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- Subject: 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.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 Satisfactorv2. 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 Capacity of the joint if the allowable tensile stress is 0.75Fy. Use A36 steel Fy=248 MPa 25 mm= DIAMETER OF BOLTS 45 45mm fmmy C. 90mm 90mm 90mm Comm P
- 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, MPaProblem 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.
- 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 CSThe 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.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
- A steel plate is to be attached to a support with three bolts. The cross-sectional area of the plate is 800 mm² and the yield strength of the steel is 260 MPa. The ultimate shear strength of the bolts is 570 MPa. A factor of safety of 1.67 with respect to yield is required for the plate. A factor of safety of 4.0 with respect to the ultimate shear strength is required for the bolts. Determine the minimum bolt diameter required to develop the full strength of the plate. Note: consider only the gross cross-sectional area of the plate-not the net area. Support Plate ↓ PA steel bar is attached to a wood support beam with four 22 mm diameter lag screws, as shown in the figure. The steel bar is 69-mm-wide by 8-mm-thick. For the steel bar, the yield strength is 235 MPa and the ultimate bearing strength is 350 MPa. The ultimate shear strength of the lag screws is 155 MPa. Factors of safety of 1.70 with respect to yield strength and 3.0 with respect to bearing strength are required for the bar. A factor of safety of 2.9 with respect to the ultimate shear strength is required for the lag screws. Determine the allowable load P that can be supported by this connection. (Note: Consider only the gross cross-sectional area of the bar-not the net area.) Support beam Answer: Pallow= 0000 P Steel bar KNProblem2. 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'