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 7, Problem 7.11.4P
To determine
(a)
The maximum service load using Load and Resistance Factor Design (LRFD).
To determine
(b)
The maximum service load using Allowed Strength Design (ASD).
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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.
A channel C250x37 mm section is welded to a 9 mm gusset plate. Welding is not permitted on the back of the channel. All steel is A36 with Fy=250 MPa and Fu=400 MPa. Use E70electrodes having and Fu=485 MPa (SMAW) process. The maximum length of lap is 250mm. The size of fillet weld is 8mm. Assume the width of slot weld is 22 mm. Size of slot weld is 13mm
Properties of C250x37
A = 4750 mm2
tw = 13.0 mm2
d = 254 mm
a. Determine the force resisted by the slot weld in kN, when the full tensile capacity is 712.5 KN (from the gross yielding capacity using ASD)
Hint: Full tensile Capacity = Force Resisted by Fillet and Slot Weld
Round your answer to 3 decimal places.
An I-section bracket is connected to the column as shown. The size of weld is 6 mm on web and 10 mm on flange. What will be the safe load (in kN) that can be carried by
the connection? Assume NA for whole weld = 110.57 x 106 mm4. Use Fe410 grade steel.
240 mm
170 mm
P
unu 00t *
280 mm
unu 00t
280 mm
Chapter 7 Solutions
Steel Design (Activate Learning with these NEW titles from Engineering!)
Ch. 7 - Prob. 7.3.1PCh. 7 - Prob. 7.3.2PCh. 7 - Prob. 7.4.1PCh. 7 - Prob. 7.4.2PCh. 7 - Prob. 7.4.3PCh. 7 - Prob. 7.4.4PCh. 7 - Prob. 7.4.5PCh. 7 - Prob. 7.4.6PCh. 7 - Prob. 7.6.1PCh. 7 - Prob. 7.6.2P
Ch. 7 - Prob. 7.6.3PCh. 7 - Prob. 7.6.4PCh. 7 - Prob. 7.6.5PCh. 7 - Prob. 7.6.6PCh. 7 - Prob. 7.7.1PCh. 7 - Prob. 7.7.2PCh. 7 - Prob. 7.7.3PCh. 7 - Prob. 7.8.1PCh. 7 - Determine the adequacy of the hanger connection in...Ch. 7 - Prob. 7.9.1PCh. 7 - Prob. 7.9.2PCh. 7 - Prob. 7.9.3PCh. 7 - Prob. 7.9.4PCh. 7 - Prob. 7.9.5PCh. 7 - Prob. 7.11.1PCh. 7 - Prob. 7.11.2PCh. 7 - Prob. 7.11.3PCh. 7 - Prob. 7.11.4PCh. 7 - Prob. 7.11.5PCh. 7 - Prob. 7.11.6PCh. 7 - Prob. 7.11.7PCh. 7 - Prob. 7.11.8PCh. 7 - Prob. 7.11.9PCh. 7 - Prob. 7.11.10P
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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.arrow_forwardProblem 7: The load that will be applied to the connection shown has a live load - to - dead load ratio of 3.0. Investigate all limit states. All structural steel is A36, and the weld is a 1/4-inch fillet weld with E70 electrodes. Note that the tension member is a double-angle shape, and both of the angles are welded as shown. Use ASD. Determine the following. 5" 5" 2L5 X 32 X 5/16 LLBB -t = ³/8" Maximum service load that can be applied without exceeding the allowable capacity on yielding on gross area of the tension member (double angle). Maximum service load that can be applied without exceeding the allowable fracture on the net area of the tension member (double angle). Maximum service load that can be applied without exceeding the allowable block shear strength. Considering the weld metal and base metal strength, calculate the maximum service load that can be applied.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. Longitudinal welds bf Given: Properties of WT12 x 38: Ag = 11.2 in² y = 3.0 in. bf = 8.99in. Use A992 Steel: Fy = 50 ksi Fu = 65 ksi LL = 3 DL tw WT12 x 38 y = centroidal distance C. What is the Governing Ultimate Tensile Capacity based on Net Fracture Round your answer to 3 decimal places. Tarrow_forward
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