V. DEMENKO MECHNCS OF MTERLS 015 1 LECTURE 14 Strength of a Bar in Transverse Bending 1 ntroduction s we have seen, onl normal stresses occur at cross sections of a rod in pure bending. The corresponding internal forces give the resultant bending moment at the section. n transverse bending, a shearing force Q occurs at a section of a rod as well as a bending moment. This force represents the resultant of elementar distributed forces ling in the plane of the section. Consequentl, in this case not onl normal, but also shearing stresses occur at cross section of a rod: M = σ xd, (1) Q = τ xd. () The appearance of shearing stresses t is accompanied b shearing Fig. 1 strains g. Hence an elementar section d undergoes angular displacement due to shear. The magnitude of normal stresses is not, however, sensibl affected b the shearing forces. hen the shearing force varies along the rod axis the formula of pure bending introduces an error in the calculation of σ x : M σ x( )= () B a simple analsis it ma be shown that this error is of the order of h/ l compared to unit, where h is the dimension of the cross section in the plane of bending and l is the length of a rod.
V. DEMENKO MECHNCS OF MTERLS 015 B definition the characteristic feature of a rod is that the dimensions its cross section are much smaller than the length. Consequentl, the quantit h/ l is ver small, and so is the error involved. Determination of Shearing Stresses Let us estimate the shearing stresses in transverse bending. n general case of external loading we have the element of the rod dx in equilibrium: Fig. solate an element of length dx from the rod. Divide the element in two parts b longitudinal horiontal plane (laer) passed at a distance from neutral laer and consider the conditions of equilibrium of the upper part: x F = 0 : σ d+ τ bdx ( ) ( σ + dσ ) d = 0. (4) x x x x * * e assume that the shearing stresses are uniforml distributed across the width of the laer b: () σ d+ τ bdx ( ) σ d dσ d = 0, x x x x * * * dm τxbdx ( ) = dσ xd, dσ x =, *
V. DEMENKO MECHNCS OF MTERLS 015 dm τ xbdx ( ) = d, * where d= Sna.. = S * * * is static(al) moment (first moment) with respect to the central axis of the portion of the area located above (or below) the laer: Since dms* τ xbdx ( ) =. dm dx = Q, we get QS * τx( ) = τ x( ) = Juravsk formula. (5) b ( ) Examples of Shear Stress Distribution in Different Cross Sections Example 1 Shear stress distribution along a height of rectangle Fig. Fig. 4
4 V. DEMENKO MECHNCS OF MTERLS 015 b ( )= b, bh =, 1 h h 1 1 h S* = * c = b + = b, 4 h 1 Q h Q bh b bh 4 h = 0 = 1 Qb 4 6 τ ( ) = = = = 0. Note, that imum stress occurs at = 0 and are distributed uniforml along axis. Example Shear stress distribution along a height of round cross section n this case we have: Fig.5 4 Comparison of Normal and Shearing Stresses in Bending 1. Points and '. Fig. 6
V. DEMENKO MECHNCS OF MTERLS 015 5 These points are most remote from the neutral axis. Shearing stresses are ero: Fig.7. Point B. σ = 0. Shearing stresses are imum: x. Points C and C'. Fig.8 There is biaxial stress state at these points: Fig. 9 Let us determine the principal stresses using general formula: σ + 1 ( ) 1,() = x σ σ ± σx σ + 4τ x. Since σ = 0, σ 1 σ1 = x + σx + 4τ x, σ = 0, σ 1 σ = x σx + 4τ x.
