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(2) PoL=280* 16' 34' PLL= 200k 18' Please select the ligutest W 12 A992 steel.

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

Two channels having the given properties shown is placed at a distance of 300 mm to back and is properly connected by a pair of lacings as shown. 305 mm Fy=248 mpa Properties of one channel A = 5690 mm² Ix = 67.3 x 10 mm* Assume K = 1.0 -260 mm- Lucre x= 17- d = 305 mm ly 2.12 x 10 mm x 17mm CIVIL ENGINEERING - STEEL DESIGN Determine the safe axial load in kN, that the column section could carry. Unsupported height of column is 6m.

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KINDLY IDENTIFY THE ZERO FORCE MEMBERS

PLEASE WITH FBD. THANK YOU

3) For the concrete beams shown, draw the approximate deflection shape and draw the locations of the reinforcing bars (top or bottom) for the following beams Fixed Pin & b & b сб $ के के bbbbbb کن के Fin roller fixed b dd free

note: SN is equal to 18, topload will now be 118kN

Compute the compressive strength of a WT 12x81 of A992 steel. The effective length with respect to the x-axis is 25 feet 7inches, the effective length with respect to the y axis is 20feet, and the effective length with respect to the z-axis is 20 feet WT12X81 A 23.9 t₁ 1.22 ly 221 d 12.5 K des 1.72 J 9.22 b₁ 13.0 y 2.70 Cw 43.8 tw 0.705 Ix 293

24 Tensile test well done with standard tensile test spaceman with diameter 25 mm & length 120 mm the table in the attachment picture shows the applied load in KN and the elongation in mm. the yield stress is Load 37.5 75 75 85 110 120 125 115 (KN) Elongation (mm) 100 0.06 0.12 1.4 4 12 18 24 29 32 O 320mm2 O 325mm2 O 239mm2 O None of the above

ZAO O "l zain IQ docs.google.com/forms O 0/10 find the tensile stress in upper plate in section 1-1 5 1 4 80 kN 80 kN Q B-bolt DIA. 10 MM 80 kN 80 kN 8-bolt DIA. 10 MM 30 MРа 33 МРа 35 MPа 37 MPa no one 10/10 find the tensile stress in upper plate in section 3-3 +

The rigid plate ABC shown in figure 2, was positioned on top of two identical steel posts, and the aluminum post was 0.1mm shorter than the steel posts at temperature of 50°C, if the allowable compressive stress in steel is limited to 120 MPa and in aluminum to 150 MPa, what is the maximum possible load P that can be supported on the rigid plate at temperature of 10°C ?

Note: SN is equal to 18, topload will be equal to 118kN

Estimate for the concrete, formworks, reinforcement requirement of Beam GG-36. Length of Beam 1 = 8200mm Length of Beam 2 = 8200mm Size of Column = 800mm x 800mm GG-36 IGG-36 GG-131 009*00t GB-4 009*00+ GG-1

Permissible As in sechion to ensure tensile Compute the design strength of beam shown it 420 MPa , fc = 30 MPa . check the maximum %3D Permissible As in sechion to ensure tensile failure. 4¢25 stimpp10 q00 8432 300 mm

Question 1 Check that the medium-term load of 38 kN applied to the spaced column shown in below complies with the design requirements of BS 5268-2. The column consists of two 38 x 150 C22 timbers 76 mm apart. All joints are glued and intermediate packs are 250 mm long. suitable connection #### x ->89² 6:38. Y T -end blocking AP 100 intermediate blocking L₂ 1₁ x W L W EFH 400 m 38kN medium term load -ICHICHID -200 600 600 600 600 600 600 200 38kN medium term load

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 =

FOR THE GIVEN PROPERTIES DESIGN d': 15 mm THE REINFORCEMENT OF : 400 mm b d = $80 mm fc = 33 MPa fy = 320 MPa Mu: 1 000 000 000 N mm A. USE B.USE OF NSCP 2001 NSCP 2015 BEAM THE BEAM 132 13 Gra mm of BELOW. #H 1606 210135 03 M 12437 P0122399/100mA 02338941 242

ure B 2m2 m C -2 m D 4 m 500 N 700 N 2 of 2 Part B Identify the zero-force members in the truss shown in (Figure 2). Check all that apply. 000 CD DE CG CF FG BC BG DF AG EF AB Submit Provide Feedback Request Answer

Q-6 While designing for a steel column of Fe250 grade the base plate resting on a concrete pedestal of M20 grade, the bearing strength of concrete (N/mm2) in LSM of design as per IS 456: 2000 is .........

Question 4 of 10 0/10 The axial assembly consists of a solid 0.93-in.-diameter aluminum alloy [E = 10400 ksi, v = 0.32, a = 10.4 x 10-6/°F]rod (1) and a solid 1.52-in.-diameter bronze [E = 16000 ksi, v = 0.19, a = 9.5 x 10-6/°F] rod (2). Assume L1=11 in., L2=22 in. If the supports at A and Care rigid and the assembly is stress free at 5°F, determine: (a) the normal stresses o1,02 in both rods at 138°F (positive if tensile, negative if compressive). (b) the displacement ug of flange B (positive if to the right, negative to the left). (c) the change in diameter Ad of the aluminum rod. (2) L2 Answers: (a) o = 0.0272 ksi, o2 = 0.0102 ksi = an (q) (c) Ad 0.015 in. 0.669 in. Assistance lised eTexthook and Media I!

1) Column AB is made of solid 410, cold worked Steel with the following specification Sy=85 KSI, E-29000 KSI Length: 9.5 in. Diameter: 425 in. Ends: Free & Fixed For the above column calculate: a) Type of column failure (show proof) b) The critical load that would cause failure. c) Allowable force with Factor of Safety of 2.5. Answer Unit Value a 13 14 14

Determine the Design Flexural Strength of W460x52 A992 Steel with the following cases: CIVIL ENGINEERING STEEL DESIGN With Continuous lateral bracing Unbraced length = 4m, Cb = 1.0 Unbraced length= 12m, Cb = 1.0 a.) b.) c.) Fy = 345 MPa Properties: d = 450mm tf = 10.8 mm 1x 212x10mm* Cw 306x10 mm" Sx** 944 x10¹mm¹ bf 142mm MY Ag - 6650mm2 Zx 1090x10'mm' Ho439mm ry = 31mm tw* 7.62mm J-211x10¹mm* ly = 6.37x10 mm rls - 38.4mm

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 80

Column Load

2. The A-36 steel Bar BC had a diameter of 40 mm before the two loads, P, were applied. If after loading, the new diameter of the bar is 39.995 mm, what was P? P -1 m B 0.75 m 1 m

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 in

Question 5 What area of steel is provided (per ft width) for No 6 @ 9"o.c.? O 0.66 in² O 0.88 in² O 0.59 in² O 2.00 in²

Problem 01 Using ASD, check the adequacy of the base plate (PL400X350×19 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. TH

cd efgh ij O klmn u op V 0 0 0 t Р What is the line for which the section will rupture in tension? All holes are the same diameter. m n j h d