DESIGN OF BEAMS AND SHAFTS
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1 DESIGN OF EAMS AND SHAFTS! asis for eam Design! Stress Variations Throughout a Prismatic eam! Design of pristmatic beams! Steel beams! Wooden beams! Design of Shaft! ombined bending! Torsion 1
2 asis for eam Design - The effects of an internal axial force are often neglected in design. - Application of shear and flexure formulas are limited to beams made of a homogeneous material that has linear-elastic behavior. - The cross-sectional area must have an axis of symmetry in the plane of the loading. - The design does provide an adequate means of obtaining both a safe and economical design. 2
3 Stress Variations Throughout a Prismatic eam P w A R A - 2 m 2 m R V M R A R A R A - P - M max x -R x V M τ 4 τ 3 τ 2 Shear stress distribution (τ) σ 5 σ 4 σ 5 σ 2 ending stress distribution (σ) 3
4 V M τ 4 τ 3 τ 2 τ max σ 5 σ 4 τ σ 2 σ 5 5 y x σ 5 σ 1 x (-σ 5, 0) σ 2 σ y (0, 0) 4 y x σ 4 x (-σ 4, τ 4 ) σ 1 τ max τ max τ σ 2 σ τ 4 τ max τ y (0, -τ 4 ) x (0, τ 3 ) 3 y x σ 1 σ 2 σ τ 3 y (0, -τ 3 ) 4
5 V M σ 5 τ 4 σ 4 τ 3 τ 2 2 σ1σ τ τ max x (σ 2, τ 2 ) 2 y x σ 2 σ 1 σ 2 σ τ 2 y (0, -τ 2 ) τ max τ τ max 1 y x σ 1 σ 1 y (0, 0) σ 2 σ x (σ 1, 0) τ max 5
6 Prismatic eam Design Section Modulus (S) σ Mc I M I / c M S S req' d M σ allow 6
7 Example 1 A beam is to be made of steel that has an allowable bending stress of σ allow 150 MPa and an allowable shear stress of τ allow 100 MPa. Select an appropriate W shape that will carry the loading shown. 200 kn 100 kn 2 m 2 m 2 m 7
8 200 kn 100 kn 50 kn 250 kn 2 m 2 m 2 m V(kN) 50 M(kN m) x(m) x (m) Allowable normal stress: σ allow 150 MPa σ S req d M S M σ allow (200x10 3 ) 1333x10 3 mm 3 150x10 6 Using the table in Appendix, the following beams are adequate: W 610x82 S 1870x10 3 mm 3 W 460x97 S 1910x10 3 mm 3 W 460x89 S 1770x10 3 mm 3 W 460x82 S 1610x10 3 mm 3 W 460x74 S 1460x10 3 mm 3 W 410x85 S 1510x10 3 mm 3 Use W 460x74 because the beam having the least weight. 8
9 200 kn 100 kn W 460x74 : A 9460 mm 2, d 457 mm, t w 9.02 mm, b f 190 mm, t f 14.5 mm, I 333x10 6 mm 4, 50 kn 250 kn 2 m 2 m 2 m V(kN) x(m) heck allowable shear stress: τ allow 100 MPa Q (221.25)(190x14.5) (214/2)(214x9.02) τ max x10-6 m 3 VQ It V(816.08x10-6 ) (333x10-6 )( ) 32.6 MPa <100 MPa, O.K V (271.7 )(150x10 3 ) 9
10 200 kn 100 kn heck shear and normal stress at 50 kn 250 kn 2 m 2 m 2 m 200 kn m σ max MPa τ max V(kN) 50 M(kN m) MPa MPa kn Β x(m) σ, MPa τ I 333x10 6 mm 4 x (m) - ending: σ σ - Shear: τ τ My V max Q It (150x10 3 )[(0.2213)(0.19x0.0145)] (333x10-6 )( ) I σ τ (-200x10 3 )(0.214) (333x10-6 ) MPa Q MPa 10