6 V. DEMENKO MECHNCS OF MTERLS 015 comparison ma be made of the absolute values of the imum normal and imum shearing stresses on cross section of a prismatic rod. For example, for a cantilever beam of rectangular section we have: whence Fig. σ M 6Fl, F τ =, bh = = na.. bh τ σ 4 = h. l This means that the imum shearing stresses and the imum normal stresses in the cross section are in about the same ratio as the cross-sectional dimension of the section and the length of the rod, i.e. the shearing stresses are appreciable less than the normal stresses for rectangle cross section. Because of the small magnitude of τ the analsis of prismatic solid rods subjected to transverse bending is made onl on the basis of normal stresses as in pure bending. Shearing stresses are not taken into consideration. is Therefore, the condition of strength in transverse bending of solid prismatic bars M σ = [ σ ]. (6) na..( ) 5 Curvature of a Beam hen loads are applied to a beam, its longitudinal axis is deformed into a curve. s an example, consider a cantilever beam B subjected to a load a F at the free end (Fig. 11):
V. DEMENKO MECHNCS OF MTERLS 015 7 Fig. 11 The initiall straight axis is bent into a curve, called the deflection curve of the beam. Let us construct a sstem of coordinate axes (x, ) with the origin located at a suitable point on the longitudinal axis of the beam. e will place the origin at the fixed support. The positive x axis is directed to the right, and the positive axis is directed upward. The beam considered is assumed to be smmetric about the x plane, which means that the axis is an axis of smmetr of the cross section. n addition, all loads must act in the x plane. s a consequence, the bending deflections occur in this same plane, known as the plane of bending. The deflection of the beam at an point along its axis is the displacement of that point from its original position, measured in the direction. e denote the deflection b the letter δ. t is important to note that the resulting strains and stresses in the beam are directl related to the curvature of the deflection curve. To illustrate the concept of curvature, consider again a cantilever beam subjected to a load F acting at the free end (Fig. 1). The deflection curve of this beam is shown at the bottom of the Fig. 1. For purposes of analsis, we will identif two points m 1 and m on the deflection curve. Point m 1 is selected at an arbitrar distance x from the axis and point m is located at a small distance ds further along the curve. t each of this points we draw a line normal to the tangent to the deflection curve, that is, normal to the curve itself. This normals intersect at point O', which is the center of curvature of the deflection curve. Because most beams have ver small deflections and nearl flat deflection curve, point O' is usuall located much further from the beam that is indicated in the Fig. 1.
8 V. DEMENKO MECHNCS OF MTERLS 015 The distance mo 1 ' from the curve to the centre of curvature is called the radius of curvature ρ, and the curvature k is defined as the reciprocal of the radius of curvature. Thus, k = 1 ρ. (7) Curvature is a measure of how sharpl a beam is bent. From the geometr of triangle Omm ' 1 we obtain ρdθ = ds (8) and (8), we get Fig. 1 in which dθ (measured in radians) is the infinitesimal angle between the normals and ds is the infinitesimal distance along the curve between points m 1 and m. Combining Eqs (7) k 1 = = d θ ρ ds. (9) t is interesting to note, that if the curvature is constant throughout the length of a curve, the radius of curvature will also be constant and the curve will be an arc of a circle. The deflections of a beam are usuall ver small compared to its length. Small deflections mean that the deflection curve is nearl flat. Consequentl, the distance ds along the curve ma be set equal to its horiontal projection dx (Fig. 1). Under this conditions the equation for the curvature becomes k 1 = = d θ ρ dx. () Both the curvature and the radius of curvature are functions of the distance x. t follows that the position O' of the center of curvature also depends upon the distance x. e will see that the curvature at a particular point on the axis of a beam depends on the bending moment at that point and on the properties of the beam itself. Therefore,
V. DEMENKO MECHNCS OF MTERLS 015 9 if the beam is prismatic and the material is homogeneous, the curvature will var onl with the bending moment. Consequentl, a beam in pure bending will have constant curvature and the beam in transverse bending will have varing curvature. The sign convention for curvature depends on the orientation of the coordinate axes. f the x axis is positive to the right and the axis is positive upward, then the curvature is positive when the beam is bent concave upward (or convex downward) and the center of curvature is above the beam. Conversel, the curvature is negative when the beam is bent concave downward (or convex upward) and the center of curvature is below the beam. This sign convention is represented on the Fig. 1. Fig. 1 6 Examples of practical problems solution Example 1 Checking problem of cantilever beam Given: [ σ ] = 160MPa, rectangle crosssection: width l = m, height h = 40 m, a = m, b = m, c = 4 m, P = 40 kn, M = 0kNm, q = 0 kn/m. R.D.: Check the strength of cantilever under specified loading. Solution 1) Calculating the internal forces Q ( x ) and M( x ), appling the Fig.14 method of sections.
0 x 4m V. DEMENKO MECHNCS OF MTERLS 015 0 x m x= 0 x= c Q ( x) = qx = 0 = 80 kn; qx x= 0 x= c M ( x) = = 0 = 160 knm Q ( x) = qc = 80 kn; c M ( x) = qc + x M x= 0= 180 x= b= 40kNm; 0 x m Q ( x) = qc P= 40 kn; c M ( x) = qc + b+ x M Px x= 0= 40 x= a= 40kNm. ) To check the strength, we will write condition of strength for critical laer of critical cross-section, i.e. the cross-section with M = 40 knm: M σ = [ σ ]. но. n our case, σ 6M 6 40 4 Since 157.5 < 160, the cantilever is strong. = = = 157.5MPa. lh 40 ) Designing the graphs σ ( ) and τ ( ) in critical cross-section with M = 40 knm and Q = 40 kn (see Fig. 15).
