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Copyright Joseph Greene 2003 All Rights Reserved 1 CM 197 Mechanics of Materials Chap 15: Design of Beams for Strength Professor Joe Greene CSU, CHICO.

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Presentation on theme: "Copyright Joseph Greene 2003 All Rights Reserved 1 CM 197 Mechanics of Materials Chap 15: Design of Beams for Strength Professor Joe Greene CSU, CHICO."— Presentation transcript:

1 Copyright Joseph Greene 2003 All Rights Reserved 1 CM 197 Mechanics of Materials Chap 15: Design of Beams for Strength Professor Joe Greene CSU, CHICO Reference: Statics and Strength of Materials, 2 nd ed., Fa-Hwa Cheng, Glencoe/McGraw Hill, Westerville, OH (1997) CM 197

2 Copyright Joseph Greene 2003 All Rights Reserved 2 Design of Beams for Strength Topics –Introduction –Basic Considerations in Beam Design –Design of Steel Beams –Design of Timber Beams –Design of Shear Connectors

3 Copyright Joseph Greene 2003 All Rights Reserved 3 Introduction Beam problems involve analysis and design of beam. Assume prismatic beam in this chapter –Straight beams with uniform cross section Given: span length, supporting conditions, and loading Selected: shape and size of the beam Design of beam includes –Selection of materials –Type of sections –Calculations of bending and shear strength (This chapter) –Calculations of beam deflections (Next chapter), lateral support of web flange, web crippling and support details Design approaches: Allowable stress (This chapter)_and Ultimate strength (Next chapter)

4 Copyright Joseph Greene 2003 All Rights Reserved 4 Basic Considerations in Beam Design Framing layout –Fig 15-1 illustrates two common methods of framing layout Part A: center-point concentration and Part B: third point concentration –When the area enclosed by the columns (bay) is not a square, the usual custom is to run the beams in the long direction, with the girders on the shorter span. Other layouts may be designed and compared with the cost. –Area of floor supported by a beam is equal to the span length multiplied by the sum of half the distance to the adjacent beams. Reactions on the beams are supplied to the girder as concentrated loads. Loads on floor supported by the beams can be: –Dead load consists of the weight of the walls, partitions, columns, floors, and roofs. Should include weight of the beam –Live load consists of the effect created by occupancy, e.g., weight of people, furniture, equipment, stored materials, etc. Building codes provide minimum live loads to be used in the design of various types of buildings Table 15-1. Minimum Live Loads

5 Copyright Joseph Greene 2003 All Rights Reserved 5 Basic Considerations in Beam Design Beam materials: Most common materials are steel and wood Beam Sections –Rolled steel shapes Most manufactured steel beams are produced by rolling a hot ingot of steel until steel shape is formed. Properties are in AISC manual –Built-up steel shapes Rolled shape is added to with additional plates or other shapes. Fig 15-2 –Timber sections Most have rectangular shapes and identified as 2x4, 4x6, 2x8, etc. Dressed surface of lumber is ½ in or ¾ in smaller in dimension than say 2x4. –Built-up timber sections Have a box shape or T-section for added stiffness. Fig 15-3. –Lateral bracing Braces are added to stiffen sections. Floor beams are arranged in bracing layout. –Web crippling Web is reinforced with stiffeners to prevent crippling. –Beam-bearing plates Walls or pier supports of beams usually rest on steel bearing plates to provide ample bearing area.

6 Copyright Joseph Greene 2003 All Rights Reserved 6 Design of Steel Beams Allowable stresses in beams are specified in design codes. –AISC specs for beams and allowable stresses,  allow,are Flex stress = 66% of yield stress;  allow = 0.66  y Shear stress = 40% of yield stress;  allow = 0.4  y –Example, structural steel, yield stress,  y,is 36 ksi (250 MPa) Flex stress = 66% of yield stress;  allow = (0.66) 36ksi = 24ksi Shear stress = 40% of yield stress;  allow = (0.4) 36ksi = 14.5ksi (100MPa) Procedure to design steel beams –Step 1: Determine the beam span, support conditions, allowable stresses. Compute external loads. Compute shear and moment diagrams. –Step 2: Determine the maximum shear force and maximum bending moment along beam. For simple loading condition use Table 13-1 For complex loadings use calculations form Chapter 14

