REVIEW FOR EXM II. J. Clark School of Engineering Department of Civil and Environmental Engineering b Dr. Ibrahim. ssakkaf SPRING 00 ENES 0 Mechanics of Materials Department of Civil and Environmental Engineering Universit of Marland, College Park Beams Introduction The most common tpe of structural member is a beam. In actual structures beams can be found in an infinite variet of Sizes Shapes, and Orientations Slide No.
Beams Introduction Figure Beam Slide No. Beams Figure Introduction b c Cable d Load e (a) a (c) (b) (d) Slide No.
Beams Introduction Figure (a) Cantilever (b) Simpl supported (c) Overhanging (d) continuous (e) Fied ended (f) Cantilever, simpl supported Slide No. 4 Normal and Shearing Stress Stresses in beams P b a w b a h P Figure 4 R O V M r + τ d σ Slide No. 5
Fleural Strains Deformation of Beam due to Lateral Loading Figure 5 P w Slide No. 6 Fleural Stress Fleural Normal Stress Distribution of Normal Stress in a Beam Cross Section Figure 9 P Centroidal ais w Neutral ais F C c F T d C c c d R V r Slide No. 7
Fleural Stress Fleural Normal Stress For fleural loading and linearl elastic action, the neutral ais passes through the centroid of the cross section of the beam Slide No. 8 Elastic Fleural Formula The elastic fleural formula for normal stress is given b and σ M c r I ma (8) M r σ I (9) Slide No. 9
Elastic Fleural Formula n alternative form of the fleural formula for maimum normal stress is given b Where S I c M r σ (0) ma S Slide No. 0 Second Moments of reas Moment of Inertia Consider an area located in the plane as shown in the figure. d O Figure 0 Slide No.
Second Moments of reas Moment of Inertia O d I I Where I moment of inertia with respect to ais I moment of inertia with respect to ais d d (a) (b) Slide No. Second Moments of reas Radii of Gration of an rea k k k z I (6a) I I z (6b) (6c) Slide No.
Second Moments of reas Parallel is Theorem O rea C C C d I I I I C C + + C C (7) Slide No. 4 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 Slide No. 5
Eamples: Elastic Fleure Formula Eample (cont d) First, we need to locate the neutral ais from the bottom edge: C 6 5 C ten ( )( 6) + ( + )( 6) M r Ma. Stress I 7 6 + 6 4 6 + 5 com ma ma Slide No. 6 Eamples: Elastic Fleure Formula C Eample (cont d) 6 Find the moment of inertia I with respect to the ais using parallel ais-theorem: 6( ) ( )( ) ( 6) + 6 + + ( 6)( ) 5 I 4 + 48 + 6 + 48 6 in 4 ( 5) (5 ) Ma. Stress (com). ksi 6 Slide No. 7
Eamples: Elastic Fleure Formula C Eample (cont d) 6 n alternative wa for finding the moment of inertia I with respect to the ais is as follows: 5 I 6 ( ) ( 5) ( ) + 6 Slide No. 8 Eamples: Elastic Fleure Formula Eample 5 Determine both the maimum fleural tensile and the maimum fleural compressive stresses produced b a resisting moment of 00 kn m if the beam has the cross section shown in the figure. 50 mm 5 mm 50 mm 00 mm Slide No. 9
Eamples: Elastic Fleure Formula Eample 5 (cont d) Locate the neutral ais from the upper edge: 5 mm 50 mm 50 mm C 50 5.5,0,70.87 7,85.90 4.6 mm ( ) + 50 5( 5 + 75) π + π 50 5 + 50 5 + ( 00) ( 5 + 50 + 50 ) 4 ( 00) 4 00 mm Slide No. 0 I Eamples: Elastic Fleure Formula Eample 5 (cont d) Calculate the moment of inertia with respect to the ais: 5 mm ( ) (50 5) ( 4.6 5) 5( 75 4.6) 50 4.6 4 ( 00) π ( 00) + ( 5 4.6 ) π + 64 4 6 4 6 4 7.4 0 mm 7.4 0 m M 00 0 (75 4.6) 0 σ ma ( ten) r 87.5 MPa 6 I 7.4 0 00 0 (4.6) 0 σ ma ( com) 7. MPa 6 7.4 0 + 50 mm 4.6 mm 00 mm Slide No.