V. DEMENKO MECHNCS OF MTERLS 015 11 Since t τ M σ ( )= ; Fig. 15 QS ( ) ( )= τ ; b hh Qb = 4= = = 1.5MPa. bh 1 4 b 1 Q 40 1 1 << s we will ignore shear stresses in stress analsis of prismatic beams. 4) Calculating imum shear stresses in the beam. n prismatic beams of constant cross-section imum shear stress will act in the cross-section with Q = 80 kn. Corresponding M = 40kNm. These internal forces are applied in Fig. 16. Fig.16
1 V. DEMENKO MECHNCS OF MTERLS 015 τ Q 80 1 1 == = = MPa. 1 4 n this cross-section, normal stresses are calculated b the formula σ 6M 6 40 4 = = = 17.5MPa. lh 40 Conclusion. n stress analsis of prismatic beams, it s possible to ignore influence of shear stresses on the results of calculation. Solution Example Design problem for cantilever beam Fig. 17 Given: Steel cantilever is loaded b linearl distributed loading with imum intensit qm = kn/m and concentrated force P = kn. lso, [ σ ] = 160MPa, a = m, b =1m. R.D.: 1) number of -beam section; ) dimensions of rectangle cross-section in h b = ; ) diameter of round solid cross-section. 1) Designing the graphs of shear forces Q( x ) and bending moments M( x ), appling method of sections. 0 x 1m M 0 x m qx x= 0 x= b Q ( x) = qx = 0 = 8.kN; ( a+ b) qx qx x x= 0 x= b ( ) = + = 0 = 4.44 knm. 6( a+ b)
V. DEMENKO MECHNCS OF MTERLS 015 1 ( x+ b) x= 0 x= a Q ( x) = qx ( + b) q + P = 18. = 5kN; ( a+ b) qx ( + b) qx ( + b) x= 0 x= a M ( x) = + Px = 4.44 = 50kNm. 6( a+ b) t is clear that critical cross-section is in rigid support since M = 50 knm. ) Calculating the sectional modulus of cross-section from condition of strength: M σ = [ σ ]; M 50 = = 1.5 [ σ] 6 160 6 m. ) Determining the dimensions of -beam section knowing that 1.5 6 m. From assortment of steel products let us find near situated -beam 4 with = 89 6 Overstress m and estimate its overstress: σ M = = = 17 MPa. 50 6 89 17 160 = 0% = 8.1% > 5%. t is more than allowable overstress 160 equals to 5%. This result requires to choose larger neighbor -beam section 4 with sectional modulus = 17 6 σ Percentage of understress m. t will be understressed since M = = = 157.7MPa. 50 6 17 160 157.7 = 0% = 1.4%, 160 The dimensions of -beam section 4 are presented in the Table: h b d t мм cm cm 4 cm 40 15 5.6 9.8 7.5 800 17 178 S cm
14 V. DEMENKO MECHNCS OF MTERLS 015 4) Designing the graphs of stress distribution σ x( ) and τ ( ) in critical cross-section of -beam section 4. Fig. 18 To draw the graph of shear stress distribution it is necessar to use shear formula (Juravsk formula): τ Q S τ ; ( )= b т Q 6 S 5 178 d 8 5.6 800 = = = 0.9 MPa. Shear stress at the point 1 belonging to the -beam flange is τ h t 40 9.8 Qbt 5 9.8 = = = 0.74 MPa. p.1 b 8 800 Shear stress at the point 1' belonging to the -beam web is τ h t 40 9.8 Qbt 5 15 9.8 = = = 16.6MPa. p.1' d 8 5.6 800 Corresponding graphs are shown in Fig. 18.
V. DEMENKO MECHNCS OF MTERLS 015 15 4) Determining the dimensions of rectangle section knowing that 1.5 6 m. M 50 6 1.5 6 bh = = = 6 [ σ ] 160 m. h Substituting = we get b 4 6 b 1.5 and 6 6 1.5 b = 7.8 m and h= b= 15.6 m. Cross-sectional area of rectangle section is 4 = bh = 7.8 15.6 = 11.8 m. The graphs of σ ( ) and τ ( ) distribution in critical cross-section are shown in Fig. 19. σ Fig. 19 M 6M 6 50 4 bh 7.8 (15.6) = = = = 158 160 MPa; τ Q 5 4 = =.1 15.6 7.8 = MPa. 5) Determining the dimensions of round section knowing that 1.5 6 m.
