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University of Technology Building and Construction Engineering Department Strength of Materials

University of Technology Building and Construction Engineering Department Strength of Materials Second Class 2014 – 2015 prepared by Professor Dr. Nabeel Al- Bayati Structural Engineering Branch. Bending Stresses in Beams. Lecture -7, Part 2. Piecewise linear normal stress variation.

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University of Technology Building and Construction Engineering Department Strength of Materials

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  1. University of Technology Building and Construction Engineering Department Strength of Materials Second Class 2014 – 2015 prepared by Professor Dr. Nabeel Al-Bayati Structural Engineering Branch

  2. Bending Stresses in Beams Lecture -7, Part 2 Lecture -7-part-2 (Bending Stresses in Beams)

  3. Piecewise linear normal stress variation. • Elemental forces on the section are • Normal strain varies linearly. Neutral axis does not pass through section centroid of composite section. • Consider a composite beam formed from two materials with E1 and E2. 7-11 Bending of Members Made of Several Materials: • Define a transformed section such that (Ratio of the moduli of elasticity). Lecture -7-part-2 (Bending Stresses in Beams)

  4. Solved Example Bar is made from bonded pieces of steel (Es = 29x106 psi) and brass (Eb = 15x106 psi). Determine the maximum stress in the steel and brass when a moment of 40 kip*in is applied. Solution Steps: Step 1: Transform the bar to an equivalent cross section made entirely of brass Step 2: Evaluate the cross sectional properties of the transformed section Step 3: Calculate the maximum stress in the transformed section. This is the correct maximum stress for the brass pieces of the bar. Step 4: Determine the maximum stress in the steel portion of the bar by multiplying the maximum stress for the transformed section by the ratio of the moduli of elasticity. Lecture -7-part-2 (Bending Stresses in Beams)

  5. To determine the location of the neutral axis, 7-12 Reinforced Concrete Beams: • Concrete beams subjected to bending moments are reinforced by steel rods. • The steel rods carry the entire tensile load below the neutral surface. The upper part of the concrete beam carries the compressive load. • In the transformed section, the cross sectional area of the steel, As, is replaced by the equivalent areanAswhere n = Es/Ec. • The normal stress in the concrete and steel Lecture -7-part-2 (Bending Stresses in Beams)

  6. Solved Example A concrete floor slab is reinforced with 5/8-in-diameter steel rods. The modulus of elasticity is 29x106psi for steel and 3.6x106psi for concrete. With an applied bending moment of 40 kip*in for 1-ft width of the slab, determine the maximum stress in the concrete and steel. Solution Steps: Step 1: Transform to a section made entirely of concrete. Step 2: Evaluate the geometric properties of the transformed section. Step 3: Calculate the maximum stresses. Lecture -7-part-2 (Bending Stresses in Beams)

  7. 7-13: Solved Examples: Example 7-21:Knowing that a beam of the cross section shown of two materials (Aluminum and steel) is bent about a horizontal axis and that the bending moment is 60 kip.ft, EAL= 10*106 psi and Es = 30*106 psi. determine the maximum stress in each material. bAL=3in Solution : 4in Transform the section to an equivalent cross section made entirely of steel Aluminum 8in steel 3in 1in 4in steel N.A. 8in steel y’ 3in Lecture -7-part-2 (Bending Stresses in Beams)

  8. Example 7-22:Knowing that a beam of the cross section shown of two materials (Wood and steel) is bent about a horizontal axis and that the (all) wood = 1200psi, (all) steel= 20000psi, EWood= 1.2*106 psi and Es = 30*106 psi. Find the allowable moment. b = 8in 3in 3in Solution : transform wood section to steel steel 1/2in steel 1 steel 2 12in Wood N.A. 8/25in =0.32in y’ steel 3 1/2in steel 6in 6in Lecture -7-part-2 (Bending Stresses in Beams)

