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 10, Problem 10.6.1P
To determine
(a)
The maximum factored concentrated load that can be supported by using LRFD.
To determine
(b)
The maximum service concentrated load that can be supported by using ASD.
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The details of an end bearing stiffener are shown in Figure . The stiffener plates are 9⁄16-inch thick, and the web is 3⁄16-inch thick. The stiffeners are clipped 1⁄2 inch to provide clearance for the flange-to-web welds. All steel is A572 Grade 50. a. Use LRFD and determine the maximum factored concentrated load that can be supported. b. Use ASD and determine the maximum service concentrated load that can be supported.
The truss below is pin connected at A and E, and is acted on by the forces shown.
E
A
D
B
Identify all of the ZERO-FORCE MEMBERS by checking the boxes below (if there are
none, leave all boxes unchecked):
BF
AF
BC
BH
CD
EG
-GH
AB
CH
DG
DH
DE
-FH
Determine the maximum service load, P, that can be applied if the live load-to-dead load ratio is 2.0. Each component is a PL 3⁄4 x 7 of A242 steel. The weld is a 1⁄2-inch fillet weld, E70 electrode. a. Use LRFD. b. Use ASD.
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Steel Design (Activate Learning with these NEW titles from Engineering!)
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- Select an American Standard Channel shape for the following tensile loads: dead load = 54 kips, live load = 80 kips, and wind load = 75 kips. The connection will be with longitudinal welds. Use an estimated shear lag factor of U = 0.85. (In a practical design, once the member was selected and the connection designed, the value of U would be computed and the member design could be revised if necessary.) The length is 17.5 ft. Use Fy=50 ksi and Fu=65 ksi. a. Use LRFD. b. Use ASD.arrow_forwardCompute the maximum acceptable tensile SERVICE LOAD that may act on a single tee section that is connected to a gusset plate using welds 12 inches long, as shown in the figure. The service live load is three times the dead load. Use A992 steel. USE LRFD ONLY, no block shear will occur. WT12 x 38 Longitudinal welds 11.2 in? y = 3.0 in. Given: Properties of WT12 × 38: Ag = Use A992 Steel: F, = 50 ksi Fu = 65 ksi bf = 8.99in. %3D %3D LL = 3 DL %3D %3D tw y = centroidal distance bf C. What is the Governing Ultimate Tensile Capacity based on Net Fracture Round your answer to 3 decimal places.arrow_forwardSOLVE FOR1 HRarrow_forward
- Compute the maximum acceptable tensile SERVICE LOAD that may act on a single tee section that is connected to a gusset plate using welds 12 inches long, as shown in the figure. The service live load is three times the dead load. Use A992 steel. USE LRFD ONLY, no block shear will occur. What is the Maximum Tensile SERVICE Load Round your answer to 3 decimal places.arrow_forwardA W14X120 is used as a tension member in atruss. The flanges of the member are connected to a gusset plate by 3/4 inch boltas shown below. Use A36 steel with Fy-36 ksi and Fu=58 ksi Determine the Yielding Capacity of the section based on LRFD (kips) Determine the Tensile Rupture capacity of the section based on LRFD Determine the Demand to Governing Capacity Ratio (based on yielding and rupture only) if the Demand load carried by the section are DL=200 kips LL=400 kips use LRFD Properties and Dimension Ag=35.30 in^2 x = 6.24 in ry= 3.74 in d=14.5 in tf=0.94 in bf=14.7 in tw=0.59 in k=1.54 d=14.5 Y k1=1.5 bf=14.7 tf-0.94 X -tw=0.59 Harrow_forwardAsap plsarrow_forward
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