7 Copyright Joseph Greene 2003 All Rights Reserved 7 Design of Steel Beams Procedure to design steel beams –Step 3: Compute minimum required section modulus from flexure formula –Step 4: Scan tables in the appendix, Table A-1, and List several possible choices of W shape that have a section modulus, S, greater than the required value, S desired. Select one with the lightest weight per foot and make sure it is greater than the desired S desired. –Step 5: Calculate shear stress on beam from Eqn 14-13 or 15-2 Check to make sure the calculated shear stress is much lower than  allow Weight of beam is usually neglected. Examples, 15-1, 15-2, 15-3, 15-4 15-1 15-2

8 Copyright Joseph Greene 2003 All Rights Reserved 8 Design of Timber Beams Allowable stresses in beams vary from wood grades. –Allowable stresses are in Table 15-2. Note: wood is not isotropic. Properties are different in different directions –Tension/compression: Wood is stronger parallel to grain than perpendicular –Shear: Wood is weaker parallel to grain than perpendicular. –Shear dictates selection of beam material Procedure to design timber beams –Step 1: Determine the beam span, support conditions, allowable stresses. Compute external loads. Compute shear and moment diagrams. –Step 2: Determine the maximum shear force and maximum bending moment along beam. For simple loading condition use Table 13-1 For complex loadings use calculations form Chapter 14

9 Copyright Joseph Greene 2003 All Rights Reserved 9 Design of Timber Beams Procedure to design steel beams –Step 3: Using largest value of the bending moment, regardless of sign, Compute minimum required section modulus from flexure formula –Step 4: Calculate the minimum required cross sectional area based upon the max shear stress is 1.5 times the average shear stress –Step 5: Scan tables in the appendix, Table A-1, and Select the lightest rectangular timber section that has –Section modulus slightly greater than the required value (Step 3) and –Area slightly greater than the required value (Step 4) Compute the percentage of the extra section modulus provided Compute the percentage of the extra area provided and Calculate the ratio of weight of beam to total design load. –Note: narrow and deep timber beams are more effective than wide and shallow beams in resisting bending moments. –Note: preferred aspect ratio (depth:width) is between 1.5 and 3. –Note: For closely spaced joints, preferred aspect ratio is between 3 and 6 –Examples, 15-5, 15-6, 15-7 15-3 15-4

10 Copyright Joseph Greene 2003 All Rights Reserved 10 Design of Shear Connectors Introduction –Beams are sometimes fabricated by joining several component parts to form a single section. Three typical examples of built-up beams are in Fig 15-4 –T-beam fabricated by nailing 2 timber planks. –Plywood and boards nailed together to form a box. –Fabricated steel beam from a steel channel bolted to »Top flange of a W shape –Concentrated forces Fig 15-5. T-beam with concentrated load at midspan. –Shear force is constant throughout beam. –Nails used to connect 2 planks are spaced at constant pitch p. »Each nail is required to carry the shear force on contact surface for each pitch length. »Longitudinal shear stress at contact level 1-1 is 11 NA y t P P/2 L/2 p V P/2 -P/2 Fig 15-4 a Fig 15-4 b Fig 15-4 c

11 Copyright Joseph Greene 2003 All Rights Reserved 11 Design of Shear Connectors Shear force on each nail –First moment of the area, Q= A’y’, of the plank on the top about the neutral axis. Shear stress is distributed uniformly in the contact surface (area = p x t) Total shear force on each nail is the area times shear stress. –Shear flow, q is VQ/I in Eqn 15-6 »the shear force in longitudinal section per unit length of beam. Maximum allowable shear force on each nail is Eqn 15-7 –Equation 15-7 can be used to determine the pitch (spacing) of the nails in Fig 15-4 a and b –Where, »Q = first moment of the area of board A »F s,allow = allowable shear force of 2 nails –In Fig 15-4 c »Q = first moment of the area of steel channel about neutral axis »F s,allow = allowable shear force of 2 bolts Example, 15-8 and 15-9 15-515-615-7


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