LECTURE Third Edition BEAMS: SHEARNG STRESS A. J. Clark Shool of Engineering Department of Civil and Environmental Engineering 14 Chapter 6.1 6.4 b Dr. brahim A. Assakkaf SPRNG 200 ENES 220 Mehanis of Materials Department of Civil and Environmental Engineering Universit of Marland, College Park LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 1 Shear and Bending Although it has been onvenient to restrit the analsis of beams to pure bending, this tpe of loading is rarel enountered in an atual engineering problem. t is muh ommon for the resultant internal fores to onsist of a bending moment and shear fore.
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 2 Shear and Bending Pure Bending Bending and Shear Fore LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. Shear and Bending Fx σ xda 0 F τ xda V Fz τ xzda 0 ( τ zτ ) M x xz M M z x zσ xda 0 ( σ ) 0 Transverse loading applied to a beam results in normal and shearing stresses in transverse setions. Distribution of normal and shearing stresses satisfies x da 0 When shearing stresses are exerted on the vertial faes of an element, equal stresses must be exerted on the horizontal faes Longitudinal shearing stresses must exist in an member subjeted to transverse loading.
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 4 Shear and Bending The presene of a shear fore indiates a variable bending moment in the beam. The relationship between the shear fore and the hange in bending moment is given b dm V (42) dx LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 5 Shear and Bending Stritl speaking, the presene of shear fore and resulting shear stresses and shear deformation would invalidate some of our assumption in regard to geometr of the the deformation and the resulting axial strain distribution. Plane setions would no longer remain plane after bending, and the geometr
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 6 Shear and Bending of the atual deformation would beome onsiderabl more involved. Fortunatel, for a beam whose length is large in omparison with the dimensions of the ross setion, the deformation effet of the shear fore is relativel small; and it is assumed that the longitudinal axial strains are still distributed as in pure bending. LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 7 Shear and Bending When this assumption is made, the load stress relationships developed previousl are onsidered valid. The question now being asked: When are the shearing effets so large that the annot be ignored as a design onsideration?
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 8 Shear and Bending t is somehow diffiult to answer this question. Probabl the best wa to begin answering this question is to tr to approximate the shear stresses on the ross setion of the beam. LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 9 Shearing Stress due to Bending Suppose that a beam is onstruted b staking several slabs or planks on top of another without fastening them together. Also suppose this beam is loaded in a diretion normal to the surfae of these slabs.
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 10 Shearing Stress due to Bending P (a) Unloaded Stak of Slabs (b) Unglued Slabs loaded Figure 22 LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 11 Shearing Stress due to Bending P () Glued Slabs Unloaded (d) Glued Slabs loaded Figure 22 (ont d)
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 12 Shearing Stress due to Bending When a bending load is applied, the stak will deform as shown in Fig. 22a. Sine the slabs were free to slide on one one another, the ends do not remain even but staggered. Eah of the slabs behaves as independent beam, and the total resistane to bending of n slabs is approximatel n times the resistane of one slab alone. LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 1 Shearing Stress f the slabs of Fig. 22b is fastened or glued, then the staggering or relative longitudinal movement of slabs would disappear under the ation of the fore. However, shear fored will develop between the slabs. n this ase, the stak of slabs will at as a solid beam. The fat that this solid beam does not
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 14 Shearing Stress exhibit this relative movement of longitudinal elements after the slabs are glued indiates the presene of shearing stresses on longitudinal planes. Evaluation of these shearing stresses will be determined in the next ouple of viewgraphs. LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 15 Development of Shear Stress Formula Consider the free-bod diagram of the short portion of the beam of Figs. 2 and 24a with a retangular ross setion shown in Fig 24b. From this figure, da t d df σ da (4) (44)
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 16 Development of Shear Stress Formula x P Figure 2 A C B D x x + x A C B D t LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 17 Development of Shear Stress Formula A (a) B (b) d M V C D M + M V + V 1 V L () V R F 1 F 2 x Figure 24 t V H τ t x
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 18 Development of Shear Stress Formula The resultant of these differential fores is F σda integrated over the area of the ross setion, where σ is the flexural stress at a distane from the neutral axis (surfae) and is given b M σ (45) LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 19 Development of Shear Stress Formula Therefore, the resultant normal fore F 1 on the left end of the segment from 1 to the top of the beam is M F1 da M ( t d) Similarl, the resultant F 2 on the right side of the element is M + M M + M F2 da ( t d) (47) 1 1 (46)
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 20 Development of Shear Stress Formula n referene to Fig. 24, a summation of fores in the horizontal diretion ields V H F 2 F M + M M 1 1 1 ( t d) M ( t d) ( t d) 1 (47) LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 21 Development of Shear Stress Formula The average shearing stress τ avg is the horizontal fore V H divided b the horizontal shear area A s t x between setion A and B. Thus τ avg V A H s M ( t x) 1 ( t d) (48)
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 22 Development of Shear Stress Formula n the limit as x approahes zero, we have M τ lim ( t d) x 0 ( t x) M lim x 0 x dm dx 1 t 1 t 1 1 1 t d ( t d) (49) LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 2 Development of Shear Stress Formula Reall that equation 42 relates the bending moment with the shear fore as V dm/dx. n other words, the shear fore V at the beam setion where the stress is to be evaluated is given b Eq. 42. The integral t d of Eq. 49 is alled the 1 first moment of the area.
