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 of the span. Neglecting its own weight, determine the size of the beam with the least cross-sectional area based on bending stress.
Problem 5.5 J3: An 8x14 beam of Douglas fir-larch No. 2 grade is 12’ long and has a concentrated load of 6 kips located 4ft from one end. Investigate the beam for shear.
Problem 5.5 J4: What should be the nominal cross-sectional dimensions for the beam of least weight that supports a total uniformly distributed load of 16 kips on a simple span and consists of Hem Fir No. 1 grade? Consider only the limiting shear stress.
Problem 5.6 J5: A 2x14 rafter cantilevers over a 4x12 support beam. If both members are of Hem Fir, No. 2 grade, is the situation adequate for bearing? The rafter load on the support beam is 3500#.
Problem 5.7 J6: Find the least weight section that can be used for a simple span of 14 ft with a total uniformly distributed load of 15 kips. Check bending, shear, deflection and find the required bearing length to support the beam. Only check deflection for TL, DTL = L/240
Problem 5.4 J1: No.2 grade Douglas fir-Larch is to be used for a series of floor...
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...
ing stress. Problem 5.4.B. A sin a span of 18 ft (5.40 placed at the third SAR A simple beam of Douglas fir-larch, select structural grade, has 18+ 15.49 m) with two concentrated loads of 4 kips (13.34 kN) each at the third points of the span. Neglecting its own weight, determine the the beam with the least cross-sectional area based on bending stress.
16.6a) A simply supported beam is to span 15 ft. It will support a uniformly distributed load of 2 kips/ft over the full span and a concentrated load of 60 kips at mid-span. What is the required plastic section modulus Zx? (Include self-weight) 16.6b) A simply supported beam is to span 15 ft. It will support a uniformly distributed load of 2 kips/ft over the full span and a concentrated load of 60 kips at mid-span. Deflection is not to...
A steel I-beam with a span of 20 feet will support both a concentrated load of 15 kips at the beam center and a uniformly distributed load of 2 kips/ft along the entire beam length. The steel beam is Grade 50 (Fy= 50,000 psi). i) Calculate for the lightest W-shape section that will safely carry the load, with the depth not to exceed 17 inches. Provide sketches. ii) The analysis is to include (a) bending strength, (b) shear strength, and...
Problem#4: A timber section with nominal 4 in x 12 in rectangular section is used on a simple span of 13 ft. The beam supports a uniformly distributed load of 355 lb/ft (including the weight of beam). Determine the (a) Maximum flexural stress due to bending (b) Maximum Shear stress
For the next two problems use the following information: A simply supported Douglas fir wood beam is designed to carry a concentrated load P of 1250 lbr in the center. The distance L between supports is 96 inches. For the beam cross sectional area given below, determine the moment of inertia and deflection. Douglas fir has the following properties: Modulus of Elasticity 1.76 x 100 psi, Density 34 lbm/ff3 Beam dimensions are: Web thickness tw 0.875 in, flange thickness t...
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
Problem# 1: Determine the location of the centroid. Determine the moment of inertia about horizontal and vertical cen 2" 2 6 Problem#1 : Select a solid, rectangular, Eastern hemlock beam for a 20 ft simple span carrying a superimposed uniform load of 325 lb/ft (15 points) Problem#2: Select the wide flange steel girder for a simple span of 36 ft subjected to a concentrated load of 215 kips at the midspan. Use A36 steel and assume that beam is supported...
2. A 30 ft long simply supported beam supports a uniformly distributed load of 2 kips/ft over the entire span. The beam and cross section are shown below. Draw the shear and moment diagrams, find the neutral axis location, moment of inertia of the composite section, the maximum bending stress on the cross section. (40 points) 10" 2 k/ft 1-3" 30'-0"
Problem 6: The beam is made of Douglas fir having an allowable bending stress of allow = 1.1 ksi and an allowable shear stress of Tallow = 0.70 ksi. Determine the width b if the height h = 26. (Show your SFD & BMD) 800 lb 80 lb/ft 800 lb -3 ft -+ -6 ft + 3 ft h = 2b b