Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
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Chapter 11, Problem 11.11PFS
To determine
To check for the given member
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Use NSCP 2015. The beam shown has continuous lateral support
of both flanges. The uniform load consisting of 50% dead load
and 50% live load. The dead load includes the weight of the
beam. Used grade 50 steel.
6 k/ft
|--0-0²-
-18'-0"-
fo
4-6-
-6'-0"
1. Considering LRFD. Which of the following most nearly gives
the design load Wu?
2. Considering LRFD. Which of the following most nearly gives
the maximum negative design moment, Mu in ft-kips?
3. Considering W12x35 beam section, The section is
4. Considering ASD the W12x35 beam section, Which of the
following most nearly gives allowable strength in ft-kips
A W36x210 can be used for Transfer Girder ‘A’ as shown in the elevation next page.Assume the LRFD loads on the ‘Transfer Girder’ are as follows:Pu= 200 kips from the column above. Assume also that the loads from self-weight of thewu = 2.0 kip/ft (girder self weight and beams framing into this transfer girder are equivalentto a uniformly distributed load). These loads are shown in the last figure.a) Check whether the W36x210 ‘Transfer Girder’ is adequate for bending and shear.Assume the unbraced length Lb =20 ft.b) Suppose that service loads, w and P, are both 50% dead load and 50% live load. Does theW36x210 satisfy a service live load deflection limit of L/480?
a) Name one benefit of a moment frame over a braced frame and name one benefit of a braced frame over a moment frame.
b) Name and describe the two major failure modes of a beam-column member comprised of a W-section.
c) What is the major difference of a beam-column from a beam or a column?
d) Why are moment magnifications needed for beam-columns? Briefly explain each of the two.
Chapter 11 Solutions
Structural Steel Design (6th Edition)
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- The frame shown below is subjected to uniformly distributed load of 2.6KP.. Determine the maximum moment of member BF (assume simple beam support) A E, 1.5m F 1.5m 351 C 1.5m H 3m Barrow_forwardPlease ans A and Darrow_forward1. According to the requirements of the AISC-Specifications for doubly symmetric members and singly symmetric members, the interaction of flexure and compression shall be limited by Equations: a) H1 -1a and H1 -1b. b) El-la and E1 -1b. c) G1 -la and G1 -1b. d) F1-la and F1 -1b. e) C1 -la and C1 -lb. 2. The beams framing into columns are commonly supported with framing angles or brackets on the sides of the columns. These eccentrically applied loads produce: a) Axial Tension. b) Shear forces. /c) Bending moments. d) Axial compression. e) Torques. 3. The required total flexural strength of a beam-column member must at least equal the sum of the: a) first-order moments only. b) Second-order moments only. c) first-order and second-order axial loads. d) first-order and second-order moments. e) Non the above. 4. The first-order moment using LRFD load combinations, due to the lateral translation of the structure only, is referred to as: a) Mnt b) Mit c) Mr d) Pnt e) Pitarrow_forward
- The two steel beams shown in Fig. P9-30 are part of a two-span beam framing system with a pin (hinge) located 4.5 ft left of the interior support, making the system statically determinate. Determine the sizes (lightest) of the two W shape beams. Assume A992 steel and continuous lateral support of the compression flanges.The beam self-weight may be neglected. Use LRFD and ASD methods.arrow_forwardpls helparrow_forwardPlease derive the three-moment factors of the following. Show complete solution:arrow_forward
- A two-story frame, is braced in y axis (Ky-1) and is connected as moment frame for x axis as shown below. Assuming elastic behavior, and all steel has Fy-50ksi, determine following for column BF. a) Kx for column. b) The governing buckling axis. c) Failure mode in column (Inelastic or elastic global buckling). d) Design strength of the column according to LRFD method (Pn). e) If the column is adequate for the service dead load of 180K and live load of 240K. Use the kx value obtained in part a and select lightest HSS rectangular to replace W12x72. Compare the weight of selected section f) E A W18 X 35 W18 X 35 F 20' B J W12 X 50 W12 X 72 W18 X 40 W18 X 50 30' G K 30' H 20¹ 15'arrow_forwardB4arrow_forwardThe steel frame (E=200 GPa) shown has a diagonal brace BD with an area of 1920 mm2. Determine thelargest allowable load P if the change in length of member BD is not to exceed 2.5 mmarrow_forward
- Find accurately length of the three beam in figure belowarrow_forwardUse proper method ASimply supported one way slab is reinforced with 0.5% tension steel and has effective depth of 130 mm for a span of 3.5. Using D. A. - Sp 16, check the slab for deflection. Use M 20, Fe 415arrow_forwardUse A992 steel, select a W-shape for the upper girder that spans left to right and picks up the loading from the three floor beams. Assume the beam is simply supported. The maximum permissible live load deflection is L/360. The service dead loads consist of a 5-inch thick reinforced concrete floor slab (normal weight concrete), a partitionload of 25 psf, and 20 psf to account for suspended ceiling and mechanical equipment. Don’t forget to check with self-weight. The service live load is 100 psf. You may use AISC Table 3-2 and/or Table 3-10 to check for bending and shear. Use LRFD only. Take Cb = 1.0. Please show all work to verify that this beam is adequate for bending, shear and deflectionarrow_forward
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