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 24 Extreme Fiber N.A Development of Shear Stress Formula The integral t d is usuall given the smbol 1 Q. Therefore, Q is the first moment of the portion of the ross-setional area between the transverse line where the stress is to be evaluated and the extreme fiber of the beam. Q t d (50) Extreme Fiber 1 LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 25 Shear on the Horizontal Fae of a Beam Element Consider prismati beam For equilibrium of beam element Fx 0 H + ( σ D σ D ) da A M D M H C da A Note, Q da A dm M D MC x V x dx Substituting, VQ H x H VQ q shear x flow
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 26 Shear on the Horizontal Fae of a Beam Element Shear flow, H VQ q shear x where flow Q da A first moment of area above 1 2 da A+ A' seond moment of full ross setion Same result found for lower area H VQ q q x Q + Q 0 first moment with respet to neutral axis H H LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 27 Example 12 Determine the first moment of area Q for the areas indiated b the shaded areas a and b of Fig. 25. 2 a 6 1.5 b 6 Figure 25 2
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 28 Example 12 (ont d) First, we need to loate the neutral axis from the bottom edge: 2 C ( 1)( 2 6) + ( 2 + )( 2 6) 2 6 + 2 6 72 from base 24 C 6 5 N.A LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 29 Example 12 (ont d) The first moments of area Q a and Q b are found as follows: 1.5 b a 6 2 5 N.A 6 2 Q Q a b ( 5 1.5)[ 2] 1.5 2 21in [ 1.5 6] 20.25 in
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 0 Example 1 A beam is made of three planks, nailed together. Knowing that the spaing between nails is 25 mm and that the vertial shear in the beam is V 500 N, determine the shear fore in eah nail. SOLUTON: Determine the horizontal fore per unit length or shear flow q on the lower surfae of the upper plank. Calulate the orresponding shear fore in eah nail. LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 1 Example 1 (ont d) Q A ( 0.020m 0.100m)( 0.060m) 6 120 10 m + 1 12 ( 0.020m)( 0.100m) 2[ 1 ( 0.100m)( 0.020m) 12 ( 0.020m 0.100m)( 0.060m) + 6 4 16.20 10 m 2 ] SOLUTON: Determine the horizontal fore per unit length or shear flow q on the lower surfae of the upper plank. 6 VQ (500N)(120 10 m ) q -6 4 16.20 10 m 704 N m Calulate the orresponding shear fore in eah nail for a nail spaing of 25 mm. F ( 0.025m) q (0.025m)( 704 N m F 92.6 N
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 2 Development of Shear Stress Formula Substituting for dm/dx of Eq. 42 and of Eq. 50 into Eq. gives dm 1 τ t d dx t 1 ( V ) Q t VQ τ t 1 (51) Q 1 t d LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. Development of Shear Stress Formula τ x Eq. 51 provides a formula to ompute the horizontal (longitudinal) and vertial (transverse) shearing stresses at eah point at a beam. These vertial and horizontal stresses are equivalent in magnitude. τ x τ x τ x x
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 4 Determination of the Shearing Stress in a Beam The average shearing stress on the horizontal fae of the element is obtained b dividing the shearing fore on the element b the area of the fae. H q x VQ x τave A A t x VQ t On the upper and lower surfaes of the beam, τ x 0. t follows that τ x 0 on the upper and lower edges of the transverse setions. f the width of the beam is omparable or large relative to its depth, the shearing stresses at D 1 and D 2 are signifiantl higher than at D. LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 5
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 6 Shearing Stress Formula At eah point in the beam, the horizontal and vertial shearing stresses are given b VQ τ t (52) Where V shear fore at a partiular setion of the beam Q first moment of area of the portion of the ross-setional area between the transverse line where the stress is to be omputed. moment of inertia of the ross setion about neutral axis t average thikness at a partiular loation within the ross setion LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 7 Shearing Stress Formula How aurate is the shearing stress formula? The formula is aurate if t is not too great. For a retangular setion having a depth twie the width, the maximum stress as omputed b more rigorous method is about % greater than that given b Eq. 52. f the beam is square, the error is about 12%. f the width is four times the depth, the error is about 100 %.
