Q.2 The simply supported steel beam shown in Fig. 2 has a span of s 24 ft. with flange having full lateral restrained from a concrete slab. The uniformly distributed applied dead load D = 1.0 kip/ft and two concentrated live loads PL = 20 kips each acts as shown in the Figure. Using the relevant Design Tables Select a suitable A992 steel (Fy = 50 ksi) W- section to satisfy LRFD Design: P₁= 20 kips 8 ft. 24 ft. FIG. 2 P₁ = 20 kips ↓ a) Strength Design (Flexure and Shear) b) Serviceability (Deflection) k 8 ft. WD = 1.0 kips/ft 38

Q.2 The simply supported steel beam shown in Fig. 2 has a span of s 24 ft. with flange having full lateral restrained from a concrete slab. The uniformly distributed applied dead load D = 1.0 kip/ft and two concentrated live loads PL = 20 kips each acts as shown in the Figure. Using the relevant Design Tables Select a suitable A992 steel (Fy = 50 ksi) W- section to satisfy LRFD Design: P₁= 20 kips 8 ft. 24 ft. FIG. 2 P₁ = 20 kips ↓ a) Strength Design (Flexure and Shear) b) Serviceability (Deflection) k 8 ft. WD = 1.0 kips/ft 38

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: A rectangular beam has width 300 mm and effective depth of 450mm .The beam reinforced with four 028…

Q: Q1: The cantilever reinforced concrete beam has a rectangular cross- section (b-250 mm) shown in the…

Q: 1. Determine the buckling strength of column C1 shown below and check the inelastic column buckling…

A: Given data: Steel A572 grade 345 material Fy=50 ksi Section for column C1, W460x177 Ag=22600 mm2…

Q: Situation #2: A properties: Width, b = 400 mm 415 MPa rectangular beam has the following Effective…

Q: 1.4 A rectangular beam has width 300 mm and effective depth of 450mm.The beam reinforced with four…

Q: A compression member adopts a section of 530UB82.0 as shown in Fig. 1. Its steel grade is Grade 350.…

Q: A W 21 x 48 beam is simply supported on a span 7 m. The beam is laterally supported at its ends and…

A: A W 21 x 48 beam is simply supported on a span 7 m. The beam is laterally supported at its ends and…

Q: The steel beam AB has a uniformly distributed load (UDL) w1 = 6 kN/m acting along its full length L1…

A: Answer To calculate the deflection at midspan of a given simply supported beam we have f =…

Q: A W21x50 structural steel wide-flange section is used as a simply supported beam with a span length…

Q: The beam shown in the figure is a two-span beam with a pin (hinge) in the center of the left span,…

A: We need to find the maximum Moment on the beam and compare it with the W 12 x 79 beam Moment…

Q: P = L1 = L2= L1 Your Answer: P 195 2.50 4.5 Answer KN m m L2 ▷ View hint for Question 1 P The…

Q: 4.4) The design shear strength D,V, is A) 81.1 kips B) 482 kips C) 529 kips D) 567 kips

Q: For the beam(300x800mm) to carried the loads shown in the figure below,design the steel…

A: Given values: Size of beam: (Width b * Depth D) = 300 * 800 mm Span = 9 m f'c = 25 MPa fy = 414 MPa…

Q: A WT305 x 41 standard steel shape is used to support the loads shown. Assume w = 42 kN/m, LAB= 1.9…

A: Given a WT305X41 steel section used to support load. It is required to find the maximum tensile…

Q: code limitation of L/240. 3/ If maximum deflecti

Q: ully welded together. The PL16×380 and PL12X500 for the

A: the flexural strength is given by F =PLbd2where . p = load on beam L = length b =…

Q: I cross-section is simply supported on a span of 4m. Find the safe uniformly distributed load the…

A: Draw the schematic diagram

Q: A simply supported beam with a span length of 48 feet is laterally supported at its ends and s…

A: Span of simply supported beam = 45 ftDead load = 400 lb/ftLive load = 1000 lb/ftSection of the beam…

Q: 4.3) The design moment strength M, is A) 880 ft-kips B) 887 ft-kips C) 1038 ft-kips D) 934 ft-kips…

Q: A WT305 × 41 standard steel shape is used to support the loads shown. Assume w = 50 kN/m, LAB = 1.4…

Q: 4. The beam shown in Fig. has lateral support only at the ends. The uniform load is a superimposed…

A: To find : Fcr

Q: A W36x210 can be used for Transfer Girder ‘A’ as shown in the elevation next page. Assume the LRFD…

A: a) Factored concentrated load on beam = 200 kips factored uniform load on beam = 2 kip/ft span of…

Q: 4.1) The value of C, is A) 1.21 B) 1.14 C) 1.30 D) 1.32

Concept explainers

Properties of materials

A property of a material describes its characteristics and distinguishes it from the others. The material properties are described it in terms of its hardness, color, appearances, and so on. The material properties generally depend on the type of material, or whether it is a metal or an alloy. There are different kinds of material properties such as mechanical properties, thermal properties, physical properties, and chemical properties. Chemical properties describe the material in terms of a chemical change. The mechanical properties come into existence when the material is acted upon by an external load. Thermal properties are associated with the temperature of the material and specific heat.

