Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN: 9781337094740
Author: Segui, William T.
Publisher: Cengage Learning
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Chapter 4, Problem 4.9.9P
To determine
(a)
To design:
Strength for LRFD.
To determine
(b)
The design Strength.
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A W12x79 of A573 Grade 60 (Fy=415 MPa) steel is used as a compression member. It is 8 m long, pinned at the top fixed at the bottom, and has additional support in the weak direction at mid-height. Properties of the section are as follows:
A = 14,500 mm^2
Ix = 258.6 x 10^6 mm^4
Iy = 84.375 x 10^6 mm^4
Calculate the effective slenderness ratio with respect to strong axis buckling using theoretical value of k.
The 350×90 section is used as column 10m long. The column is fixed at one end and hinge at the other end. Brace at mid-height about the weak axis (y-axis). W350×90
GIVEN: bf 250mm,d=350mm, Ix=266×10^6mm^4, Iy=44.54×10^6mm^4.
Determine the critical effective slenderness ratio of the column. And determine the Pn. Using NSCP 2015.
An unsymmetrical flexural member consists of a ½ x 12 top flange, a ½ x 7 bottom flange, and a 3/8 x 16 web.
b.) If A572 Grade 50 is used, what is the plastic moment Mp for the horizontal plastic neutral axis (already answered)
QUESTION: SHOW THE SOLUTION FOR GETTING THE y for the top and bottom flange and the web. NOT THE ȳ . and draw the figure for better understanding. SOLVE THE CIRCLED RED PARTS OF THE IMAGE.
Chapter 4 Solutions
Steel Design (Activate Learning with these NEW titles from Engineering!)
Ch. 4 - Prob. 4.3.1PCh. 4 - Prob. 4.3.2PCh. 4 - Prob. 4.3.3PCh. 4 - Prob. 4.3.4PCh. 4 - Prob. 4.3.5PCh. 4 - Prob. 4.3.6PCh. 4 - Prob. 4.3.7PCh. 4 - Prob. 4.3.8PCh. 4 - Prob. 4.4.1PCh. 4 - Prob. 4.4.2P
Ch. 4 - Prob. 4.6.1PCh. 4 - Prob. 4.6.2PCh. 4 - Prob. 4.6.3PCh. 4 - Prob. 4.6.4PCh. 4 - Prob. 4.6.5PCh. 4 - Prob. 4.6.6PCh. 4 - Prob. 4.6.7PCh. 4 - Prob. 4.6.8PCh. 4 - Prob. 4.6.9PCh. 4 - Prob. 4.7.1PCh. 4 - Prob. 4.7.2PCh. 4 - Prob. 4.7.3PCh. 4 - Use A992 steel and select a W14 shape for an...Ch. 4 - Prob. 4.7.5PCh. 4 - Prob. 4.7.6PCh. 4 - Prob. 4.7.7PCh. 4 - The frame shown in Figure P4.7-8 is unbraced, and...Ch. 4 - Prob. 4.7.9PCh. 4 - Prob. 4.7.10PCh. 4 - Prob. 4.7.11PCh. 4 - Prob. 4.7.12PCh. 4 - Prob. 4.7.13PCh. 4 - Prob. 4.7.14PCh. 4 - Prob. 4.8.1PCh. 4 - Prob. 4.8.2PCh. 4 - Prob. 4.8.3PCh. 4 - Prob. 4.8.4PCh. 4 - Prob. 4.9.1PCh. 4 - Prob. 4.9.2PCh. 4 - Prob. 4.9.3PCh. 4 - Prob. 4.9.4PCh. 4 - Prob. 4.9.5PCh. 4 - Prob. 4.9.6PCh. 4 - Prob. 4.9.7PCh. 4 - Prob. 4.9.8PCh. 4 - Prob. 4.9.9PCh. 4 - Prob. 4.9.10PCh. 4 - Prob. 4.9.11PCh. 4 - Prob. 4.9.12P
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- A W12x79 of A573 Grade 60 (Fy = 415 MPa) steel is used as a compression member. It is 7 m long, pinned at the top fixed at bottom, and has additional support in the weak direction 3 m from the top. Properties of the section are as follows: A = 14,500 mm^2 Ix = 258.6 x 10^6 mm^4 Iy = 84.375 x 10^6 mm^4 1. Calculate the critical slenderness ratio of the member. a. 26.208 b. 45.882 c. 36.706 d. 27.529 2. calculate the nominal axial load capacity of the column a. 3104 kN b. 4851 kN c. 4213 kN d. 5344 kN 3. calculalte the service axial dead load if the service axial live load is twice as that of the dead load. Use LRFD. a. 1354 kN b. 992 kN c. 1093 kN d. 634 kNarrow_forward1. An unsymmetrical flexural member consists of a 70 × 600 top flange, a 70 x 400 bottom flange, and a 12 × 180 web. a, Determine the Section Modulus. b. Determine the distance from the bottom of the shape to the horizontal plastic neutral axis. ( Take note: Bottom Distance i need, not the top distance) c. If A572 Grade 50 steel is used, what is the plastic moment MPy for the horizontal plastic neutral axis?arrow_forwardCheck the adequacy of the column shown in red colour with the 200UC59.5 section. The material is 300 Plus. Dead and live loads for all floors are the same as shown in Figure 1. All the support conditions are shown in the Figure. For bending about the minor axis, y, there is an intermediate secondary member that prevents its buckling and the column is simply supported. In case of buckling about the major axis, x, there is no intermediate lateral support and the column is pinned at the top and fixed at the bottom.arrow_forward