16 V. DEMENKO MECHNCS OF MTERLS 015 d 6 = π 1.5 m. d 6 1.5 = = 14.7 π.14 m. Cross-sectional area of round section is ( ) d.14 14.7 4 = π = = 169.7 m. 4 4 The graphs of σ ( ) and τ ( ) distribution in critical cross-section are shown in Fig. 0. τ σ Fig. 0 M = = = 160 MPa;.14 14.7 50 ( ) Q S 4Q 4 5 b 4 169.7 6) Comparing the cross-sectional areas: = = = = MPa. < < 7.5 < 11.8 < 169.7. Note, that the -beam section is the most effective in strength-to-weight ratio.
V. DEMENKO MECHNCS OF MTERLS 015 17 Example Problem of allowable external load for two-supported beam Given: steel simple beam with hollow rectangle cross-section is loaded b distributed load q and concentrated moment M in the left support. H = cm, B = 6cm, h = 4cm, b = cm, [ σ ] = 00 MPa, a = 1m. R.D.: the largest working (allowable) load [q]. Н h b В Fig. Note, that allowable value of internal bending moment is determined from condition of strength: [ M] [ σ ]. On the other hand, the method of section connects the external forces with internal bending moments. That is wh we begin from calculating the internal forces to find critical cross-section. Solution 1) Calculating the reactions in supports: а) M = 0: Fig. 1 a M + RB a q a = 0, 9 5 5 RB a= qa qa = qa RB = qa ; 4 a b) MB = M + qa a qa R a = 0,
18 V. DEMENKO MECHNCS OF MTERLS 015 1 7 R a= qa + qa qa R = qa. 4 c) Checking: 5 7 F = R + R q a= qa+ qa qa = 0. 4 4 B The reactions are correct. ) Designing the graphs of shear forces Q( x ) and bending moments M( x ), appling the method of sections. 0 x a ( ) x= 0 0 x= a Q x = qx = = qa, 0 x a qx 1 ( ) = x= 0 = 0 x= a = M x qa. 5 1 1 7 Q ( x) = qa ( + x) RB = qa ( + x) qa= qx qa x= 0 = qa x= a = qa. 4 4 4 4 Finding of extremum bending moment value due to intersecting the shear force graph with x-axis in the second portion. 1 1 Q( x0) = qx0 qa= 0 x0 = a coordinate of the intrsection. 4 4 qx ( + a) qx ( + a) 5 ( ) B x= 0 M x = + R x= + qax = 4 1 15 = qa x= a= qa x= 0.5a = qa. n result, = M qa (see Fig. 1). ) Calculating the sectional modulus of the cross-section (see Fig. ): BH bh 6 4 1 1 1 1 1 600 64 = = = = = 98 cm. H 5 60
V. DEMENKO MECHNCS OF MTERLS 015 19 4) Calculating the allowable loading [q]: M σ = [ σ ], M [ σ ] 6 6 [ σ ] 00 98 9800 [ qa ] = [ σ ] [ q] = = = = 9.8kN/m. a a a Example 4 Checking problem for cantilever in plane bending Given: cast iron cantilever beam of T-section is loaded b the forces: distributed load q =knm, concentrated moment M = 0kNm and concentrated force P = kn. Cross-sectional dimansions: a = 6cm, b = 1cm, c = cm, d = 1cm. llowable stresses: [ σ ] = 00MPa, [ σ ] = 400 MPa, length of the portion l = 1m. t c R.D.: check the cantilever strength and select optimal orientation of the cross-section relative to plane of loading. a B b c d c B Fig. Fig.4 Solution 1) Calculating the internal forces and moments in the cantilever cross-sections 0 x 1m ( ) x= x= l Q x = P+ qx 0 = = 0kN;