Shear Forces and Bending Moments in Beams Variation of Shear and Moment Forces In 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. Slide No. Figure 4 Shear Forces and Bending Moments in Beams Variation of Shear and Moment Forces a w b L P O Slide No.
Shear Forces and Bending Moments in Beams Sign Convention Figure 5 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 Slide No. 4 Shear Forces and Bending Moments in Beams Sign Convention 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. 5b), and a positive moment tends to bend the segment concave upward (Fig. 5c) Slide No. 5
Shear Forces and Bending Moments in Beams Eample 6 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. 5 kn/m 0 kn B 4 m m m Slide No. 6 Shear Forces and Bending Moments in Beams Eample 6 (cont d) free-bod diagram for the beam is shown Fig. 6. The reactions shown on the diagram are determined from equilibrium equations as follows: + M 0; R (8) ( 5 4)( 6) 0( ) 0 R + R 50 kn B B 0 kn F 0; R B + 50 5 ( 4) 0 0 Slide No. 7
Shear Forces and Bending Moments in Beams Eample 6 (cont d) 0 kn 5 kn/m Figure 6 50 kn B R 50 kn R B 0 kn 5 kn/m S M + V 50 5 for 0 < < 4 0 for 0 < < 4 V + M S 0; M + 50 5( ) M 50 7.5 0; V + 50 5 0 Slide No. 8 F Shear Forces and Bending Moments in Beams Load, Shear Force, and Bending Moment Relationships In cases where a beam is subjected to several concentrated forces, couples, and distributed loads, the equilibrium approach discussed previousl can be tedious because it would then require several cuts and several free-bod diagrams. In this section, a simpler method for constructing shear and moment diagrams are discussed. Slide No. 9
Shear Forces and Bending Moments in Beams Load and Shear Force Relationships Slope of dv d Shear Diagram w( ) Distributed Load Intensit () Slide No. 0 Shear Forces and Bending Moments in Beams Load and Shear Force Relationships V Change in Shear V V V rea under Loading Curve between V dv w( ) d and (4) Slide No.
Shear Forces and Bending Moments in Beams Shear Force and Bending Moment Relationships Slope of Moment Diagram dm V (40) d Shear Slide No. Shear Forces and Bending Moments in Beams Shear Force and Bending Moment Relationships M Change in Moment M M M M rea under Shear diagram between dm V d and (4) Slide No.
Shear Forces and Bending Moments in Beams Shear and Moment Diagrams Loading Shear Diagram, dv w d Moment Diagram, dm V d Slide No. 4 Shear Forces and Bending Moments in Beams Shear and Moment Diagrams Loading Shear Diagram, dv w d Moment Diagram, dm V d Slide No. 5
Shear and Moment Diagrams Eample 9 Draw the shear and bending moment diagrams for the beam shown in Figure a. 40 lb/ft 600 lb ft 0 ft 000 lb in Figure a Slide No. 6 Shear and Moment Diagrams Eample 9 (cont d) Support Reactions The reactions at the fied support can be Figure b calculated as follows: + F 0; R 40( ) 600 0 R 080 lb + M 0; M + 40() ( 6) + 600( 0) + 000 0 M 5,880 lb in 40 lb/ft 600 lb ft M 5,880 lb ft R 080 lb 0 ft 000 lb ft Slide No. 7
Shear and Moment Diagrams 5,880 lb ft Eample 9 (cont d) Shear Diagram Using the established sign convention, the shear at the ends of the beam is plotted first. For eample, when 0, V 080; and when 0, V 600 40 lb/ft M V 080 40 for 0 < < 080 lb V ( ) M 080 40 5,880 for 0 < < Slide No. 8 Shear and Moment Diagrams Eample 9 (cont d) 40 lb/ft 5,880 lb in ft 080 lb V M V ( ) 600 for < < 0 ( )( - 6) + 080 for < 0 M 5,880 40 < Slide No. 9
