kmarks Develop Window Help ara University Mail 8 camino.instructure.com in (22) Erik Keifer | Linkedin. 2 m 21 ++ 5 kN 2 ft 1m- 2. Problem No. 2: The horizontal platform ABCD weighs 60 lb and supports a 240-lb load at its center. The platform is normally held in position by hinges at A and B and by braces CE and DE. For this problem, assume that the brace DE is removed. B B a) Draw the free body diagram for the platform ABDC. How many unknowns are there in total? How many equations of equilibrium do you have? Is this problem statically determinate? Is the platform stable? b) Assume that the hinge at A does not exert any axial thrust (There is no reaction force in z direction at the hinge A). Determine the reactions at the hinges and the force exerted by the brace CE. 32 3 ft 300 lb (C. C Practice Assignment 12 D Tu

kmarks Develop Window Help ara University Mail 8 camino.instructure.com in (22) Erik Keifer | Linkedin. 2 m 21 ++ 5 kN 2 ft 1m- 2. Problem No. 2: The horizontal platform ABCD weighs 60 lb and supports a 240-lb load at its center. The platform is normally held in position by hinges at A and B and by braces CE and DE. For this problem, assume that the brace DE is removed. B B a) Draw the free body diagram for the platform ABDC. How many unknowns are there in total? How many equations of equilibrium do you have? Is this problem statically determinate? Is the platform stable? b) Assume that the hinge at A does not exert any axial thrust (There is no reaction force in z direction at the hinge A). Determine the reactions at the hinges and the force exerted by the brace CE. 32 3 ft 300 lb (C. C Practice Assignment 12 D Tu

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter9: Composite Construction

Section: Chapter Questions

Problem 9.2.1P: A W1422 acts compositely with a 4-inch-thick floor slab whose effective width b is 90 inches. The...