0 V. DEMENKO MECHNCS OF MTERLS 015 0 x 1m qx ( ) x= x l M x = Px 0 = 0 = 15 knm. ( ) Q x = P+ ql =0 kn; ( + l) q x M ( x) = P( x+ l) + M x= 0 = 5 x= l = 15 knm. ) Calculating the neutral axis position. For this, let us select -axis as original (see Fig. 5). B B Fig. 5 b) determine the neutral axis position using the formula Vertical coordinate c a) Determine first moment of cross-sectional area relative to -axis: d c S = ad ( c ) = 1 = 6 ( 1 ) = 1 cm ; c S = bc d = = 1 1 = 70cm. c S + S 1+ 70 = = =.7cm and =.7cm, B = 8.7 cm. + 6 ( 1 ) + 1 ) Calculating the central moment of inertia. ( ) ( ) d c a d c = + ( d ca ) = c 1 1 6 1 = + ( 1 ) 6.7 = 15.44 cm 4 ; 1
V. DEMENKO MECHNCS OF MTERLS 015 1 ( ) bd c c = + bd ( c) d = c 1 ( ) 1 1 = + 1 ( 1 ) 1+.7 = 78.68 cm 4. 1 = + = 15.44+ 78.68= 404.1cm 4. c c c 4) Calculating the imum normal stresses in and B points (see Fig. 6): σ M = = c 15.7 = = 19.1 8 404.1 MPa, Fig. 6 σ B MB = = ( ) 15 1.7 404.1 8 c = 08.5MPa. 5) Checking the strength of the beam in two possible cases of its orientation to determine its optimal orientation relative to the plane of loading. Let us compare stresses in and B points for two possible cases of cross-section orientation. For orientation (see Fig. 7) tens σ = σ = 19.1MPa. Since [ σ ] t = 00MPa the most tensile point is strong. Simultaneousl, comp σb = σ = 08.5 MPa. Since [ σ ] c = 400 MPa the most compressed point B is also strong. Therefore, this orientation of cross-section corresponds to the condition of strength (see Fig. 7).
V. DEMENKO MECHNCS OF MTERLS 015 Fig. 7 For orientation comp σ = σ = 19.1MPa. Since [ σ ] c = 400 MPa the most compressed point is strong. Simultaneousl, tens σb = σ = 08.5 MPa. Since [ σ ] c = 00 MPa the most tensile point B becomes unstrong (Fig. 8). Fig. 8 n result, this orientation of cross-section does not satisf the condition of strength. Conclusion. To be situated rationall, larger acting stresses must correspond to larger allowable stresses.
V. DEMENKO MECHNCS OF MTERLS 015 Example 4 Design problem for round simple beam Given: M = 4kNm, q = kn/m, [ σ ] = 160MPa. R.D.: Diameter of round cross-section. Solution 1) Calculating the support reactions: (a) M = 0 : B R 4+ q M = 0; R (b) M = 0 : 4 = =.5 kn; 4 RB 4 M q 1 = 0; RB 4+ 1 = =.5 kn. 4 Fig. 9 (c) Checking: F = 0:.5+.5 = 0. The reactions are correct. ) Designing the graphs of shear forces Q( x ) and bending moments M( x ), appling the method of sections. І 0 x m ( ) x= 0 x= Q x = R qx=.5 x =.5 =.5kN. n cross-section with Q = 0, M graph grows to imum value. Using equation ( ) Q x 0 = 0 let us find 0 x :.5 x = 0 x = 1.17 m. 0 0 x ( ) x= 0 x= 1.17 x= M x = R x q =.5x 1.5x = 0 =.05 = 1kNm. 0 x 1m Q = R q =.5 =.5kN.
