Select the lightest adequate W-section with a nominal depth of 14 in or less, for a 34ft. beam with a total uniformly distributed load of 1000 lb/ft. Select the beam based only on bending stress then check to see if deflection is O.K. LL Allowable deflection = L/360 TL Allowable deflection = L/240 Fb = 2/3 FY Fy = 50,000 lb/in2 LL is 70% of total load. No figure given.
Find
M max =
Sr =
Selection =
delta LL allow =
delta LL max =
and is
Deflection O.K.?
Select the lightest adequate W-section with a nominal depth of 14 in or less, for a...
Mmax, Sr, Selection, delta LL allow, delta LL max, Deflection ok? Select the lightest adequate W-section with a nominal depth of 14 in or less, for a 34ft. beam with a total uniformly distributed load of 1000 lb/ft. Select the beam based only on bending stress then check to see if deflection is O.K. LL Allowable deflection = L/360 TL Allowable deflection = L/240 Fb = 2/3 FY Fy = 50,000 lb/in2 LL is 70% of total load
Select the lightest adequate W-section with a nominal depth of 14 in or less, for a 34ft. beam with a total uniformly distributed load of 1000 lb/ft. Select the beam based only on bending stress then check to see if deflection is O.K. LL Allowable deflection = L/360 TL Allowable deflection = L/240 Fb = 2/3 FY Fy = 50,000 lb/in2 LL is 70% of total load M max = Select one: a. 4.25 k-ft b. 144.5 k-ft c. 101.1...
Razuesto Information Select the lightest adequate W-section with a nominal depth of 14 in or less, for a 34ft, beam with a total uniformly distributed load of 1000 lb/ft Select the beam based only on bending stress then check to see if deflection is OK LL Allowable deflection L/360 TL Allowable deflection L/240 Fb = 2/3 FY Fy 50,000 lb/in? LL IS 70% of total load Question 18 Noted Post of 100 M max Select one: 0.4.25 k-ft b. 144.5k...
Problem 1 (100 pts) Select the lightest W24 beam section with Fy = 50 ksi using LRFD for the following span and loading. The unbraced length of the compression flange is 30 ft (Lb = 30'). Consider Cb > 1. The given dead load does not include beam weight. Verify that the selected beam has adequate shear strength. Maximum allowable deflection due to live load is L720. Maximum allowable deflection due to total load is L/360. WD = 1.2 k/ft...
Problem 2: the most economical (lightest) W shape to support a uniformly distributed load of 4k/ft (This load includes the weight of the beam) on a simply supported span of 25 ft as shown. Assume the yield stress of the steel to be 50 ksi. The deflection limit (Allowable) is Select 360 w 4 kips/ft 25'-0" R 50 kips RB 50 kips
Please refer AISC 15th edition 2. Select the lightest W shape to carry a uniformly distributed dead load of 0.5 kips/ft and a live load of 1.0 kips/ft on a simply supported span of 42 ft. Adequate lateral support is provided. The live load deflection is limited to 360, Use A572 Grade 50 steel and LRFD. (credit weight 30)
Select a suitable wide flange section to support the beam loads shown in the figure below. Use ASTM A992 steel and allowable stresses of 0.66 Sy for bending and 0.40 Sy for shear. Be sure to generate a list of candidate beams from which, you will choose the lightest. Determine the factor of safety in bending for the beam you select. Find the deflection at the end of the overhang using the beam you selected. What is the total weight...
Please refer AISC 15th edition 2. Select the lightest W shape to carry a uniformly distributed dead load of 0.5 kips/ft and a live load of 1.0 kips/ft on a simply supported span of 42 ft. Adequate lateral support is provided. The live load deflection is limited to 360, Use A572 Grade 50 steel and LRFD. (credit weight 30)
structures homework, please answer all parts of the question thank you 2. Wood Beam Design Design a 2x dimensioned lumber floor joist to carry the given dead + live floor load (neglect joist selfweight). Assume the floor meets conditions of 4.4.1 so CL=1.0. Also Ct, Cfu, and Ci = 1.0. Find the short term deflection of your chosen beam under live load only (100% LL is short term). Compare your LL deflection with the code limit of L/360. SPAN DATASET:...
Problem 5.4 J1: No.2 grade Douglas fir-Larch is to be used for a series of floor beams 8ft on center, spanning 10ft. If the total uniformly distributed load on each beam, including the beam weight, is 3050 lb, select the section with the least cross-sectional area based on bending stress. Problem 5.4 J2: A simple beam of Hem Fir, No. 1 grade, has a span of 14ft with two concentrated loads of 5 kips each placed at the third points...