Using A36 steel, determine the safe load P based on the capacity of the bottom strut with a cross section as shown. Use LRFD. Let k 1.0. C9x20 I 3.4m 4.5m 90mm a) Calculate the radius of gyration at strong axis of the built-up section. b) What is the critical slenderness ratio? c) Find the critical buckling stress. d) Compute the nominal capacity of the bottom strut. e) Determine the safe load P.

Using A36 steel, determine the safe load P based on the capacity of the bottom strut with a cross section as shown. Use LRFD. Let k 1.0. C9x20 I 3.4m 4.5m 90mm a) Calculate the radius of gyration at strong axis of the built-up section. b) What is the critical slenderness ratio? c) Find the critical buckling stress. d) Compute the nominal capacity of the bottom strut. e) Determine the safe load P.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

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

Q: A W21 × 48 is used as a simply supported, uniformly loaded beam with a span length of 50 feet and…

A: GIVEN- W21×48L=50ftFy=60ksi

Q: Determine whether the column shown in Figure is adequate to support 850 kN dead load 2400 kN live…

A: Solution: Given: S 275 steel HE 450 B section length = 9 m braced at midspan both ends pinned, K = 1…

Q: AWT205 x 30 structural steel section (see Appendix B for cross-sectional properties) is used for a…

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

Q: The beam shown in Figure is laterally braced only at the ends. The 30-kip load is a service live…

A: ASD is based on the conventional stress calculation of structure with respect to the allowable…

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: 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: B A 150 mm 50 mm

Q: snip

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

Q: A WT205 x 30 structural steel section (see Appendix B for cross-sectional properties) is used for a…

Q: The three suspender bars are made of A992 steel and have equal cross-sectional areas of 450mm².…

A: 3 suspended bars are given, of equal cross-sectional area as shown in the figure above and it has…

Q: A W12 X 65 is used as a simply supported, uniformly loaded beam with a span length of 50 feet and…

A: It is required to determine the upper limit for compact. It is required to determine the upper…

Q: 5m 3m 3m 3m 3m Solve for the safest factored load Pu the compression members AB and BC can carry in…

A: For W16x50 Area = 14.7 sq-inch Assuming A992 steel So, yield strength (fy) = 50 ksi So, Tmax =…

Q: A W610 x 82 beam of A 36 steel has end reaction of 302 kN and is supported on a plate with a length…

A: Yield Strength :Yield strength of an element is a maximum stress an element or component can bear…

Q: A WT305 x41 standard steel shape is used to support the loads shown. Assume W= 50 kN/m, LAB = 1.1 m.…

A: To Determine Maximum Compressive and tensile stress.

Q: A simply supported plate girder carries the factored loads shown in the Figure below, self-weight is…

A: Web=750×10 mmFlange= 380×40 mmFlanges are fully retrained laterally and restrained against…

Q: The below figure represents a section of a pre- stressed beam. For a no-tension design where o is…

A: Given- Permissible stress in concrete=σcwidth of beam =bdepth=dIt is needed to find the total moment…

Q: F1-17. The strut is glued to the horizontal member at surface AB. If the strut has a thickness of 25…

Q: Q-11: A mild steel T section has the following cross-sectional dimensions: Total depth = 200 mm…

A: To find the plastic moment capacity of the section and shape factor of the T section given Thickness…

Q: The given frame is unbraced, and bending is about the x axis of each member. The axial service loads…

A: Dead load (PD) = 908 KNLive load (PL) = 1816 KNYield strength (Fy) = 350 MPa ≈ 50.76 ksiMoment of…

Q: Design all Tension members of the truss. Use A36 steel type. Dead Loads and Live loads are shown in…

Q: H HAVANIS 160kN 140KN 120KN 10.8m 8.1m D = 100KN 130kN 150KN 8 panels @ 6m ea. = 48m Determine the…

Q: AW18 × 40 standard steel shape is used to support the loads shown on the beam. Assume P = 20 kips, w…

A: Given Data: Load=p=20 kips w=5.8 kips LAB=4.0 ft LBC=4.0 ft LCD=13 ft

Q: A compression member is built up from a W14x90 and a W10x49, both of A992 steel. a. Compute rx and…

A: We need to first obtain the Moment of inertia about X and Y axis We can determine the moment of…

Q: A WT305 x 41 standard steel shape is used to support the loads shown. Assume w = 56 kN/m, LAB = 1.0…

A: To Determine Maximum Compressive and tensile bending stress.