4 V. DEMENKO MECHNCS OF MTERLS 015 ( ) ( ) ( 1) M x = R x+ q x + = 0 x 1m ( ) ( ) x= 0 x= 1 =.5 x+ 6 x+ 1 = 1 = 1.5kNm. ( ) Conclusion. M =.5kNm. ( ) Q x = R =.5 kn. B x= 0 x= 1 M x = R x=.5x = 0 =.5 knm. ) Determining the diameter of round beam from the condition of strength in critical cross-section with M =.5kNm. σ M M.5 6 = [ σ] = = 15.6 [ ] 6 σ 160 6 πd 15.6 = d = = = 5.4 π π Example 5 Checking problem for -beam B m = 15.6 cm. m. Given: Cross-section -beam 50, [ σ ] = 160MPa, P = 0 kn, M = 40kNm, q = 0 kn/m. R.D.: Check the strength of the beam. Solution 1) Calculating the support reactions: (а) M = 0 : B R l+ q 6 9+ P M = 0; Fig. 0 R 0 6 9+ 0 40 = = 14 kn. (b) M = 0 : RB l P 7 M q 6 1 = 0;
(c) Checking: F = 0 : V. DEMENKO MECHNCS OF MTERLS 015 5 RB The reactions are correct. 0 7+ 40+ 0 6 1 = = 86 kn. 14+ 86 0 6 0= 0. ) Calculating the shear forces Q( x ) and bending moments M( x ) appling the method of sections. 0 x m 0 x 4m x= 0 x= Q ( x) = qx= 0x = 0 = 40kN; qx x= 0 x= M ( x) = = x = 0 = 40kNm; ( ) ( ) x= 0 x= 4 Q ( x) = q x+ + R = 0 x+ + 14 = 94 = 14 kn; ( + ) q x M ( x) = + Rx= ( x+ ) + 14x x= 0 = 40 x= 4 = 176kNm; 0 x m V V 0 x m Q ( x ) = q 6+ R = 0 6+ 14= 14 kn; M ( x ) = 6 qx ( + ) + R ( x + 4) = = 10( x+ ) + 14( x+ 4) = 176 = 18kNm; x= 0 x= V Q ( x ) = R = 86 kn; V B x= 0 x= M ( x) = M + R x= 40+ 86x = 40 = 18kNm. The graphs are shown in Fig. 0. B
6 V. DEMENKO MECHNCS OF MTERLS 015 ) Checking the strength of -beam 50. From assortment (GOST 89-7) find its sectional modulus = 1589 cm. n our case, M = 18 knm. Condition of strength is M 18 6 1589 σ = = = 17. MPa < [ σ ] = 160 MPa. Conclusion. The beam is strong. Example 6 Checking problem for cast iron beam Given: cast iron of T-section is loaded b q = kn/m, P = kn, M = 6kNm. (b) M = 0 : Fig. 1 Cross-sectional dimensions are shown in Figs. and for two possible orientations of the section. llowable stresses for cast iron are [ σ ] = 40MPa, [ σ ] = 80MPa. c R.D.: find optimal orientation of the beam relative to the plane of loading from two possible versions shown in Figs. and check its strength. Solution 1) Calculating the support reactions. (а) M = 0 : B R 6+ q 7+ P 4 M = 0; R 7+ 4 6 = = 8 kn; 6 t RB 6 P + q 1 M = 0; RB 6+ 1 = = 1 kn. 6
(c) Checking: F = 0: V. DEMENKO MECHNCS OF MTERLS 015 7 + 8 + 1= 0. Reactions are correct. ) Calculating the shear forces and bending moments in cross-sections of the beam. 0 x m 0 x m ( ) x= 0 x= Q x = qx= x = 0 = 6kN; qx x= 0 x= M ( x) = = 1.5x = 0 = 6kNm; Q ( x ) = q + R = + 8= kn; x= 0 x= M ( x) = qx ( + 1) + R x= 6( x+ 1) + 8x = 6 = knm; 0 x m V V 0 x m Q ( x ) = R = 1 kn; B x= 0 x= M ( x) = R x= x = 0 = knm; B V Q ( x ) = R = 1 kn; V B x= 0 x= M ( x) = R ( x+ ) M = x+ 6 = 4 = knm. B Fig. Fig.
8 V. DEMENKO MECHNCS OF MTERLS 015 ) Selection of cross-section optimal orientation in plane bending comparing two orientations shown in Figs. and. Note, that the T-section is non-smmetrical relative to neutral axis. Largest stresses appear in the most remote points of the section (points and B). Since for cast iron [ σ] < [ σ ], the largest compressed stresses must appear in the laers situated at the t c larger distance from neutral axis i.e. at B distance. That is wh the orientation shown in Fig. is optimal in this case of loading. 4) Checking the strength of the cross-section shown in Fig. (of optimal orientation). Let us write conditions the strength for and B points: M comp B σb = σ = [ σ ] c; M tens σ = σ = [ σ ] t. c c For this, let us calculate central moments of inertia c calculating preliminar vertical coordinate of the cross-sectional centroid. e will use 0 -axis as the origin (see Fig. 4). Coordinate of centroid is S 0 11+ = = c = 1+ 1 1.5+ 6 6 = = cm. 1 + 6 Central moment of inertia is: Fig.4 1 1 1 = + a + + a = c 1 1 6 + ( 1.5) 6+ 1 1 + 18 = = 4 cm 4. Then
