MPa) 12: 12/1 DELL end F12 insert prt sc F10 home F11 F6 F8 F9 F3