Shear and Moment Diagrams Eample 9 (cont d) 40 lb/ft 600 lb M 5,880 lb ft ft R 080 lb 0 ft 000 lb ft 080 V (lb) (+) 600 M (ft lb) (-) -000 Slide No. 40 Shear and Moment Diagrams Eample 0 Draw complete shear and bending moment diagrams for the beam shown in Fig. 8000 lb 000 lb/ft B C D Figure a ft 4 ft 8 ft Slide No. 4
Shear and Moment Diagrams Eample 0 (cont d) The support reactions were computed from equilibrium as shown in Fig..b. 8000 lb 000 lb/ft B C D ft 4 ft 8 ft Figure a R,000 lb R C,000 lb Slide No. 4 Shear and Moment Diagrams Eample 0 (cont d) 000 lb/ft B C D,000 lb ft 4 ft 8 ft,000 lb,000 8,000 lb V (lb) (+) (+) 5.5 ft (-),000 lb 0,50 M (ft -lb) (-) (-),000 64,000 8000 lb Slide No. 4
Shear and Moment Diagrams Eample 0 (cont d),000,000,000.8 5.5,000 Slide No. 44 Shearing Stress in Beams Shearing Stress due to Bending P (a) Unloaded Stack of Slabs (b) Unglued Slabs loaded Figure Slide No. 45
Shearing Stress in Beams Shearing Stress due to Bending P (c) Glued Slabs Unloaded (d) Glued Slabs loaded Figure (cont d) Slide No. 46 Shearing Stress in Beams Shearing Stress Formula t each point in the beam, the horizontal and vertical shearing stresses are given b VQ τ It (5) Where V shear force at a particular section of the beam Q first moment of area of the portion of the cross-sectional area between the transverse line where the stress is to be computed. I moment of inertia of the cross section about neutral ais t average thickness at a particular location within the cross section Slide No. 47
Shearing Stress in Beams Eample Determine the first moment of area Q for the areas indicated b the shaded areas a and b of Fig. 5. a 6.5 b 6 Figure 5 Slide No. 48 Shearing Stress in Beams Eample (cont d) First, we need to locate the neutral ais from the bottom edge: C 5 N. C ( )( 6) + ( + )( 6) 6 + 6 7 from base 4 6 Slide No. 49
Shearing Stress in Beams.5 Eample (cont d) b a The first moments of area Q a and Q b are found as follows: 6 5 N. 6 Q Q a b ( 5.5)[ ].5 in [.5 6] 0.5 in Slide No. 50 Shearing Stress in Beams Shearing Stress Formula How accurate is the shearing stress formula? t t t t d Great t is small % error d t % error d t 00% error, worst case 4d t Slide No. 5
Shearing Stress in Beams Variation of Vertical Shearing Stress in the Cross Section N. V Ma Stress Slide No. 5 Shearing Stress in Beams Eample The transverse shear V at a certain section of a timber beam is 600 lb. If the beam has the cross section shown in the figure, determine (a) the vertical shearing stress in. below the top of the beam, and (b) the maimum vertical stress on the cross section. Slide No. 5
Shearing Stress in Beams Eample (cont d) 8 in. in. 4 in. 8 in. Slide No. 54 Shearing Stress in Beams 4 in. 8 in. Eample (cont d) From smmetr, the neutral ais is located 6 in. from either the top or bottom edge. 8( ) 4( 8) in. 8 in. N.. 8 in. 4 in. 8 in. in. (a) τ (b) τ I Q 8 Q N Q ma VQ 98. in ( )( 5) + [ ( )(.5) ] 94.0 in 8( )( 5) + [ ( )( 4) ].0 in 6000( 94) 4.7 psi It 98.( 4) ma 6000( ) 7. psi It 98.( 4) VQ 4 Slide No. 55
Composite Beams Steel Bending of Composite Beams These are called composite beams. The offer the opportunit of using each of the materials emploed in their construction advantage. Concrete luminum Steel Slide No. 56 Composite Beams Transformed Section b N. E n E b Figure 9 b n b Slide No. 57
Composite Beams Transformed Section C σ N.. M I σ Figure 0. Distribution of Fictitious Normal Stress on Cross Section Slide No. 58 Composite Beams Stresses on Transformed Section. To obtain the stress σ at a point located in the upper portion of the cross section of the original composite beam, the stress is simpl computed from M/I.. To obtain the stress σ at a point located in the upper portion of the cross section of the original composite beam, stress σ computed from M/I is multiplied b n. Slide No. 59
