4.4-1 AHSS10 × 8 × 3⁄16 is used as a compression member with one end pinned and the other end fixed against rotation but free to translate. The length is 12 feet. Compute the nominal compressive strength for A500 Grade B steel (Fy = 46 ksi). Note that thisis a slender-element compression member, and the equations of AISC Section E7 mustbe used.

Q: Compute its design strength based on rupture in kiloNewtons..* 412 485 453 389 438 This is a…

A: Given, Fu = 400 MPa  Fy = 248 MPa D = 16 mm  Dh = D + 2

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Q: ASSIGNMENT #5 BIAXIAL BENDING Problem: Check the beam shown for compliance with the NSCP…

A: Section: W410 x 100 A992 steel fy = 345 MPa DL = 30% LL = 70%

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Q: Determine the available strength of the compression member shown in Figure , in each of the…

A: Given data HSS 10×6×12 section Length, L=15ft=180in. ASTM A500, Grade C Steel Fy = 50 ksi Ag = 13.5…

Q: 1. For the Pratt bridge truss and the loading shown below, determine the average normal stress in…

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Q: 8. HSS 10 x 8 x1/4 with Fy-50 ksi is selected to be used as a compression member with KL=10ft.…

A: Since you have asked multiple question, we will solve the first question for you. If youwant any…

Q: (3) 2 12" WOL= 2 KLF, WLL = 3 KLf 16/ 24" 81 NWC. NWC, fé = 4000 psi fy= 60ksi #4 tied bars. Please…

A: Problem (3): Given: b = 12 in h = 24 in WDL = 2 k/ft WLL = 3 k/ft The span between the supports, L =…

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Q: Prob 01: An 18” depth beam is bolted to a 24” depth girder is connected similar to that in figure…

A: Given:- The depth of the beam = 18 inch = 457.2 mm The depth of the girder = 24 inch = 609.6 mm The…

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A: This question is related to the Bending Stresses from Strength of Material. In this question we…

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Q: As shown in Figure 7, members AC and BD are used to brace the pin-connected structure against a…

A: Factored load, 12 kips. A36-steel properties: fu = 58 ksi fy = 36 ksi

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A: Given, WT205 x 30 section Length = 8.8 m or 8800 mm

Q: Determine the maximum axial compressive service load that can be supported if the live load is twice…

A: Given:- HSS 8×4×14ASTM A500 Grade BTo find:-maximum axial compressive service load using,a)…

Q: Determine whether the compression member shown in Figure is adequate to support the given service…

A: Given, D=110 kips L=280 kips

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Q: HOMEWORK-1 - 3m· For the truss shown in the figure; a) Calculate the bar forces for F=4000 kg.

A: a.given that,  F=4000kg 1-2=4m 1-3=3m 2-3=42+32=5m so, we have to calculate the bar force.

Q: Determine the design moment capacity of the section. (kN-m)

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Q: Problem 4: Use A992 steel and select a W shape. The service dead load is 142 kips, and the service…

A: First we summarise the given data : Service dead load = 142 kips Service live load = 356 kips To…

Q: snip

A: The factored axis load is Pu=1.6×28=44.8kipsthe factored transversed load and bending moment…

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Q: 10/.. find the shear stress in bolts (4 80 kN 80 kN 8-bolt DIA. 10 MM 80 kN 80 kN 8-bolt DIA. 10 MM…

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Q: An HSS10 × 8 × 3/16 is used as a compression member with one end pinned and the other end fixed…

A: The given data is shown below: HSS10×8×316Length=3.65mA500 Grade BFy=318 MPa

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A: Given, W14 section PDL = 990 kN PLL = 680 kN

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Q: Delermine the available axial compressive strength by each of the following methods: a. Use AISC…

A: Given Data -

Q: Situation 2: A rectangular beam has the following properties. Assume 12 mm stirrups. Assume a single…

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Q: A W14X120 is used as a tension member in atruss. The flanges of the member are connected to a gusset…

A: Answer  Calculation of yielding capacity of the section based on LRFD We know that yielding capacity…

Q: me column shown what is the nominal compressive strength (Pn) ? Pn=? W12X50 10 W12X50 TITT X…

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Q: The tension member shown below is a C 310 x 30.8 of A572M gr 345 steel materials. Will it safely…

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Q: A W 24 x 103 is used as a beam that is simply supported at its ends. The properties of the section…

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Q: Two square rods AB and BC are welded together at B and loaded as shown. Determine the average normal…

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Q: resistance of the section. The p ion.

A: given:

Q: 4- The reduction factor (0) is equal to 0.9 if the: Es 2 Ety + 0.003 Es S Ety + 0.003 Es S Ety 5-…

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Q: Problem 01 Using ASD, check the adequacy of the base plate (PL400X350×19 mm) centered on a W12X152…

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Q: HOMEWORK-1 3m For the truss shown in the figure; Design members 1-2 and 2-3 assuming that out of…

A: a.given that,  F=4000kg AB=4m AC=3m BC=42+32=5 m so, we have to calculate the bar force.…

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A: Solution Take design load= 2100 kN Effective length= l= 6.5 m( Both end pinned connection)

Q: or panel of the welded I-section shown below (Fy = 50 ksi). PL 1/2" x 96" - PL 2" x 16"

A: configuration shear strength is determined according to section G2 of AISC  360 data we havr given…