V. DEMENKO MECHNCS OF MTERLS 015 9 tens 6 8 σ = σ = = 5.5MPa < [ σ ] t = 40 MPa; 4 comp 6 6 8 σb = σ = = 71MPa < [ σ ] c = 80 MPa. 4 Conclusion. Since both condition of strength are satisfied the beam is strong. Example 7 Problem of allowable external load Given: steel -beam No4 cantilever is loaded b the force P. The length l = m, [ σ ] = 160MPa. R.D.: calculate allowable value if the P-force from condition of strength. Solution 1) Calculating the internal forces in the cantilever appling the method of sections. 0 x m ( ) Q x = P; Fig. 5 ( ) x= 0 x= M x = Px = 0 = P. Corresponding graphs are shown in Fig. 5. ) Calculating the allowable force value [Р]. Critical section is situated in rigid support and M = P. From condition of strength: [ M ] σ = = [ σ] [ M] = [ σ ]. From assortment, (GOST 89-7) for -beam 4 we find its sectional modulus = 89 cm. fter this, allowable internal bending moment becomes 6 6 [ M ] = 160 89 = 46.4 knm and allowable force is
0 V. DEMENKO MECHNCS OF MTERLS 015 [ P] [ M ] 46.4 = = = 15.4 kn. Example 8 Problem of optimal supports placement in plane bending Given: simple beam with overhang is loaded b uniforml distributed load q. Position of the right support is determined b the k-factor. R.D.: Calculate the k-factor value which provides the largest allowable value [q]. Solution 1) Calculating the support reactions. (а) M = 0 : ql RB kl =0 ; R B = ql k. (b) M = 0 : B Fig. 6 l ql kl R kl = 0 ; R ql 1 = k. k (c) Checking: ql ql 1 ql kql+ ql F = ql k = ql = 0. k k k The reactions are correct. ) Calculating the shear forces Q( x ) and bending moments M( x ) appling the method of sections. 0 x kl
V. DEMENKO MECHNCS OF MTERLS 015 1 ql 1 ql 1 ql 1 Q ( x) = R qx= k qx x= 0 = k x= kl = k qkl. k k k n this portion, to 0: M graph has imum value which is calculated equating shear force ( ) ql 1 l 1 k qx = x = k. k k Q x 0 = 0; 0 0 0 qx ql ( 1 k) ql 1 ( ) = x= 0 = 0 x= kl = x= x = M x R x k 0 x ( 1 ) ) Calculating the k-factor. 0 k k l x= 0 x= l(1 k) Q ( x) = qx = 0 = ql(1 k ); (1 ) ( ) = qx x= 0 = 0 x= l(1 k) = ql k M x. t is evident that [q] will be the largest in value if imum bending moments in both portions will be equal to teach other: M = M ; ql 1 ql (1 k) 1 = k k = k (1 k ) k 1 = (1 1 k k k); k = ; 1 1 k = kk ( 1); k k + = 0; k 1 1 1, =± ; k,4 = 1±.. Therefore, 1 k = or k = 1 1, since another values of k contradict the problem data.
V. DEMENKO MECHNCS OF MTERLS 015 Example 9 Design problem for cantilever Given: steel cantilever is loaded b the following loadings: q = 0 kn/m, P =1 kn, M = 5kNm. The cantilever lengths: a = 1. m, b = 1m, c = 1. m. [ σ ] = 00 MPa. R.D.: 1) diameter of solid round section; ) dimensions of rectangle cross-section; ) the number of -beam section. Solution 1) Calculating the shear forces Q ( x ) and bending moments M ( x ), appling the method of sections. Fig. 7 0 x 1.m 0 x 1m x= 0 x= 1. Q ( x ) = P = 1 kn; M ( x) = Px= 1x = 0 = 14.4kNm; x= 0 x= 1 Q ( x) = P+ qx= 1+ 0x = 1 = 8kNm; 1 Q ( x0) = 1+ 0x0 = 0 x0 = = 0.6 m; 0 qx x= 0 x= 0.6 x= 1 M ( x) = Px ( + a) = 1( x+ 1.) x = 14.4 = 18 = 16.4 knm; 0 x 1,m. Q ( x) = P+ qb = 8kN;
V. DEMENKO MECHNCS OF MTERLS 015 b M ( x) = P( x+ a+ b) qb x+ M = 1 ( x+.) 0 ( x + 0.5) 5 x= 0 = = 14,4 = 18.kNm. x= 1. Conclusion. Critical cross-section is determined taking into account imum value of the bending moment: M =18.kNm. Corresponding shear force Q = 8kN. ) Calculating the sectional modulus of the cross-section using condition of strength. M σ = [ σ ]; M 18. = = 91 [ σ ] 6 00 6 m. ) Knowing 91 6 m let us find corresponding -beam number from assortment. Closest less -beam section is 14 with = 81.7 6 m. t will be evidentl overstressed: Overstress is σ M = = =.8MPa. 18. 6 81.7.8 00 = 0% = 11.4% > 5%. Since overstress is non- 00 permissible we will find in assortment larger -beam section 16 with = 9 6 Understress shown in the Table: m. ts understress is σ M = = = 167 MPa. 18. 6 9 00 167 = 0% = 16.5%. The dimensions of selected -beam are 00 h b d t S mm cm cm 4 cm cm 160 81 5.0 7.8 0. 87 9 6.
