LETURE Third Edition BEMS: SHER ND MOMENT DGRMS (FORMUL). J. lark School of Engineering Department of ivil and Environmental Engineering 1 hapter 5.1 5. b Dr. brahim. ssakkaf SPRNG 00 ENES 0 Mechanics of Materials Department of ivil and Environmental Engineering Universit of Marland, ollege Park LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 1 Moment of nertia There are man important topics in engineering practice that require evaluation of an integral of the second moment of area or moment of inertia of the tpe d (1)
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. Moment of nertia The integral of Eq. 1 is referred to as the moment of inertia for an area. Methods used to determine the area moment of inertia will be discussed briefl in this chapter. Full treatment of this important topic can be found in almost ever standard statics book. LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. Moment of nertia onsider an area located in the plane as shown in the figure. d O Figure 10
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 4 Moment of nertia lso consider the element of area d of coordinates and. The second moment, or moment of inertia of the area (see Fig. 10) with respect to the ais, and the second moment, or moment of inertia with respect to the ais are defined, respectivel, as provided in the net viewgraph: LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 5 Moment of nertia O d Where moment of inertia with respect to ais moment of inertia with respect to ais d d (a) (b)
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 6 Moment of nertia The quantities and are referred to as rectangular moments of inertia, since the are computed from the rectangular coordinates of the element d. While each integral is basicall a double integral, it is possible in man applications to select elements of area d in the shape of thin horizontal or vertical strips. LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 7 Polar Moment of nertia The second moment, or polar moment of inertia of an area with respect to an ais perpendicular to the plane of the area is denoted b the smbol J. For eample, the second moment of area shown in Fig. 11 with respect to a z-ais that is perpendicular to the plane of the area at O of the -coordinate sstem is:
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 8 z r O Figure 11 d Polar Moment of nertia J z r d ( + ) d + + d d () LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 9 z Polar Moment of nertia The second moment or polar moment of area with respect to z ais is given b d r O J + z (4)
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 10 Polar Moment of nertia The quantit J z is known as the polar second moment of the area and was used in hapter 7 in the calculation of the stress in circular shafts transmitting torques. For circular shaft of radius r, π r J 4 z (5) LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 11 Radius of Gration of an rea The radius of gration of planar area has units of length and is a quantit often used for the design of columns in structural mechanics. Provided the areas and moments of inertia are known, the radii of gration are determined from the following formulas:
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 1 Radii of Gration of an rea k k k z (6a) z (6b) (6c) LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 1 Eample 1 Determine the moment of inertia for the rectangular area shown in Fig. 11 with respect to (a) the centroidal ais, (b) the ais passing through the base of the rectangle, and (c) the pole or z ais perpendicular to the - plane and passing through the centroid.
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 14 Eample 1 (cont d) Figure 1 d h h b b LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 15 Eample 1 (cont d) a) We select as an element of area a horizontal strip of length b and thickness d. Therefore, d d ( b d ) ntegrating from -h/ to +h/, we obtain h h b b h h b d h h + 8 8 1 1 bh
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 16 Eample 1 (cont d) b) We select as an element of area a horizontal strip of length b and thickness d. Therefore, ( b d) d d ntegrating from 0 to h, we obtain h h b b d 0 0 b h LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 17 Eample 1 (cont d) c) We select as an element of area a vertical strip (Fig. 1) of length h and thickness d. Therefore, d d ( h d ) ntegrating from -b/ to +b/, we obtain b b h h b b h d b b + 8 8 1 1 hb
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 18 Eample 1 (cont d) Figure 1 h h d b b LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 19 Eample 1 (cont d) Using Eq. 4, the polar moment of inertia about the z ais is computed as follows: J z bh 1 bh 1 + hb + 1 ( h + b )
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 0 Beam Section Properties The maimum normal stress due to bending, Mc M σ m S section moment of inertia S c section modulus beam section with a larger section modulus will have a lower maimum stress onsider a rectangular beam cross section, 1 bh S 1 1 bh 1 h c h 6 6 Between two beams with the same cross sectional area, the beam with the greater depth will be more effective in resisting bending. Structural steel beams are designed to have a large section modulus. LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 1 Properties of merican Standard Shapes