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O 88 130% v - + I Annotate T| Edit Trial expired Unlock Full Version ENGR 263 + A A A T O 4.10 Member AB is the beam under consideration. As shown in the illustration of the loading condition, member AB is an overhanging beam that supports a uniformly distributed roof load of 500 lb/ft. It also carries concentrated loads from a rooftop HVAC unit (4000 lb), an interior hanging display support (2000 lb), and a marquee sign (3000 lb). Marquee overnang Displau SLIPPort II Raof Kiots Ol I W = 500 Ib/Ft A B 4000 b 2000 000 I Steel beam !! I1 3 FE 5 FE - Ft RoOFtop HVAC unit 10 FE 4 Ft Marquee sign< R R2 Steel beam (negligible weight) Free-body diagram Dispiay Support Loading condition
4. The tension member shown below is a C 310 x 30.8 of A572M gr 345 steel materials. Will it safely support a service dead load of 265 kN and a service live load of 555 kN. L 50 102.5 102.5 50 65 65 65 O O 22 veny diameter bolts C310 x 30.8 Iso A36M stool matorial and select a double anglo tonsion member to rosist
Pabacboadcom Question Campletion Stutn MOD LE 4 A Movtng to another question will save this response Question 1 A hwo span beam subjected to shear and flexure only is reinforced as follows: @ MIDSPAN 2-016 mm 3-016 mm SECTION @ FACE OF SUPPORTS TOP BARS BOTTOM BARS 5-016 mm 2-016 mm Given: Stirrup diameter, dg - 10 mm Concrete fe=21 MPa Steel rebar fy 415 MPa Stirrup fy= 275 MPa Beam sizebxh=250 mm x 450 mm Assume all bars laid out in single layer. Calculate the following: Tensile steel ratio in positive bending at midspan = (in 5 decimal places) Design Moment strength of section at midspan for positive bending = kN-m (nearest whole number) kN m (nearest whole number) Nominal Moment strength of section at face of support for negative bending =
Note: Hi, pls need help. Any solutions are highly appreciated. I will rate this right away. Thank you in advance! What is the required diameter of sag rods in mm? Considering sag rods and tie rods at midspan (rounded value) A 20m Fink Truss are spaced at 'S' meters o.c. the purlins are made up of C10X20 sections and are spaced 1.1967m on centers. The truss has a height of 4m. Roof LL is 'RLL' N/m2 of roof surface and roof covering is assumed to be 'RDL' N/m2 of roof surface. Determine whether the purlins are adequate to carry such loadings. Given: Fy=248 MPa w=29.76 Kg/m A=3787.09 mm^2 Ix=32.84×10^6 mm^4 Iy=1.17×10^6 mm4 Zx=317.91×10^3 mm3 Sx=258.92×10^3 mm3 Zy=44.25×10^3 mm3 Sy=21.47×10^3 mm3 S= 6 m RLL=900 N/m^2 RDL= 900 N/m^2
"messenger_media?thread_id%=D100002917920802&attachment_id%3891210971774069&message_id%3Dmid.%24CAAAA Question 3 For the truss shown, determine the forces in members (a) BD; and (b) BF. 4 m4 m4 m-4 m- A B C 6 m E F G H O00 20 kN 20 kN 20 kN A BD-40kN (C); BF-33.33kN (T)
Part 2: Identify the read forces 4m 11 ang 600 4m 60 60 4m 10KN Rea'd: All member forces (use methorl of joints)
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sted ingid 250MM B k steel bar bolt AUR rigid bar 12 mm thick 8 mm Brass 350mm Jointd 16 mene thick 350mm JointB 250 mm Xg = 20x10% Eg = 90 G Pa -6 A, = 12x10/0 Es = 200 G Pa thickness> 16 mm 8 mm bolt %3D 12 mmi thick shear strength f bolf bearing streng th of holt= 100 la 50 MPa %3D Assume no failure will take place in steel or brass. Temperature chauge brass steel on Temperature change on Determine thai can be applied to system - the max
TABLE 8.5 Allowable Spans in Feet and inches for Low-Slope Raftern, Slope 3 in 12 or Less (No Celling Siat Spacing IN) Allowable Extreme Fiber Stress in Bending F (pai). 1200 500 600 700 800 900 1000 1100 1300 1400 1500 1600 1700 1800 1900 11-7 13-0 0.94 13-7 109 11-9 0.94 9.7 14-2 124 12-4 14-9 154 15-11 16-5 191 14-2 16-11 17-5 2.28 15-1 9-2 10-0 17-10 10-10 0.55 9-5 12-4 0.67 O80 10-8 1.56 13-3 12.0 2.09 033 7-11 0.29 6-6 0.44 140 12-10 1.73 13-9 247 156 10-0 058 8-2 113 0.82 9-2 8-8 14-8 1.35 10-10 0.70 107 10-0 121 10-5 0.99 19-6 1.50 1.65 11-7 181 11-11 1.97 12-4 214 12-8 2x6 160 0.38 7-1 0,48 7-8 0.39 8-8 11-3 1.22 20-11 L73 18-2 0.24 12-1 0.33 10-6 0.29 8-7 024 15-5 0.33 13-4 0.31 13-3 0.44 11-6 0.48 15-3 0.57 16-3 0.67 17-1 0.77 17-11 0.88 18-9 L10 20-3 1.35 21-7 148 22-3 2.09 1.75 22-11 23-7 2.28 24.0 1.61 144 067 13-3 0.94 14-10 1.09 15-6 124 16-3 140 16-10 121 13-9 0.99 24-10 1.56 17-6 12.0 0.55 12-5 0.48 10-1 0.39 0.80 14-0 191 18-9 2.47 19-10 20-5 19-4 0.70 I1-6 2x8 16.0…
Problem. A bolted connection shown consists of two plates 300 mm x 12 mm connected by 4 - 22 mm diameter bolts. 12 mm bolts T12 mm te12 mm P 300 75 75 75 Edge distances = 75 mm %3D dnole for tensile and rupture = dp +3 mm %3D dnole for bearing strength for Lec = d + 1.5 mm Fy = 248 MPa Fu = 400 MPa %3D En 330 MPa %3D Use LRFD design method. Determine the design strength due to tensile rupture strength of plates. (kN)
A:17 docs.google.com a 35 MPa 37 MPa no one 10/10 find the tensile stress in upper plate in section 3-3 5 80 kN 80 kN 8-bolt DIA. 10 MM 80 kN 80 kN 8-bolt DIA. 10 MM 10 MPa 12.5 MPa 15 MPa 8.5 MPa no one 10/.. find the bearing stress between bolts and plate
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 150 mm x 90 mm x 12 mm angular section is welded to a gusset plate as shown. Area of the angle is 2736 sq.mm, allowable shear Fv is 150MPa, Allowable tensile stress Ft = 0.6Fy with Fy = 250 MPa. Questions: a) Design force P b) Total required length of weld using 12 mm fillet weld c) Value of “b”