Composite Beams Eample 7 steel bar and aluminum bar are bonded together to form the composite beam shown. The modulus of elasticit for aluminum is 70 GPa and for streel is 00 GPa. Knowing that the beam is bent about a horizontal ais b a moment M 500 N- m, determine the maimum stress in (a) the aluminum and (b) the steel. Slide No. 60 Composite Beams Eample 7 (cont d) M Steel 0 mm luminum 40 mm 0 mm Slide No. 6
Composite Beams Eample 7 (cont d) First, because we have different materials, we need to transform the section into a section that represents a section that is made of homogeneous material, either steel or aluminum. We have n E s Ea 00 70.857 Slide No. 6 Composite Beams Eample 7 (cont d) 0 mm n 85.7 mm Steel 0 mm luminum luminum 40 mm luminum 0 mm Figure a 0 mm Figure b Slide No. 6
Composite Beams I C Eample 7 (cont d) N 0 Consider the transformed section of Fig. b, therefore ( 85.7 0) + 40( 0 40) ( 85.7 0) + ( 0 40) ( ) ( 85.7 0)(.5 0) 85.7.5 0 + ( 40 + 0.5) 4 9 4.5 mm from top 85.4 0 mm 85.4 0 m Slide No. 64 Composite Beams Eample 7 (cont d) 85.7 mm C.5 mm N.. C 0 mm 40 mm 0 mm Slide No. 65
Composite Beams Eample 7 (cont d) a) Maimum normal stress in aluminum occurs at etreme lower fiber of section, that is at -(0+40-.5) -7.65 mm. + 66.5 MPa (T) ( 7.65 0 ) M 500 6 66.5 0 Pa 9 I 85.4 0 σ al Slide No. 66 Composite Beams Eample 7 (cont d) b) Maimum normal stress in stelel occurs at etreme upper fiber of the cross section, that is. at +.5 mm. (.5 0 ) M 500 n (.867) I 85.4 0.8 MPa (C) σ St 9.8 0 6 Pa Slide No. 67
Composite Beams Reinforced Concrete Beam Figure d b d - b C N.. σ F n s (a) (b) (c) Slide No. 68 Composite Beams Reinforced Concrete Beam The ratio n is given b Modulus of Elasticit for Steel E n Modulus of Elasticit for Concrete E The position of the neutral ais is obtained b determining the distance from the upper face of the beam (upper fiber) to the centroid C of the transformed section. s c Slide No. 69
Composite Beams Reinforced Concrete Beam The neutral ais for a concrete beam is found b solving the quadratic equation: b b + n n d b 0 s s (6) d d - C n s Slide No. 70 Beam Deformation Deflection of Beams P Figure w w P (a) w >> w (b) P >> P Slide No. 7
Beam Deformation Methods for Determining Beam Deflections Three methods are commonl used to find beam deflections: ) The double integration method, ) The singularit function method, and ) The superposition method Slide No. 7 Beam Deformation The Differential Equation of the Elastic Curve for a Beam d EI M d ( ) (8) E modulus of elsticit for the material I moment of inertia about the neutral ais of cross section M() bending moment along the beam as a function of Slide No. 7
Beam Deformation Sign Convention M - negative M - positive d negative d d negative d Figure 5. Elastic Curve Slide No. 74 Beam Deformation Sign Convention L.H.F M V M V (a) Positive Shear & Moment R.H.F M V Figure 6 V (b) Positive Shear (clockwise) (c) Positive Moment (concave upward) M Slide No. 75
Beam Deformation Relation of the Deflection with Phsical Quantities such as V and M deflection slope moment ( M ) shear ( V ) load ( w) d d d EI d dm d EI (for EI constant) d d 4 dv d EI (for EI constant) 4 d d (9) Slide No. 76 Beam Deformation Load Shear (V) Moment (M) Slope (θ) P / PL 6EI Deflection () (+) L PL 4 PL 48EI P (+) (-) P / PL 6EI Figure 7 Complete Series of Diagrams for Simpl Supported beam Slide No. 77