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. ommonl Used Second Moments of Plane reas Figure 14a LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. ommonl Used Second Moments of Plane reas Figure 14b
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 4 ommonl Used Second Moments of Plane reas Figure 14c LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 5 Parallel is Theorem O rea d + + (7)
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 6 Eample Repeat Part (b) of Eample 1 using the parallel-ais theorem. From Part (a), Using Eq. 7, + bh h + 1 ( hb) bh 1 bh + bh 1 bh LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 7 Eamples: Elastic Fleure Formula Eample Determine the maimum fleural stress produced b a resisting moment M r of +5000 ft lb if the beam has the cross section shown in the figure. 6 6
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 8 Eamples: Elastic Fleure Formula Eample (cont d) First, we need to locate the neutral ais from the bottom edge: 6 5 ten ( 1)( 6) + ( + )( 6) M r Ma. Stress 7 6 + 6 4 6 + 5 com ma ma LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 9 Eamples: Elastic Fleure Formula Eample (cont d) 6 Find the moment of inertia with respect to the ais using parallel ais-theorem: 6( ) ( )( ) ( 6) + 6 + + ( 6)( 1) 5 1 1 4 + 48 + 6 + 48 16 in 4 ( 5) (5 1) Ma. Stress (com).1 ksi 16
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 0 Eamples: Elastic Fleure Formula Eample (cont d) n alternative wa for finding the moment of inertia with respect to the ais is as follows: 6 5 6 ( ) ( 5) ( 1) + 16 LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 1 Eamples: Elastic Fleure Formula Eample 4 pair of channels fastened back-to-back will be used as a beam to resist a bending moment M r of 60 kn m. f the maimum fleural stress must not eceed 10 MPa, select the most economical channel section listed in ppendi B of the tetbook.
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. Eamples: Elastic Fleure Formula Eample 4 (cont d) σ M S, However, we have two channels, hence M σ S 60 10 S 10 6 ( 10 ) M S σ 50 10 6 m Select 54 0 channel 50 10 mm From Table B - 6 of Tetbook : LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. Eample 4 (cont d) Select
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 4 Eamples: Elastic Fleure Formula Eample 5 Determine both the maimum fleural tensile and the maimum fleural compressive stresses produced b a resisting moment of 100 kn m if the beam has the cross section shown in the figure. 50 mm 5 mm 150 mm 100 mm LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 5 Eamples: Elastic Fleure Formula Eample 5 (cont d) Locate the neutral ais from the upper edge: 5 mm 50 mm 150 mm 50 5 1.5,0,70.87 17,85.90 14.6 mm ( ) + 150 5( 5 + 75) π + π 50 5 + 150 5 + ( 100) ( 5 + 150 + 50 ) 4 ( 100) 4 100 mm
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 6 Eamples: Elastic Fleure Formula Eample 5 (cont d) alculate the moment of inertia with respect to the ais: 5 mm ( ) (50 5) ( 14.6 5) 5( 175 14.6) 50 14.6 4 ( 100) π ( 100) + ( 5 14.6 ) π + 64 4 6 4 6 4 17.4 10 mm 17.4 10 m M 100 10 (75 14.6) 10 σ ma ( ten) r 87.5 MPa 6 17.4 10 100 10 (14.6) 10 σ ma ( com) 7. MPa 6 17.4 10 + 50 mm 14.6 mm 100 mm LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 7 Eample 6 SOLUTON: Based on the cross section geometr, calculate the location of the section centroid and moment of inertia. Y ( + d ) cast-iron machine part is acted upon b a kn-m couple. Knowing E 165 GPa and neglecting the effects of fillets, determine (a) the maimum tensile and compressive stresses, (b) the radius of curvature. ppl the elastic fleural formula to find the maimum tensile and compressive stresses. Mc σ m alculate the curvature 1 M ρ E
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 8 Eample 6 (cont d) SOLUTON: Based on the cross section geometr, calculate the location of the section centroid and moment of inertia. 1 rea, mm 0 90 1800 40 0 100 000, mm 50 0, mm 90 10 4 10 114 10 114 10 Y 8 mm 000 ( d ) ( 1 bh + d ) 1 ( 1 90 0 + 1800 1 ) + ( 1 0 40 + 100 18 ) + 1 1-9 4 868 10 mm 868 10 m LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 9 Eample 6 (cont d) ppl the elastic fleural formula to find the maimum tensile and compressive stresses. Mc σ m M c kn m 0.0m σ σ +76.0 MPa 9 4 868 10 mm M c kn m 0.08m σ B B σ 9 4 B 11. MPa 868 10 mm alculate the curvature 1 M ρ E kn m 165 ( GPa 9 4 )( 868 10 - m ) 1-0.95 10 m 1 ρ ρ 47.7 m
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 40 For an specified transverse cross section of a beam, the method for determining fleural stresses discussed previousl is adequate if the objective is to determine the fleural stresses on that section. However, if the maimum fleural stress is required in a beam subjected to a LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 41 loading that produces a resisting moment that varies with position along the beam, it is necessar to have a method for determining the maimum resisting moment. Similarl, the maimum transverse shearing stress will occur at a section where the resisting shear V r is maimum.