11 30.44 MPa y x MPa σ average σ x σ y MPa 2 2 R σ x - σ y ( ) (τ xy ) 2 2 (64.27) 2 (30.44) 2 ± MPa σ avg MPa (-64.27, 71.11) τ σ - R average (-135.4, 0) 2 (-128.5, ) x R O y (0, 30.44) 1 (6.84, 0) σ σ average R τ max MPa < 100 MPa, O.K. σ max MPa < 150 MPa, O.K. (-64.27, ) (σ x - σ y )/2 (-128.5)/
12 omments Q ya Q (221.25)(190x14.5) (214/2)(214x9.02) mm x10-6 m τ max VQ It V(816.08x10-6 ) (333x10-6 )( ) 271.7V τ avg V max A web V ( x0.0145)( ) V 4.7 % off τ avg V max A web V (0.457)( ) V 10.7 % off Note: dimension in mm I 333x10 6 mm 4, 12
13 Example 2 The laminated wooden beam shown supports a uniform distributed loading of 12 kn/m. If the beam is to have a height-to-width ratio of 1.5, determine its smallest width. The allowable bending stress is σ allow 9 MPa and the allowable shear stress is τ allow 0.6 MPa. Neglect the weight of the beam. 12 kn/m 1.5a A 1 m 3 m a 13
14 12 kn/m A 1 m 32 kn 3 m V (kn) M (kn m) 16 kn a x 0 x1.33 m x (m) a heck allowable shear stress τ allow 0.6 MPa VQ τ It V τ 1.5 A 20x10 0.6x (a 2 ) a m 183 mm - x (m) Use a 185 mm -6 14
15 12 kn/m A 1 m 32 kn 3 m V (kn) kn x 0 1.5a a 185 mm heck allowable normal stress σ allow 9 MPa x1.33 m x(m) σ max Mc I (10.67x10 3 )(278/2) (1/12)(185)(278 3 ) M (kn m) 4.48 MPa MPa < 9 MPa, O.K. - x (m) Use a 185 mm -6 15
16 Example 3 The wooden T-beam shown is made from two x 30 mm boards. If the allowable bending stress is σ allow 12 MPa and the allowable shear stress is τ allow 0.8 MPa, determine if the beam can safely support the loading shown without having the deflection more than L/240. The spacing of nails specified to hold the two boards together if each nail can resist 1.50 kn in shear are: between use 125 mm, and D use 250 mm. Take E 12 GPa, and the formula for deflection at the the mid-span: 4 3 5wL PL υ ( ) 768EI 48EI w0.5 kn/m P1.5 kn 30 mm 2 m 2 m D 30 mm 16
17 0.5 kn/m 1.5 kn 1.5 kn 2 m 2 m V (kn) 1.5 D 1 kn x (m) M (kn m) 2.0 x (m) 17
18 Section property y ' y 30 mm y ya A (100 )( ) (215 )( 200 2( ) 30 ) 30 mm mm y ' mm I Σ ( I Ad 2 ) ( I Ad 2 ) ( I Ad 2 ) web flang [(1/12)(30)(200 3 ) (30x200)( ) 2 ] [(1/12)(200)(30 3 ) (200x30)( ) 2 ] x10 6 mm x10-6 m 4 18
19 0.5 kn/m 1.5 kn - Stress distribution at point 1.5 kn 2 m 2 m D 1 kn V (kn) kn τ -, MPa (60.12x10-6 )(0.03) x(m) heck allowable shear stress: τ allow 0.8 MPa τ max V max Q It (1.5x10 3 )[( )(0.03x0.1575)] MPa MPa < 0.8 MPa, O.K. 30 I 60.12x10-6 m 4 19
20 0.5 kn/m 1.5 kn - Stress distribution at point 1.5 kn 2 m 2 m M(kN m) 2.0 D 1 kn x (m) 20 kn m kn τ, MPa 5.24 σ, MPa 200 heck allowable normal stress: σ allow 12 MPa σ max Mc I (2x103 )(0.1575) (60.12x10-6 ) kn m 5.24 MPa 5.24 MPa < 12 MPa, O.K. 30 I 60.12x10-6 m 4 20
21 w0.5 kn/m P1.5 kn υ L 4 5wL 768EI 1.5 k N L/22 m L/22 m 3 PL ( ) 48EI D kn I 60.12x10-6 m 4 heck maximum deflection: υ allow L/240 υ L 4 5wL 768EI 3 PL ) ( 48EI 4 5(.5)4 ) 9 768(12 10 )( m, ( ) < ) 3 (1.5)4 9 48(12 10 )( ( m, O. K. ) ) 21