Q: The rigid frame shown is unbraced. The members are oriented so that bending is about the strong…

A: Given- Axial strength of column AB=?

Q: The rigid frame shown is unbraced. The members are oriented so that bending is about the strong…

A: Answer We need to first find G factors at Joint A and joint B in order to calculate 'Kx' GA = 1…

Q: A column is formed by welding two A36 MC sections back-to- back. The column is 32 feet long. With…

A: Following are the given information. A=6.45 in2, d=10 in bf=3.32 in, x=0.990 in Ix=102 in4, Iy=6.40…

Q: Find the moment of resistance of the section. The permissible stresses are 10 N/mm2 in tension and…

A: The values and diagram given in the question are as follows: The permissible stress in…

Q: Pls help me answer. Thank you.

A: Here's an analysis of the failure mode and the importance of horizontal stirrups in deep beams:Deep…

Q: A rectangular post (40 mm x 80 mm) made of wood, the length of which is 5 m, supports the BC beam.…

A: Consider the given data l=6 mh=5 mF.S=1.50ζy=2…

Q: A WT305 x 41 standard steel shape is used to support the loads shown. Assume w = 52 kN/m, LAB = 2.0…

Concept explainers

Braced cuts

The braced cut is a method of providing lateral support to the soil in foundation trenches, to prevent the trenches from deformation by the caving in of soil and resisting it from moving laterally. During shallow excavation in firm soil, generally, slopes are provided to resist the caving in of soil. During shallow excavation in soft soil and deep excavation, the chances of caving in of soil increase. Hence, it becomes necessary to make a provision to resist the deformation of trenches due to caving in of soil. The braced cut is the method used for such conditions. The excavation using braced cuts is a bit uneconomical as compared to the traditional open-cut excavation method but provides better safety than the cut excavation method.

Question

Using A36 steel, determine the safe load P based on the capacity of the bottom strut with a cross section as shown.
Use LRFD. Letk 1.0.
C9x20
3.4m
-→ I
4.5m
90mm
a) Calculate the radius of gyration at strong axis of the built-up section.
b) What is the critical slenderness ratio?
c) Find the critical buckling stress.
d) Compute the nominal capacity of the bottom strut.
e) Determine the safe load P.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

Step 2

Step 3

Step 4

Step 5

Trending now

This is a popular solution!

Step by step

Solved in 5 steps with 15 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

Topic: BEAM BEARING PLATES AND COLUMN BASE PLATES - STEEL DESIGN Please solve your Solution in a handwritten Note: It should be handwritten pleaseeee
In steel design use nscp 2015 i need a complete solutions ty..
Determine the slenderness ratio of the section.Determine the elastic buckling stress of the section.Determine the critical stress of the section.Determine the allowable strength of the section.Determine the design strength of the section.
Rigid member BED is supported using steel rods AB, CD, and EF as shown below. E-200 GPa for steel. If a force P is applied at E, determine: a) The stress in each rod b) The length of each rod after applying the force LI (mm) LI (mm) P (kN) AAB (mm2) ACD (mm?) AEF (mm) 395 350 20 250 250 450 L1 L2 Note: BE ED P.
3-3 A rectangular section with width b=200mm and depth h=450mm is reinforced with 4 16 of HRB400, The design bending moment is M=100KN •m. Is the section safe or not? F00
2. Calculate the internal forces in the bar GF and GD in a roof truss as shownbelow. For each bar, indicate if the internal force is tensile or compressive. Usethe section method (snittmetoden). Point A has simple support and D has rollersupport.
Figure 1 shows the cross-section of a heavily welded I-section steel column under axial compression. The yield stress of the steel section is 300 MPa. (a) Check the slenderness of the flanges and webs. (b) Determine the effective and gross areas of the cross-section. (c) Determine the design axial capacity of the cross-section. 380 E 10 425 -8 Figure 1. Cross-section of the rectangular hollow section steel column (Units: mm)
The given beam has a continuous lateral support. if the live load is twice the dead load, what is the maximum total service load, in kips/ft, that can be supported?
Pls I need answer asap, I will like it.
solve and give the solution