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 4 Moments in Beams Variation of Shear and Moment Forces The variation of shear (V) and moment (M) forces as a function of the position of an arbitrar point along the beam s ais can be obtained b using the method of section. Here, it is necessar to locate the imaginar section at an arbitrar distance from the end of the beam rather than at specified point. LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 4 Variation of Shear and Moment Forces n general, the internal shear V and bending moment M variations will be discontinuous, or their slope will be discontinuous at points where a distributed load changes or where concentrated forces or couples are applied.
1 LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 44 Figure 14 Variation of Shear and Moment Forces O a w b L P LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 45 Variation of Shear and Moment Forces Shear and bending moment functions must be determined for each segment of the beam located between an two discontinuities of loading. For eample, sections located at 1,, and will have to be used to describe the variation of V and M throughout the length of the beam in Fig. 14.
1 LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 46 Variation of Shear and Moment Forces These functions will be valid onl within regions from L O to a for b 1 a P a to b for, and w b to L for O Figure 14 LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 47 Sign onvention Before presenting a method for determining the shear and bending moment as functions of and later plotting these functions (shear and moment diagrams), it is first necessar to establish a sign convention so define positive and negative shear force and bending moment acting in a beam.
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 48 Sign onvention Figure 15 V V M M L.H.F R.H.F V V (a) Positive Shear & Moment (b) Positive Shear (clockwise) M (c) Positive Moment (concave upward) M LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 49 Sign onvention With reference to Fig. 15a, on the left-hand face (L.H.F.) of the beam segment, the internal shear force V acts downward and the internal moment M acts counterclockwise. n accordance with Newton s third law, an equal and opposite shear force and bending moment must act on the righthand face (R.H.F.) of a segment.
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 50 Sign onvention Perhaps an eas wa to remember this sign convention is to isolate a small beam segment and note that positive shear tends to rotate the segment clockwise (Fig. 15b), and a positive moment tends to bend the segment concave upward (Fig. 15c) LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 51 Procedure for nalsis The following procedure provides a method for constructing the shear and moment functions (formulas) for a beam. Support Reactions Determine all the reactive couples and forces acting on the beam and resolve the forces into components acting perpendicular and parallel to the beam s ais. Shear and Moment Functions (Formulas) Specif separate coordinates having an origin at
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 5 Procedure for nalsis (cont d) The beam s left end and etending to regions of the beam between concentrated forces and/or couples and where there is no discontinuit of distributed loading. Section the beam perpendicular to its ais at each distance and from the free-bod diagram of one of the segments, determine the unknowns V and M at the cut section as a function of. On the free-bod diagram, V and M should be shown acting in their positive sense (see Fig 15). V is obtained from F 0 and M is obtained b summing moments about point S located at the cut section, M S 0 LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 5 Eample 7 beam is loaded and supported as shown in the figure. Using the coordinate aes shown, write equations for shear V and bending moment M for an section of the beam in the interval 0 < < 4 m. 15 kn/m 0 kn B 4 m m m
LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 54 Eample 7 (cont d) free-bod diagram for the beam is shown Fig. 16. The reactions shown on the diagram are determined from equilibrium equations as follows: + M 0; R (8) ( 15 4)( 6) 0( ) 0 R + R 50 kn B B 0 kn F 0; R B + 50 15 ( 4) 0 0 LETURE 1. BEMS: SHER ND MOMENT DGRMS (FORMUL) (5.1 5.) Slide No. 55 Eample 7 (cont d) 15 kn/m 0 kn Figure 16 50 kn R 50 kn 15 kn/m S M + F V 50 15 for 0 < < 4 0 for 0 < < 4 V + M S 0; M + 50 15( ) M 50 7.5 B R B 0 kn 0; V + 50 15 0