Deflection b Integration Eample Boundar Conditions (a) Slope 0 at 0 Deflection 0 at 0 (b) Slope at L/ 0 Deflection 0 at 0, and L Figure 8 (c) Slope at rollers? Deflection at rollers 0 (d) Slope 0 at 0 Deflection 0 at 0 and L Slide No. 78 Deflection b Integration Eample beam is loaded and supported as shown in the figure. a) Derive the equation for the elastic curve in terms of w, L,, E, and I. b) Determine the slope at the right end of the beam. c) Find the deflection at L. Slide No. 79
Deflection b Integration Eample (cont d) w L B FBD wl L wl M 6 wl R Slide No. 80 Deflection b Integration Eample (cont d) Find an epression for a segment of the distributed load: Equation of Straight Line w w w w w L w L w L w L ( ) w L L w w L (a) Slide No. 8
Deflection b Integration Eample (cont d) w wl M 6 wl V R wl wl ( ) ( w w ) + M s 0; M + w 0 6 or M w ( w w ) wl wl w M ( ) + (b) 6 Slide No. 8 Deflection b Integration Eample (cont d) The solution for parts (a), (b), and (c) can be completed b substituting for w into Eq. b, equating the epression for M() to the term EI(d /d ), and integrating twice to get the elastic curve and epression for the slope. Note that the boundar conditions are that both the slope and deflection are zero at 0. d i.e.; EI EI M ( ) d Slide No. 8
Singularit Functions Definition singularit function is an epression for n written as 0, where n is an integer (positive or negative) including zero, and 0 is a constant equal to the value of at the initial boundar of a specific interval along the beam. Slide No. 84 Singularit Functions Properties of Singularit Functions B definition, for n 0, 0 n 0 ( ) 0 n when 0 when < Selected properties of singularit functions that are useful and required for beamdeflection problems are listed in the net slides for emphasis and read reference. 0 (6) Slide No. 85
Singularit Functions Selected Properties 0 n 0 ( ) 0 n when n > 0 and when n > 0 and < 0 0 (7) 0 0 0 when n > 0 and when n > 0 and < 0 0 (8) Slide No. 86 Singularit Functions Integration and Differentiation of Singularit Functions n n+ 0 d 0 + C when n > 0 n + (9) d d n n 0 n 0 when n > 0 (0) Slide No. 87
Singularit Functions Tpical Singularit Functions P M w 4 R L R R M ( ) R P L L w + M 0 for 0 < < L () Slide No. 88 Singularit Functions Moment due to Distributed Loads w 0 w 0 w 0 L L L w 0 M w 0 w 0 M w 0 6( L ) M w 0 k n+ Figure. Open-ended-to-right distributed loads Slide No. 89
Singularit Functions Figure Moment due to Distributed Loads L L w 0 w 0 -w 0 The moment at section due to distributed load alone is w0 w0 0 + M w 0 Slide No. 90 Singularit Functions Figure 4 Moment due to Distributed Loads L L w 0 w 0 -w 0 The moment at section due to distributed load alone is w0 w0 + M w 0 Slide No. 9
Singularit Functions Moment due to Distributed Loads w L w ( L ) L w 0 L w 0 w w0 The moment at section due to distributed load alone is w0 M w 0 6 w0 -w + 0 -w Figure 5 ( ) 6( ) w + w 0 0 Slide No. 9 Singularit Functions Moment due to Distributed Loads Note that in Fig. 4, the linearl varing load at an point is w 0 w w0 ( ) w The moment of this load for From similar triangles : w w 0 n point is w0 M ( ) ( ) w0 ( ) ( ) 6 Slide No. 9
Deflection b Superposition Figure 9 Method of Superposition P L P w w a + P P P w u b + + w t Slide No. 94 Deflection b Superposition Figure 7 Eample 7 Use the method of superposition, determine the deflection at the free end of the cantilever beam shown in Fig. 7 in terms of w, L, E, and I. wl w L B L C Slide No. 95
