SteelL-6m,0-25mm,2.5m4.5mE=200GPG0-15mm,E-70GPBAPRODTQ1) (20Marks): Find the maxallowable load (P) which can besupported by three rods asshown in Figure (1) so that thestresses are not exceeding(150MPa) in Steel and (100MPa)in Aluminum, assume that thefloor is to be rigid but it is notnecessary to remain horizontalalso the force at (B) is equal tothe force at (C)?AluminumL-4m,Figure (1)

if you have any doubts please comment

Answered: Steel L-6m, 0-25mm, 2.5m 4.5m E=200GPG…

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

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter5: Beams

Section: Chapter Questions

Problem 5.5.15P

See similar textbooks

Similar questions

H.W A Solid Steel bar of dlameter 6omm and length 350 mm is Placed inside an aluminum cylinder öf Inside diameter Femm and out side d lameter Ilomm, a comPressive load of looo lais applied on the assembly, find the stresses in the bar and the cylinder ? take Est = 200GPa and EAl = F0GPan %3D Cover Plate 0.25mm d- do 35omm A lum. cylinder Steel bar
7 7a 7b 7c Alaterally supported beam was designed for flexure. The beam is safe for shear & deflection. The most economical section is structural tubing however the said section is not readily available at the time of the construction. If you are the engineer in charge of the construction what alternative section will be the best replacement? Why? The section is 8" x 8" x 7.94 mm thick: Use Fy=248 MPa: E=200,000 MPa AISC wall thickness Ix 106 S x 103 Jx 103 mm4 mm3 mm4 rx =ry Area Ag (mm2) mm Designation Weight/m 8x8 7.94 47.36 6,039 79.25 37.84 371.99 60.35 expla'n briefly your cho'ce. (transform your comparative analys's 'nto a narative form to support your cho'ce) 8x8 14.29 80.61 10,258 76.2 59.52 585.02 99.06 8x8 72.7 9,290 76.96 54.53 539.13 90.32 8x8 9.53 56.09 7,161 78.48 44.12 432.62 70.76 mm 12.7 Zx 103 mm3 437.53 714.48 650.57 512.92
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
I need a clear answer by hand, not by keyboard and fast answer within 20 minutes. Thank you | dybala For the beam loaded as shown, find the maximum tensile and compressive bending stresses and maximum shearing stress.. IN.A=4*10mm* 80mm 100mm 60mm 60mm 1.5KN 2kN/m 1mal A |-----|- | Im ++ Im->> 3m 3kN/m
(1) Given A992 steel Ag = 10 in %3D KLy- 80, ry what is $ Pa=? = 60 the' column streugth design
A built-up section shown, 30 feet long, is fixed at both ends and braced in the weak direction at the midheight. A992 steel is used. Determine the following: 11. Ix 12. Ky 13. effective slenderness ratio 14. Fe 15. LRFD design strength 16. ASD allowable strength 18 4 W10 × 49 A 14.4 in² bf 10 in d. 10 in tw 0.340 in tf 0.560 in Ix= 272 in¹ Iy=93 4in¹ W10×54 A 15-3 in² bf 10 in d = 10 lin tw0-370 in tf 0.615 in Ix= 303 in 4 Ty = 103 in 1
8. H / The beam shown below has across seclion of channel Shape with width 360mm.and haghs h= loomm, the web thickness ist= 14mm.Deler- Mine ihe Max imum tensile aud Compressive stressin the beam due to unifom Load? 360Mm 3.6KN/M 14m tiom Please Sir 5T CASIC f-991ES PL DATURAL u CALC (-)
Determine the safe load of the column section shown, if it has a yield strength of 25 MPa. E = 200000 MPa. Use NSCP Specifications. Fyz248 mpa Properties of Channel Section d = 305 mm t₂ = 7.2 mm A = 3929 mm² t₁ = 12.7 mm Ix=53.7 x 10mm¹ x = 117 mm Properties of W 460 x 74 A = 9450 mm² b = 190 mm ly= 1.61 x 10 mm x = 17.7 mm tw = 9.0 mm rx = 188 mm ry = 41.9 mm d = 457 mm tr = 14.5 mm Ix = 333 x 10 mm Iy = 16.6 x 10mm* 7.21 When the height of column is 6 m. When the height of column is 10 m. Assume K= 1.0 457 CIVIL ENGINEERING- STEEL DESIGN
أ.د. رمضان المبروك A PL170mm x 10mm tension member is welded to a gusset plate as shown in Figure below. The steel is A36. Assume that Ae = Ag and compute the following. %3D fy = 250 MPa & f, = 400 MPa %3D %3D PL 170mmx10mm a. The design strength for LRFD. . The allowable strength for ASD. roblem (4):
7 Enercise:- 30 long with is subjected to a Assuming deformation is entirely elastic and manimum allowable elongation tensile 30000y that the is 0.3 mm while manimum. allowable reduction in, digmeter is 0.015mm. choose amory below only. candidate meeting requirent E (GPa) A cylindrical rod 100 mm diameter of 12 mm load of 30000 N. Aluminium alloy Bronze Titanium allow 70 100 110 U(Poiss ons vahi) 0.33 0.34 0.3
The A992 structural steel bar and a 2014-T6 aluminium bar shown below are secured to rigid supports at ends A and B and are connected together by a 1” diameter tool steel pin in double shear. The pin slides freely into place at room temperature. Determine the shearing stress on the pin caused by a 60 degree F drop in temperature and comment on its ability to carry the applied load. L L - 1" dia. pin A Steel area = 1 in.² Aluminum area = 2 in.² B
Determine the design tensile strength of plate (200x8 mm) connected to 10-mm thick gusset using 20 mm bolts as shown in the figure, if the yield and the ultimate stress of the steel used are 250 MPa and 410 MPa, respectively. Add 1mm around the bolt for the hole. Use LRFD method. Plate 8-mm thick 2 3 40+ 30 301 T 200 mm Gusset 10-mm thick 3af 30 2_3 *40 40+ 50,54 +40

SEE MORE QUESTIONS

Recommended textbooks for you

Steel Design (Activate Learning with these NEW ti…

Civil Engineering

ISBN:

9781337094740

Author:

Segui, William T.

Publisher:

Cengage Learning