Question

100%

Q.2 The simply supported steel beam shown in Fig. 2 has a span of s 24 ft. with flange having
full lateral restrained from a concrete slab. The uniformly distributed applied dead load D =
1.0 kip/ft and two concentrated live loads PL = 20 kips each acts as shown in the Figure.
Using the relevant Design Tables Select a suitable A992 steel (Fy = 50 ksi) W- section to
satisfy
LRFD
Design:
P₁=20 kips
8 ft.
24 ft.
P₁ = 20 kips
↓
FIG. 2
a) Strength Design (Flexure and Shear)
b) Serviceability (Deflection)
k
8 ft.
WD = 1.0 kips/ft
3

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: Mention the given Data

VIEW

Step 2: Selection of section

VIEW

Step 3: Flexure and shear check

VIEW

Step 4: Deflection check

VIEW

Solution

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 5 steps with 4 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

RCDA reinforced concrete T- beam, section in Figure 2 (a), is subjected to an ultimate uniformload and concentrated loads as shown in the Figure 2(b). The beam is reinforced with 6-28 mm diameter tension bars at the point of maximum positive moment (at midspan -bars at the bottom), and 3-28 mm diameter tension bars at the point of maximum negativemoment (at the supports - bars at the top) .. The design properties used are: f ‘c = 21MPa and fy = 414 MPa, ; flange width = 650 mm, flange thickness is = 90 mm; webthickness = 250 mm; effective depth = 410 mm.Need to answerQ. Draw the shear and moment diagrams for the given loads;
Please correct solution
AW14 x 43 column in a braced frame has a yield stress of 50 ksi and a height of 10 ft. The column is pinned at the top and bottom and has no intermediate bracing. Determine the available strength of the column considering the critical stress for flexural buckling. From AISC Manual Table 1-1, the properties of a W14 x 43 are: A = 12.6 in^2 tw = 0.305 in bf/2tf = 7.54 h/tw = 37.4 rx = 5.82 in ry = 1.89 in Ix = 428 in4 ly = 45.2 in4 G= 11,200 ksi J = 1.05 in^4 Cw=1950 in^6 What is the nominal axial strength considering flexural buckling stress of the column section (in kips). O 310 O 468 O 447 O 375 O 420 O 436 O 517 O 280
5. The beam-column has a height of 7.5m and is a member of a braced frame. It is subjected to an axial load of DL-65KN and LL= 1OOKN. It is pinned at both ends. A W12X35 section was used and was made up of A-36 steel with Fy-250MPA. Compute the interaction value for beam- column based on NSCP 2015 code provisions. Ngelect the weight of the section. A = 6645 mm2 Zx- 839x10^3 mm3 d- 317.50mm Sx- 747x10^3 mm3 tw= 7.62 Rx= 133.35 bf= 166.62 ly= 10x10^6 mm4 tf- 13.21 Sy= 122x10^3 mm3 Ix= 119x10^6 mm4 Ry3 39.12mm 3.5 Note: Use C=D1.00%--0.2
4
Determine collapse load Wu for continuous beam ABCDE loaded and supported as shown in figure . For span AC plastic moment of resistance is 50 % more than that of CE .
The answer is between 18.25 and 20.25
A compression member AB (HE 400 B) is pinned at both ends and braced in the weak direction as shown in the figure. Material and geometrical properties are given below. For the given loading and ASD load combination (PG+Pq=3500 KN), determine whether the section is satisfactory or not. Material: 5355 F355 MPa F 510 MPa E= 200000 MPa Section properties (HE 400 B): A: 19780 mm² d: 400 mm h: 298 mm t: 13.5 mm. tr: 24 mm b: 300 mm i,: 170.8 mm İy: 74 mm B X -BRACING IN THE WEAKDIRECTION
Compute the total elongation of the tapered steel bar shown in Fig. 5 subject to an axial load P (= 80 kN). The bar is of constant thickness 12 mm and tappers from W¡ mm at left to w2 mm at the other end. Given E=200 GPa for steel. Values of w1 and w2 for different Roll numbers are stated in Table below. What axial stress exists at mid-length section? Last digit of W1 W2 Roll No. (mm) (mm) P P 1, 4, 7 2, 5, 8 40 120 50 140 12 m 3, 6, 9 & 0 60 160
For the structure shown in Figure, with a uniformly distributed service dead load wD= 1.5 kips/ft and a uniformly distributed service live load wL= 2 kips/ft applied to beam BC, answer the following questions: 1. Find the minimum area of steel required to serve as a cross section for member AB. Assume welded connections and shear lag factor U = 0.75 2. Find the lightest standard rolled W section to adequately resist the bending moments in beam BC. Assume that the top flange is laterally braced at points B, C, and at midspan. Compute the value of the bending coefficient cb.
A WT305 × 41 standard steel shape is used to support the loads shown. Assume w = 48 kN/m, LAB = 1.6 m, LBC = 3.2 m, and LCD = 1.6 m. Consider the entire 6.4 m length of the beam and determine:(a) the maximum tensile bending stress at any location along the beam (enter a positive answer), and(b) the maximum compressive bending stress at any location along the beam (enter a negative answer). (c) Calculate the reaction forces at B and E. Forces are positive upwards.
A W420x85 steel beam is fully restrained with a uniformly distributed live load of 25 kN/m. The beam has a span of 10m. Use A36 steel. Properties of W40 x 85 A = 10839 mm2 %D bf = 180 mm tf = 18 mm %3D tw = = 11 mm Ix = 310 x 106 mm4 %3D d = 420 mm (Note: Since this is a built-up section, for checking of web slenderness, use h = d - 2t;) %3D 1) Calculate the design shear strength of the section. Check if the section is adequate in shear. USE LRFD and U = 1.2D + 1.6L. %3D 2) Is the maximum deflection at the midspan of the beam due to live load adequate? 3) Calculate the maximum uniform live load that can be carried by the beam.