22 0.5 kn/m 1.5 kn 1.5 kn 2 m 2 m V(kN) x(m) I x10-6 m 4 Nail spacing: s 125 mm, and s D 250 mm; (F s ) allow 1.5 kn - Segment D 1 kn F nail τa shear VQ ( It ) 200 (s x0.03) τ -, MPa τ F nail (1.5x10 3 )[(0.0575)(0.2x0.03)] (0.125x0.03) (60.12x10-6 )(0.03) kn < 1.5 kn, O.K. 22
23 0.5 kn/m 1.5 kn 1.5 kn 2 m 2 m - Segment D D 1 kn V (kn) 1 m x(m) I x10-6 m 4 Nail spacing: s 125 mm, and s D 250 mm; (F s ) allow 1.5 kn F nail τa shear VQ ( It ) 200 (s D x0.03) τ -, MPa τ F nail (1x10 3 )(0.0575)(0.2x0.03) (60.12x10-6 )(0.03) (0.25x0.03) kn < 1.5 kn, O.K. 23
24 Example 4 From the beam shown, determine the largest load P that can be applied to the beam. Requirements specified: σ allow 12 MPa τ allow 0.8 MPa Nail 150 mm φ 3 mm τ allow 200 MPa P 2 m 2 m D 72.5 mm 30 mm mm I 60.12x10-6 mm 4 24
25 V (N) M (N m) P P/2 2 m 2 m P/2 P/2 -P/2 P 30 mm 72.5 mm D P/ mm I x10-6 mm 4 x (m) -P/2 x (m) Q neck A neck y 30 ( )(72.5 ) m Maximum shear force on Nail: Nail 150 mm, φ 3 mm, τ allow 200 MPa F s (τ allow ) nail (A nail ) (200x10 6 )(πx ) 1414 N VQ 1414 ( )( t s) It P ( )(0.345) )(0.150) 6 ( ) P 3286 N 25
26 V (N) M (N m) P 2 m 2 m P/2 P/2 -P/2 P 30 mm 72.5 mm D mm I x10-6 mm 4 x (m) -P/2 x (m) Q max A max y ( )( ) m ending: σ allow 12 MPa Shear : τ allow 0.8 MPa σ Mc I P(0.1575) P 4580 N VQ τ It P 3 ( )( ) ( )(0.03) 0 6 P 3879 N 26
27 σ allow 12 MPa τ allow 0.8 MPa Nail 150 mm φ 3 mm τ allow 200 MPa P 2 m 2 m D Summary: Nail, shear 3.28 kn, use 3 kn eam, bending 4.58 kn eam, shear 3.88 kn 27
28 Example 5 From the beam shown, determine the largest load w that can be applied to the beam. onditions specified: Wood σ allow 12 MPa in tension σ allow 8 MPa in compression τ allow 0.8 MPa For each nail, (F nail ) allow mm spacing w 72.5 mm 30 mm 4 m 2 m D mm I 60.12x10-6 mm 4 28
29 w 1.5w 4 m 4.5w 2 m D 72.5 mm mm 30 mm M (N m) 1.125w I 60.12x10-6 mm 4 x (m) -2w eam : ending : σ allow 12 MPa in tension, σ allow 8 MPa in compression - ompression - Tension 6 (2w)(0.1575) 6 ( w)( ) w kn/m w kn/m 6 (1.125w)(0.725) (2w)(0.725) w kn/m w kn/m 29
30 w 1.5w 4 m 4.5w 2 m D 72.5 mm mm 30 mm V (N) 1.5w 2 w I 60.12x10-6 mm 4 x (m) M (N m) 1.125w -2.5w x (m) -2w eam: τ allow 0.8 MPa VQ τ It (2.5w)( ) ( )(0.03) w 1.55 kn/m Shear on nail : (F nail ) allow 1.5 kn spacing 100 mm VQ F s (τ allow ) nail (A s ) ( )( t s) It (2.5w)( ) 1.5 ( t)(0.1) 6 ( )( t) w 1.04 kn/m 30
31 σ allow 12 MPa in tension σ allow 8 MPa in compression τ allow 0.8 MPa (F nail ) allow 1.5 kn Nail spacing 100 mm w 4 m 2 m D Summary: eam bending, compression 1.53 kn/m tension 4.07 kn/m eam shear, 1.55 kn/m Nail shear 1.04 kn/m, use 1 kn/m 31
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