- Two structural steel angles back to back are used as a compression member that is 4.5 m long. The angles are separated at intervals by spacer blocks and connected by bolts as shown in Figure Q4, a configuration which ensures that the double-angle shape acts as a unified structural member. Assume the pinned connections at each end of the column, and use E=200 GPa for the steel. A single angle has the following properties, A = 1600 mm, Ix = 1.64 x 106 mm4, Iy =0.787 x 106 mm4, x = 19.7. Determine the Euler buckling load for the double-angle column if the spacer block thickness is 5 mm (10 marks) 20 mm (10 marks)arrow_forwardA built-up section consisting of W 350 x 90 with two 12 – mm plates welded to form a box section as shown in figure. The section is used as a column 10 meters long. The column is fixed at both ends, and braced at mid height about the weak axis (Y-axis). Use Fy = 248 MPa. a. Determine the effective slenderness ratio of the column with respect to lateral buckling about the x-axis. b. Determine the effective slenderness ratio of the column with respect to lateral buckling about the y-axis. c. Determine the axial load capacity of the column in kN.arrow_forwardA W12x79 of A573 Grade 60 (Fy = 415 MPa) steel is used as a compression member. It is 8 m long, pinned at the top fixed at bottom. The properties are as follows: Ag = 14,968 sq.mm. rx= 133.65 mm, Ty = 77.47 mm 1. Calculate the critical slenderness ratio if there is an additional lateral support in the weak direction at mid-height. a. 36.14b. 41 90c. 103.39d. 51.63 2. If the columns sustains a service axial live load of 990 kN calculate the safe service axial dead load in kN by LRFD. a. 1,578b. 1,136c. 1,620d. 1,827 3. Calculate the allowable axial strength (kN) by ASD. a. 3,777b. 2,348c. 2,620d. 3,529 4. Calculate the design axial strength (kN) of the colurn by LRFD.arrow_forward
- A W530x92 supports a uniformly distributed load of 12 kN/m (including the weight of the beam) and equal concentrated load P at quarter points (L/4, L/2, and 3L/4). The compression flange is laterally supported at quarter points and at the ends. The beam is 12m simply supported. Using A36 steel, the beam has the following properties: d = 533 mm bf = 209 mm tf = 15.6 mm tw = 10.2 mm k = 32 mm ry = 44.9 mm Sx = 2070 x 10^3 mm^3 Zx = 2360 x 10^3 mm^3 Es = 200,000 MPa Ix = 552 x 10^6 MPa Find the maximum allowable load P so that it will not exceed its maximum allowable moment capacity. Use conservative assumption for Cb.arrow_forward1. A W8x15, A36 steel is used as a simply supported beam uniformly loaded over a span of 5.9 m. If the top flange is continuously braced against lateral displacement by the floor, calculate the ultimate uniform load the beam can sustain against bending about the major axis. Write your answer in kN-m with 2 decimal places only. Use Zx = 223 x 103 mm3 2. A W8x31, A36 steel is used as a simply supported beam uniformly loaded over a span of 5.8 m. If the top flange is continuously braced against lateral displacement by the floor, calculate the allowable uniform load the beam can sustain against bending about the major axis by ASD. Write your answer in kN-m with 2 decimal places only. Use Zx = 498 x 103 mm3 PLEASE ANSWER THESE TWO BEFORE I GIVE POSITIVE FEEDBACK. THANKS!arrow_forwardThe beam-column in the figure below is a member of a braced frame. A second-order analysis was performed with factored loads and reduced member stiffnesses to obtain the moments and axial force shown. Use LRFDand determine whether this member is adequate. Suppose that P = 165 kFor W10 x 60 with L, = 15 ft and C6 = 1.0: ф,M, = 257 ft-kips, ф,M, = 280 ft-kips;for Lc = 15 ft: cPr = 556 kips.arrow_forward
- Situation 17. A w12x79 of A573 Grade 60 (Fy-415MPa) steel is used as a compression member. It is 8m long pinned at the top fixed at bottom with additional lateral support at mid height in the weak direction. The properties are as follows Ag=14,500 sq.mm = 258.6x10^6 mm^4 Iy=84.375x10^6 mm^4 Calculate the critical slenderness ratio with buckling about strong axis. calculate the critical slenderness ration with buckling about weak axis. calculate the flexural buckling stress Fcr in MPAarrow_forward12 A W12x79 of A573 Grade 60 (Fy = 415 MPa) steel is used as a compression member. It is 6.8 m long, fixed at the top and bottom, and has additional support in the weak direction at mid-height. Properties of the section are as follows: A = 14,500 mm^2 Ix = 258.6 x 10^6 mm^4 Iy = 84.375 x 10^6 mm^4 Calculate the critical slenderness ratio with respect to weak axis buckling using theoretical value of k.arrow_forwardDetermine the plastic moment strength (kN-m) of a C15x50, A36 steel, about the minor axis d = 381 mm bf = 94.39 mm tw = 18.19 mm tf = 16.51 mmarrow_forward
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