4 V. DEMENKO MECHNCS OF MTERLS 015 Drawing the graphs of acting stress distribution σ x( ) and τ x ( ) in critical cross-section. Calculating the shear stresses in specific points of the section: pp. 1, f, w,. Note, that f point belongs to the flange and w point belongs to the web. General Juravsk formula is t is evident that τ = 0 since S * = 0. QS * τ ( )=. b For f -point the formula becomes: τ p. f Fig. 8 h t 160 7.8 Qbt 8 7.8 = = = 0.54 MPa. b 8 87 For w -point the formula is the following: τ p. w h t 160 7.8 Qbt 8 81 7.8 = = = 8.8MPa. d 8 5 87 Maximum shear stresses act at the points of neutral axis: T QS 6 8 6. p.1 d 8 5 87 τ = τ = = = 11.4MPa.
V. DEMENKO MECHNCS OF MTERLS 015 5 Estimating the stress state of cross-section specific points and their strength. Note, that -beam section is an example of thin-walled section, that is wh we will estimate the stress state of specific points taking into account not onl acting normal stresses but also shear ones. Point. Point 1 Deformation tension. Stress-state uniaxial. Condition of strength is σ = 167 MPa, τ = 0. σ [ σ ], 167 MPa < 00 MPa. Conclusion: point is strong. τ = 11.4MPa, σ = 0. Deformation pure shear, Stress state biaxial. To prove this concept, let us calculate principal stresses: σ + σ 1 ( ) 4 Fig. 40 σ = τ. Condition of strength is written appling third strength theor: α β σ1() = ± σα σβ + τ ; 1= σ τ ; σeq = σ1 σ [ σ ], σeq = τ + τ [ σ ],.8 MPa < 00 MPa. Conclusion: point is strong. Point w Fig. 9 Fig. 41
6 V. DEMENKO MECHNCS OF MTERLS 015 h h h h M t M t h M t t σ = a σ p. w = = = = = h h σ h 80 7.8 = 167 = 19. MPa. 80 τ = τ =8.8MPa. a p. Let us calculate the principal stresses if σ = 19.MPa, σ = 0, τ = 8.8MPa: w α 6 6 α β 4 α 140. σα + σβ 1 19. 1 σ1 ( ) = ± ( σ σ ) + τ = ± = = 69.6MPa± 70.15MPa. n this, σ1 = 19.75 MPa, σ = 0, σ = 0.55MPa. Principal planes are situated at the α 0 angle: τ 8.8 tgα0 = α = = 0.16, α 0 = 6'' (clockwise rotation). σ σ 0 19. β α Conclusion. Stress state at this point is two-dimensional (plane). 4) Calculating the dimensions of rectangle cross-section using condition of strength. bh 6 = 91 m. 6 B α h n = b we have that 4 6 b 91 and 6 6 91 b = 5.15 m. h= b=. m. 4 = hb = 5.15. = 5.05 m. Let us draw the graphs of stress distribution in critical cross-section.
V. DEMENKO MECHNCS OF MTERLS 015 7 σ Fig. 4 M = = = 00MPa; 5.15. 6 18. τ ( ) 1/ QS Q 8 b 4 5.05 = = = =.6MPa. 5) Calculating the diameter of round section using condition of strength. d 6 = π 91 m. d 6 91 = = 9.75 π.14 m. d.14 9.75 4 4 = π = = 74.6 m. 4 4 Draw the graphs of stress distributions in critical cross-sections.
8 V. DEMENKO MECHNCS OF MTERLS 015 τ σ Fig. 4 M = = = 00MPa;.14 9.75 18. ( ) 1/ QS 4Qx 4 8 b 4 74.6 6) Comparing the cross-section areas: = = = = 1.4MPa. < < : 4 4 4 0. < 5.05 < 74.6 m. Note, that the -beam section has the largest strength- to-weight ratio.