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 8 Shearing Stress Formula How aurate is the shearing stress formula? t t t t d Great t is small % error d 2 t 12% error d t 100% error, worst ase 4d t LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 9 Variation of Vertial Shearing Stress in the Cross Setion N.A V Max Stress
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 40 Shearing Stresses τ x in Common Tpes of Beams For a narrow retangular beam, VQ V 2 τ x 1 b 2 A 2 V τmax 2 A For Amerian Standard (S-beam) and wide-flange (W-beam) beams VQ τave t V τ max Aweb LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 41 Example 14 A mahine part has a T-shaped ross setion and is ated upon in its plane of smmetr b the single fore shown. Determine (a) the maximum ompressive stress at setion n-n and (b) the maximum shearing stress.
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 42 Example 14 (ont d) 4 in 0.5 in 1.5 kips 12 in n 2 in 15 in n 0.5 in LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 4 Example 14 (ont d) V (kip) M (kip-in) 12 in (-) (+) n 1.5 1.5 kips 1.5 kips 12 in 15 in n n M n-n 18 n 22.5 M V
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 44 Example 14 (ont d) First, we need to loate the neutral axis. Let s make our referene from the bottom edge. 4 in 0.5 in 2 in 0.5 in C ten ( 1)( 2 0.5) + ( 2 + 0.25)( 4 0.5) 2 0.5 + 4 0.5 2.5 1.8 0.667 in M r Max. Stress x max om 1.8 in 1.8 in max LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 45 1.8 in Example 14 (ont d) 4 in Next find the moment of inertia about the neutral axis: 0.5 in C N.A. 2 in 0.5 in x ( ) 4( 0.667).5( 0.167) 4 0.5 1.8 + 1.417 in (a) Maximum normal stress is a ompressive stress: σ M n ( ) 18 1.8 1.417 max max n 2. ksi (C)
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 46 Example 14 (ont d) 1.8 in (b) Maximum shearing stress: The maximum value of Q ours at the neutral axis. Sine in this ross setion the width t is minimum at the neutral axis, the maximum shearing stress will our there. Choosing the area below a-a at the neutral axis, we have 4 in 0.5 in 0.5 in N.A. 2 in 1.8 Q ( 0.5)( 1.8) C 2 VQ 1.5 ( 0.840) τ max t 1.417 ( 0.5) 0.840 in 1.778 ksi LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 47 Further Disussion of the Distribution of Stresses in a Narrow Retangular Beam Consider a narrow retangular antilever beam subjeted to load P at its free end: 2 P τ x 1 2 A 2 Px σ x + Shearing stresses are independent of the distane from the point of appliation of the load. Normal strains and normal stresses are unaffeted b the shearing stresses. From Saint-Venant s priniple, effets of the load appliation mode are negligible exept in immediate viinit of load appliation points. Stress/strain deviations for distributed loads are negligible for tpial beam setions of interest.
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 48 Example 15 SOLUTON: Develop shear and bending moment diagrams. dentif the maximums. A timber beam is to support the three onentrated loads shown. Knowing that for the grade of timber used, σ all 1800 psi τ all 120psi determine the minimum required depth d of the beam. Determine the beam depth based on allowable normal stress. Determine the beam depth based on allowable shear stress. Required beam depth is equal to the larger of the two depths found. LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 49 Example 15 (ont d) SOLUTON: Develop shear and bending moment diagrams. dentif the maximums. Vmax kips M max 7.5kip ft 90kip in
LECTURE 14. BEAMS: SHEARNG STRESS (6.1 6.4) Slide No. 50 Example 15 (ont d) 1 bd 12 1 2 S bd 6 1 2 (.5in. ) d 6 ( 0.58in. ) d 2 Determine the beam depth based on allowable normal stress. M σ max all S 90 10 lb in. 1800 psi 2 d d 9.26in. ( 0.58in. ) Determine the beam depth based on allowable shear stress. V τ max all 2 A 000lb 120psi 2 (.5in. ) d d 10.71in. Required beam depth is equal to the larger of the two. d 10.71in.