  9. 3in 1 0.5in 13-4.9 =8.1 8.1-0.5 =7.6 13 2 N.A. 4.9-0.5 =4.4 =0.32in y’=4.9 0.5in 3 6in Lecture -7-part-2 (Bending Stresses in Beams)

  10. Home Work : Solve the previous example (7-22) by transform steel section to wood Example 7-23: For the beam shown: Draw bending moment diagram along beam Draw stress distribution due to Max Positive Moment only. where EWood= 14kN/mm2 and ESteel = 210 kN/mm2. y M=65kN.m w=10kN/m D B C A hinge y 1.5m 1.5m 4m 400mm steel 10mm Wood 360mm 10mm steel 200mm Section y-y Lecture -7-part-2 (Bending Stresses in Beams)

  11. Solution : w=10kN/m Member CD:RC= RD =10*4/2 = 20 kN MC = 0 , MD = 0, M(mid-span CD) = 10*42/8 = 20kN.m C D RC RD 4m 40kN.m w=10kN/m RA=50kN RC=20kN RD=20kN 40kN.m C A B Member ABC: M=65kN.m RA RC=20kN VB=35kN 1.5m VC=20kN 1.5m VA=50kN  MC = 0 , 40+ 65 + 10 *3 * 1.5 – RA *3 = 0, RA = 50kN x  MA = 0 , MA+ 65 - 10*3*1.5 + 20*3 = 0, MA = 40kN.m S.F.D VD=20kN Shear VA = RA = 50kN VB = 50 – 10 *1.5 = 35 VC = 50 – 10 *3 = 20 VD = 50 – 10 * 7 = -20 = RD Moment x = 0 to 1.5m MA = - 40kN.m MBA = - 40 – 10*1.5*1.5/2 + 50* 1.5 = 23.75kN.m x = 1.5m to 3m MBC = 23.75 – 65 = - 41.25kN.m MC = 0 Mmax (Positive) = +23.75kN.m +23.75kN.m +20kN.m A B C D -40kN.m - 41.25kN.m Lecture -7-part-2 (Bending Stresses in Beams)

  12. Transform steel sections to wood 6000mm 100mm 100mm Wood 2 10mm 4 4 N.A. Wood Wood 360mm 3 y’ 10mm Wood 1 200mm 3000mm 6000mm 100mm 100mm 10mm Wood 2 151.06 4 4 Wood Wood N.A. 360mm 3 y’=228.94 3000mm 10mm Wood 1 200mm Lecture -7-part-2 (Bending Stresses in Beams)

  13. 400mm A=13.634MPa A 10mm Steel B B=12.73 C=0.848 c2=151.06 C 141.06 N.A. N.A. Wood 360mm 218.94 c1=228.94 E=19.76 D D=1.317 E 10mm Steel F=20.66 F Stress distribution at point B due to max. positive moment 200mm Lecture -7-part-2 (Bending Stresses in Beams)

  14. Example 7-24: Circular sections of aluminum and steel having maximum bending moment = 80kN.m., where EAluminum= 70*103MPa and ESteel = 21*104 MPa. Find max. bending stress for steel and aluminum and draw stress distribution of the section. 300mm Aluminum Solution : Steel 150mm Transform steel sections to aluminum 300mm 300mm 300mm b=225mm Aluminum Aluminum Aluminum a=75mm R1=150 Steel N.A. Steel to Aluminum R2=75 150mm Steel to Aluminum 3*150=450mm 3*150=450mm 150mm Lecture -7-part-2 (Bending Stresses in Beams)

  15. 300mm A=26.827MPa A B B=13.413 C=40.24 Aluminum 150mm C 75mm N.A. N.A. Steel 75mm 150mm D E=13.413 E D=40.24 F F=26.827 Stress distribution of the composite section Lecture -7-part-2 (Bending Stresses in Beams)

  16. Thank you for listening Prof. Dr. Nabeel Al-Bayati Lecture -7-part-2 (Bending Stresses in Beams)

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