EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 5 Beams for Bending

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1 MA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 5 Beams for Bending

cases for beams Types of loading: Concentrated load P i

2 Introduction esign of beams for mechanical or civil/structural applications Transverse loading in most cases for beams Types of loading: Concentrated load P i istributed load w (N/m) Uniformly distributed Non-uniformly distributed

3 Common Beam Configurations Statically eterminate Beams Simply supported Overhanging beam Cantilever beam Statically Indeterminate Beams Continuous beam Beam fied at one end & simply supported at the other Fied beam

quilibrium Analysis for Beam under Transverse Loads ample: a simply supported beam with two transverse loading forces of P 1 & P 2 and uniformly distributed load w At an arbitrary cross-section C,

4 quilibrium Analysis for Beam under Transverse Loads ample: a simply supported beam with two transverse loading forces of P 1 & P 2 and uniformly distributed load w At an arbitrary cross-section C, draw body diagram (FB) for each side: Internal shear force V (or V ) that creates shear stress in cross-section C Bending moment M (or M ) that creates normal stress in cross-section C σ = My I and σ ma = Mc I I moment of inertia for cross-section C y istance from the neutral surface c Maimum distance from the neutral surface

beam at that point as indicated ( right down ) The bending moment at any point is positive (+)

5 Sign Conventions for Internal Shear & Bending Moment (1) The shear at any point of a beam is positive (+) when eternal forces (loads and reactions) acting on the beam tend to shear off the beam at that point as indicated ( right down ) The bending moment at any point is positive (+) when the eternal moments acting on the beam tend to bend the beam at the point as indicated ( ends up )

6 Sign Conventions for Internal Shear & Bending Moment (2) Positive shear if trying to twist the element clockwise Positive bending moment if trying to bend the element concave up ( ends up )

7 Shear and Bending-Moment iagrams (1) istribution of shear (force) & bending moment along the ais will help easily determine the maimum shear & normal stresses that will occur, which is critical to structure design ample 1 For the simply supported beam under load P at the midpoint F raw FB for entire. 0.5L P 0.5L P F quilibrium for gives: F y = 0 R + R = P M = 0 0.5PL = R L R = R = 0.5P R F R

8 Shear and Bending-Moment iagrams (2) y raw FB for the two sections when left of F : quilibrium gives: F y = 0 V = R = 0.5P M = 0 M = 0.5P As to sign of V and M : R = 0.5P M P F R = 0.5P R = 0.5P = 0.5P V = 0.5P V positive V P F M - positive M R = 0.5P

9 Shear and Bending-Moment iagrams (3) raw FB for the two sections when right of F : quilibrium gives: F y = 0 V = P - R = 0.5P M = 0 M + V = P 0.5L Therefore M = 0.5PL 0.5P = 0.5P(L-) As to sign of V and M : y R = 0.5P R = 0.5P P V = 0.5P R = 0.5P P F M = 0.5P(L-) V - negative, M - positive M V R = 0.5P

10 Shear and Bending-Moment iagrams (4) 0.5L P 0.5L F To the left of F V = 0.5P To the right of F V = -0.5P V 0.5P 0-0.5P Shear iagram 0.5L L To the left of F M = 0.5P To the right of F M = 0.5P(L-) M 0.25PL 0 Moment iagram 0.5L L

11 Shear and Bending-Moment iagrams (5) Based on the bending moment diagram drawn, for a prismatic beam with uniform crosssection and moment of inertia I and largest dimension from neutral ais c, maimum aial normal stress for this beam can be determined: At mid-point F of beam = 0.5L, M ma = 0.25PL M 0.25PL 0 ma 0.5L P 0.5L Moment iagram M ma c I F 0.5L L (0.25PL) c I PLc 4I

12 For a simply supported beam with localized bending moment M 0 applied at center point F as illustrated, please draw the shear & bending moment diagram. Class ercise M 0 0.5L F L FB for the entire beam is drawn: quilibrium for gives: F y = 0 R + R = 0 R M = 0 R L + M 0 = 0 Therefore, R = -M 0 /L R = M 0 /L M 0 F R

13 Class ercise FB for section to the left of the center point is drawn. quilibrium gives: F y = 0 V = R M = 0 R = M V = -M 0 /L (negative due to V trying to rotate counter clockwise) M = -M 0 /L (negative due to bending concave down or ends down ) R = M 0 /L M 0 V R = M 0 /L V 0 -M 0 /L M 0 M Shear iagram R = M 0 /L 0.5L L Moment iagram -0.5M 0 0.5L L

14 Class ercise FB for section to the right of the center point is drawn: quilibrium gives: F y = 0 M = 0 V = R V(L-) = M V = -M 0 /L (negative due to V trying to rotate counter clock wise) M = M 0 (L-)/L (positive due to bending concave up or ends up ) V 0 -M 0 /L M 0.5M 0 M 0 R = M 0 /L R = M 0 /L M 0 V Shear iagram R = M 0 /L 0.5L L Moment iagram -0.5M 0 0.5L L

