Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
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Chapter 9, Problem 9.18PFS
To determine
The lightest satisfactory
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Material properties are:
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A simplified line diagram of Girder G2 is shown below:
Determine the tributary area for each concentrated load for G-2.
30-0¹
B4
B4
69
Colum
Dea
30-0
- Mark
89
PD & PL
30 square feet
400 square feet
700 square feet
1200 square feet
90-0
BI
-Ser
PD & PL
a=
1. Determine the force in each member of the loaded truss. All triangles are 3-4-5. Indicate
whether it is compression or tension. Use method of joint.
C 40KN
B
XX
H
G
-4 panels at 8 m-
A
32 KN
22 KN
F
D
E
For the Integrated Project (office building), design the Floor Girders in the gravity-only system,
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1. Use Table 3-2 from the Steel Manual to select the most economical compact W-shape
section.
Demands:
Stories 1 and 2 Edge: 114.4 k.ft
From Table 3-2, use W12x50
D/C = should be less than 0.95: D/C =
Stories 1 and 2 Interior: 226.4 k.ft
Use W12x50: D/C =
Roof Edge: 33.3 k.ft
Use W12x40: D/C =
Roof Interior: 58.4 k.ft
Use W12x40: D/C =
Chapter 9 Solutions
Structural Steel Design (6th Edition)
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- A plate girder must be designed for the conditions shown in Figure P10.7-4. The given loads are factored, and the uniformly distributed load includes a conservative estimate of the girder weight. Lateral support is provided at the ands and at the load points. Use LRFD for that following: a. Select the, flange and web dimensions so that intermediate stiffeners will he required. Use Fy=50 ksi and a total depth of 50 inches. Bearing stiffeners will be used at the ends and at the load points, but do not proportion them. b. Determine the locations of the intermediate stiffeners, but do not proportion them.arrow_forwardb Irom Fig. 9-10 in the text. 9-17. Select the lightest satisfactory W shape section if Fy provided at the ends only. Determine C,. (Ans. W24 × 117 LRFD and ASD) 50 ksi. Lateral bracing is PL = 40 k PL = 40 k D = 1 k/ft U PES %3D 10 ft 10 ft 10 ft FIGURE P9-17 9-18. Repeat Prob. 9-17 if lateral bracing is provided at the concentrated loads as well as at the ends of the span. Determine C,.arrow_forwardFor the plate girder shown below, check the flexural strength, then shear strength on end and second panels. Lateral bracing is provided at each applied load (P). The girder is ASTM A1085 Gr. A (Fy=50 ksi). Using AISC-15th ed LRFD provisions. Pp=60 Kips, PL=30 Kips. 20" 0.375" -105" -105" -285" 105" 105-arrow_forward
- 2. Use LRFD method for the following welded shape: The flanges are 1 in × 10 in, the web is 5/16 in x 50 in, and the member is simply supported, uniformly loaded. Lateral support of the compression flange is provided at the ends and at the midspan. (a) Determine the design flexural strength if A572 - Grade 50 steel is used. (b) If the span is 34 ft and supports the loads wp = 4.5 k/ft (not including the beam weight) and w₁ = 4.5 k/ft, verify the moment capacity and select transverse stiffen- ers as needed.arrow_forwardThe given girder has beams framing into it at the ends and at every L/3 point. The beam carries a service live load of 20 kips as shown and superimposed uniformly distributed service dead load of 10 kip/ft. Select the lightest A992 W-section that can carry the load. Do not check for deflection. P, = 20 kips LL PLL W. = 10 kip/ft DL = 20 ft WDL L/3 BEAMS FRAMING INTO GIRDER Larrow_forwardA3 meter high W14 x 43 columns in a 5-Story steel braced frame is pinned at the top and bottom and has no intermediate bracing. Determine the available strength of the column. Use A992 steel. a) LRFD b) ASDarrow_forward
- A bracket of length 1.5m is rigidly attached to the simply-supported beam at its end. The beam is subjected to the combined effect of M,, M, and compressive force P caused by the factored loads as shown. There are lateral supports only at the ends. The member is made of HE300A section and steel grade is S275 (Fy-275MPA, Fu= 430 MPa). The loads acting are determined from LRFD combinations. Determine whether the member can safely carry the applied loads. Multiply first-order moment diagrams by a factor of 1.2 to account for second-order effects. No need to check shear limit state. You do not need to additionally consider the beam self-weight. E Use the following simple expression to calculate the value of L:: L, = T*rts 0.7*Fy y P = 400kN 1.5 m F = 150kN 2 т 4 т X: Lateral support НЕЗ00Аarrow_forwardA plate girder cross section consists of two flanges, 11⁄2 inchesx 15 inches, and a 5⁄16-inch 3 66-inch web. A572 Grade 50 steel is used. The span length is 55 feet, the service live load is 2.0 kips/ft, and the dead load is 0.225 kips/ft, including the weight of the girder. Bearing stiffeners are placed at the ends, and intermediate stiffeners are placed at 69-20 and 12,-9,, from each end. Does this girder have enough shear strength? a. Use LRFD. b. Use ASDarrow_forwardQ4. [TYPE C] A loaded K-truss is supported by a roller at A and a pin at G as shown. Determine the force in members LM, MK, KC, and BC of the K-truss, and state whether they are in tension (T) or compression (C), or whether it is a zero-force member (ZFM). 500 lb 2 ft 2 ft A © 2023 Shanbe Huang 3 ft B K 3 ft M N C 3 ft D 3 ft P 3 ft H E F 3 ft Garrow_forward
- Tutorial 7 Вeams 1. A simply supported beam must carry the factored ultimate loads (two 15kN point loads) shown in the figure. Do not consider the self-weight of the beam. The beam is laterally and torsional supported only at A, B, C and D. For a 203 x 133 x 25 I- section: (Grade 350 W steel) (a) Determine the class of the section. (b) Determine whether the section has sufficient flexural resistance. 15 kN 15 kN 4,0 m 2,0 m 4,0 marrow_forward0O REDMI NOTE 8 PRO O AI QUAD CAMERA Q - 4. For the given truss structure below; BY USING THE METHOD of SECTIONS; determine the internal forces of the elements BD, CD & CE. Also determine whether these forces are TENSILE or COMPRESSIVE. 3 in 3 m -4 m -4 m -4 m- 1.75 kN 2KN 1.5 kN 2kN I kN 9.75 kN G 6 m 6 m 6m 6 marrow_forward5.5-3 A simplrted beam is subjected to a uniform service dead load of 1.0 kip cluding the weight of the beam), a uniform service live load of 0 kipft, and a concentrated service dead load of 40 kips. The beam feet long, and the concentrated load is located 15 feet from the left The beam has continuous lateral support, and A572 Grade 50 steel is used. Is a W30 x 108 adequate a. Use LRFD. b. Use ASD.arrow_forward
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