Deflection b Superposition w L Eample 7 (cont d) wl B C L wl Loading Loading B C L L Figure 8 + w B L L C B B C + Straight Line B C δ C θ C ( δ C ) ( θ C ) ( δ C ) ( θ C ) Slide No. 96 Deflection b Superposition Eample 7 (cont d) Using the solutions listed in Table a. Cases and (Tetbook Table B-9) with P wl δ ( δ ) + ( δ ) ( δ ) + ( δ ) + L( θ ) C C P(L) EI ( L) wl EI C C 4 wl wl + L 8EI 6EI B 4 wl wl 7wL + L 8EI 6EI 4EI B 4 Slide No. 97
Deflection b Superposition Slopes and Deflection Tables Table a Slide No. 98 Staticall Indeterminate Beams Introduction In all of the problems discussed so far, it was possible to determine the forces and stresses in beams b utilizing the equations of equilibrium, that is F 0 M 0 F 0 (9) Slide No. 99
Staticall Indeterminate Structures Staticall Indeterminate Beam When the equilibrium equations alone are not sufficient to determine the loads or stresses in a beam, then such beam is referred to as staticall indeterminate beam. Slide No. 00 Staticall Indeterminate Transversel Loaded Beams Illustrative Eample using Superposition Determine the reactions at the supports for the simpl supported cantilever beam (Fig.5) presented earlier for the integration method. w B L Slide No. 0
Staticall Indeterminate Transversel Loaded Beams Illustrative Eample using Superposition Method (cont d) First consider the reaction at B as redundant and release the beam from the support (remove restraint). The reaction R B is now considered as an unknown load (see Fig. 9) and will be determined from the condition that the deflection at B must be zero. Slide No. 0 Staticall Indeterminate Transversel Loaded Beams Illustrative Eample using Superposition Method (cont d) L w R B w B + B L ( B ) w (a) (b) (c) B R B ( B ) RB Figure 9. Original Loading is Broken into Two Loads Slide No. 0
Staticall Indeterminate Transversel Loaded Beams Illustrative Eample using Superposition Method (cont d) In reference to Table a cases and (Table B9 of Tetbook): R L EI B ( ) + and ( ) 4 wl EI B R B B w 8 The deflection at B in the original structural configuration must equal to zero, that is ( ) + ( ) 0 B B R B w (7) (8) Slide No. 04 Staticall Indeterminate Transversel Loaded Beams Slopes and Deflection Tables Table a Slide No. 05
Staticall Indeterminate Transversel Loaded Beams Illustrative Eample using Superposition Method (cont d) Substituting Eq. 7 into Eq. 8, gives 4 R + B L wl 0 (9) EI 8EI Solving for R B, the result is R B + 8 wl (40) Slide No. 06 Staticall Indeterminate Transversel Loaded Beams Illustrative Eample using Superposition Method (cont d) From the free-bod diagram for entire beam (Figure 40), the equations of equilibrium are used to find the rest of the reactions. + F 0; R + R wl 0 B R wl R B (4) Slide No. 07
Staticall Indeterminate Transversel Loaded Beams Illustrative Eample using Superposition Method (cont d) R M L/ wl w B R R B Figure 40. Free-bod Diagram for the Entire Beam Slide No. 08 Staticall Indeterminate Transversel Loaded Beams Illustrative Eample using Superposition Method (cont d) But RB wl from Eq. 40, therefore 8 5 R wl wl wl (4) 8 8 L + M 0; - M RBL + ( wl) 0 M RBL + wl wl L wl 8 wl (4) 8 Slide No. 09
Staticall Indeterminate Transversel Loaded Beams Illustrative Eample using Superposition Method (cont d) From Eqs.40, 4, and 4, R M 0 8 wl 5 wl 8 wl 8 Which confirms the results found b using the integration method. R R B Slide No. 0