15 Load Shear Relationship for Beam (1) For a simply supported beam AB subjected to distributed load w, choose an arbitrary element section CC with width of Δ, FB for section CC is shown. quilibrium along vertical y ais F y = 0 V (V+ΔV) wδ = 0 ΔV = wδ When 0 dv d w The slope of shear diagram (V vs. ) equals negative local distributed load per unit length

16 Load Shear Relationship for Beam (2) Integrating between C and Or V dv d V V w C V C dv C wd wd area under w - curve V Shear iagram Note: Not applicable when there is concentrated load as shear curve will NOT be continuous V C 0 V R A w V C wδ V R B

17 Shear - Bending Moment Relationship (1) quilibrium for moment about C Therefore, or M C' 0 M + M + w 2 M = V w ( )2 2 M = V 1 2 w When 0 dm d = M + V The slope of moment diagram (M vs. ) equals local shear force V

18 Shear - Bending Moment Relationship (2) dm V dm Vd d Integrate between C and Or M M M C M C area C Vd under shear curve Note: Not applicable when localized bending moment is applied between C and as M will NOT be continuous V V C 0 V R A M M M C 0 Shear iagram w C V M Moment iagram C

19 Class ample (1) Please draw the shear & bending moment diagram for a simply supported beam with uniformly distributed load w. w L Free body diagram for the entire beam is drawn. quilibrium for gives: F y = 0 R + R = wl M = 0 wl 0.5L= R L R = R = 0.5wL R w R

20 Class ample (2) efine coordinate system For shear at, V = R = 0.5wL It is positive. At any point, shear force is V ( ) V wd 0. 5wL 0 For bending moment, at, M = 0 At any point, the bending moment is 0 0 w V 0.5wL 0-0.5wL M 0.125wL 2 0 Shear iagram Moment iagram 2 0.5wL w d 0.5wL 0.5 M ( ) M Vd w R = 0.5wL 0.5L w 0.5L R = 0.5wL

21 esign of Prismatic Beams for Bending (1) esign of a long (or large-span) beam is often controlled by consideration of normal stress, which largely depends on M ma As a result, the largest normal stress: σ ma = M mac I σ ma = M ma S If allowable stress is known, the minimum allowable value for section modulus S = I/c is: S min = M ma σ allowable

22 esign of Prismatic Beams for Bending (2) esign goal: for the same materials and same support/brace mechanism, the beam design with the smallest crosssectional area (weight per unit length) that meets safety standard should be selected. General steps 1. etermine the allowable stress (either from design specification or from ultimate strength/safety factor). 2. raw the shear and bending moment diagram and determine M ma. 3. Calculate the minimum allowable section modulus S min. 4. Specify the dimension based on the cross-section shape and relation with S min.

23 Class ample For a wide-flange (W-shapes) prismatic beam to support the P = lb of load as shown, the allowable normal stress is psi. Please select an appropriate wide-flange. Maimum bending moment M ma PL 15000lb (8 12in) 1440kip in 8 ft P Maimum aial normal stress M ma m S allowable Minimum section modulus S M lb in 3 ma 60 2 allowable 24000lb in in 3

24 Class ample For a wide-flange (W-shapes) prismatic beam to support the P = lb of load as shown, the allowable normal stress is psi. Please select an appropriate wide-flange. 3 S 60in For the various wide-flange beams that has S in the range: The best is W16 40 due to smallest cross-section area (or weight per length) P 8 ft Shape S (in 3 ) A (in 2 ) W W W W W W

25 Nonprismatic Beams Prismatic beam: Beam with constant (or uniform) crosssection along its length Normal stress at critical sections (i.e., with largest bending moment M ma ) should, at most, be equal to allowable stress overdesigned at other parts Nonprismatic beam: Beam with varying cross-section Local dimension (or section modulus) could be varied based on local bending moment so that material/cost could be saved Prismatic beam w Nonprismatic beam w

26 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 5 Beams for Bending Homework 5.0 Read tetbook chapter 5.1, 5.2, and 5.3 and give an honor statement confirm reading

27 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 5 Beams for Bending Homework 5.1 For the beam, please draw the shear and bending moment diagrams and determine the equations of the shear and bending moment curves w L

28 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 5 Beams for Bending Homework 5.2 For the beam, please draw the shear and bending moment diagrams and determine the equations of the shear and bending moment curves a P F L-a

29 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 5 Beams for Bending Homework 5.3 For the beam and loading, please (a) draw the shear and bending moment diagram, and (b) calculate the maimum normal stress due to loading on a transverse section at F. Knowing the cross-section for the beam is square with b=h=10 cm 2 kn/m F 10 kn 1 m 1 m 1 m

30 MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 5 Beams for Bending Homework 5.4 For the beam and loading shown, design the width b of the beam knowing the material used has allowable stress of 10 MPa and the beam has height of h = 10 cm. 2 kn/m F 1 m 2m

EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 6 Shearing Stress in Beams & Thin-Walled Members

EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 6 Shearing Stress in Beams & Thin-Walled Members Beams Bending & Shearing EMA 3702 Mechanics & Materials Science Zhe Cheng (2018)