Chapter 9 BIAXIAL SHEARING

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Sherman Cain
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1 9. DEFNTON Chapter 9 BAXAL SHEARNG As we have seen in the previous chapter, biaial (oblique) shearing produced b the shear forces and, appears in a bar onl accompanied b biaial bending (we ma discuss about pure oblique shearing onl eventual in a cross section, but not in the entire bar). For this reason the tpes of loading which produce oblique shearing are the one presented in chapter 8 (Fig.8. and Fig.8.3): a. The loads applied parallel to G ais will produce the shear force (Fig.8.), respectivel those parallel to G ais will produce the force. b. The resultant shear force from section (Fig.8.3) will be decomposed into the components: cosα and sinα So, in both case of loading, each shear force will generate on turn shear ( tangential ) stresses in the vertical elements (the webs) of the cross section and in the horiontal elements (the flanges) of the cross section: : S ( ) ; b S b ς ( ς ) S ( η) S ( ) : ; bη b and finall, from superposing: CROSS SECTONS SYMMETRC WTH RESPECT TO PERPENDCULAR AXES (PRNCPAL AXES) 9.. Rectangular cross section As we have seen at straight shearing (Chapter 8) the shear force produced a tangential stress having a parabolic distribution along the rectangle side which is parallel to the shear force. Corresponding to this observation, the shear force will produce a shear stress having also a

2 parabolic distribution, but now, along the rectangle side that is parallel to the shear force. Fig.9. As we see in figures 9. and 9., the tangential stress along the rectangle side of length h has the maimum value along G ais: T, 5 ma A while the tangential stress has the maimum value along G ais: T ma,5 A T T Fig.9. From the distribution of the tangential stresses presented in a perspective view in figure 9. we ma conclude that the maimum values of both shear stresses ma and ma are superposed onl in the cross section centroid, where: + +,5, 5 ma A A

3 9.. The double T () section, made from narrow rectangles As we know from straight shearing, the shear force will produce the shear stress parallel to G ais, having a parabolic distribution and a maimum value in the neutral ais G (figure 9.3). From the same shear force, in both flanges shear stresses will eist, parallel to G ais, having a linear distribution and maimum values in the web vicinit (Fig. 9.3). Fig.9.3 Similarl the shear force will produce the shear stress parallel to G ais (Fig. 9.3), with a parabolic distribution and a maimum value in the neutral ais G (practicall, the maimum values of are in the vicinit of the web, where the cross section width is much smaller, being the flange thickness). n the cross section web produced b will be ero (the static moment S 0). For this reason the shear force ma be considered to be divided into equal parts / to each flange. We ma admit a maimum value for the shear stress produced b corresponding to the narrow rectangle:,5 A ma,5 bt Finall, the maimum shear unit stresses are: : in the section corresponding to G ais ma ma + : in the section corresponding to G ais ma ma ma

4 9.3 CROSS SECTONS SYMMETRC WTH RESPECT TO A SNGLE PRNCPAL AXS 9.3. Simple conne cross section f the cross section is a monosmmetrical double T () section, the shear stresses distribution is similarl to the one presented in the previous paragraph, in figure 9.3. The difference is that the shear force, which pass through the shear center C (whose position is unknown, et), is now decomposed into different components s i and (figure 9.4) inversel proportional to the distances h s and h i measured from C to the median lines of the flanges: h s i and hi + hs i hs h + h i s Fig.9.4 As this proceeding impose first the determination of the shear center position (presented in the net paragraph) we can t calculate the shear stresses from as for the double smmetrical section. We ma calculate these shear stresses appling Juravski s formula: S b n the above formula the static moment and the moment of inertia are written with respect to the neutral ais G.

5 Finall, the maimum shear unit stresses are calculated in the same manner adding the stress from to the one from. f the monosmmetrical section has two webs (figure 9.5) we shall proceed in the same manner, writing Juravski s formula for, separatel from each shear force and. The shear stresses distribution for this cross section is presented in Fig Fig Sections of an arbitrar shape n this paragraph we shall see an interesting section, presented in what concern the oblique shearing. This is the angle with equal legs (Fig. 9.6). Fig.9.6

6 As Juravski s formula was demonstrated for the compound solicitation of shearing with bending, the formula is applied in the principal sstem of ais G of the cross section. So, whatever is the tpe of loading (fig.8. or 8.3), the shear forces will act in the shear center C (situated to the intersection of the median lines of each leg) parallel to G and G ais, the principal ais passing through the centroid G, but rotated with 45 o with respect to the central sstem of ais G (figure 9.6). Now, the median lines aren t parallel with neither nor. That s wh both shear stresses and will have a parabolic distribution along each leg. The maimum shear stress ma will be in the angle corner, as the neutral ais G, when calculating, intersects the cross section median line in its corner. Similarl to this, the maimum shear stress ma will be in the points of intersection (points A and B) the neutral ais G (necessar when calculating ), with the cross section median line (figure 9.6). To prove the parabolic distribution of both shear stresses and, we consider a calculus level s. The distances from the centroid of the hatched area (figure 9.6) of length s to the principal sstem of ais G, are: d+e a-s/ d s s d a and d a e The static moments with respect to each principal ais, are: s S t s d t s a s S t s d t s a e As we ma see from the above epressions, the variable s is at the second power in both static moments, what prove the parabolic distribution of and. Finall, in a certain section s the compound shear stress will be: S S + + t t

7 9.4 THE POSTON OF THE SHEAR (TORSONAL) CENTER The shear center C is the significant point from the cross section, in which the shear stresses are reduced. ts eact position is et unknown, what we know until now is that the shear center is located somewhere on the smmetr ais, for the monosmetrical sections (if the cross section has smmetr ais the shear center C is in their intersection, so it is identicall with the centroid G) Channel ( U ) section We consider a channel (U shape) section with thin walls (Fig. 9.7). We admit that the shear force pass through the shear center C, so the section is subjected onl to uniaial shearing. The torsion moment is null: ( M ) 0 t C We represent the shear stresses distribution in the cross section walls. As we know, from we represent the tpical parabolic distribution of in the cross section web, while in flanges the linear distribution of is drawn. Along each median line the resultants of these shear stresses, R and R, are also represented (figure 9.7). Fig.9.7

8 Assuming G is the neutral ais (the forces plan is parallel to G), the significant values of are: S h h h bht dh + + bt d d d 4 d 8 S d S 3 3 t d bht bh The resultants of the shear stresses R and R are calculated as the volume of each diagram (taking account that, on the narrow rectangle thickness, is constant, having a uniform distribution): R i A t i With this formula, each resultant is: 3 ( ) bh t dh R + h h d + 3 But: R bt 3 b ht dh bt h dh bh t + + bt + 4 n the above epression of the moment of inertia the second term is ver small and it was neglected. This moment of inertia is replaced in the epression of R and finall we obtain: R We write the condition that the torsion moment with respect to the shear center C, is null: ( t ) M C 0 : R η + R h 0 R h η R After replacing R and R we obtain the distance η measured from the median line of the web to the shear center (Fig. 9.7): b h η 4 t

9 9.4. section with unequal flanges We want to compute the shear center position for the cross section presented in paragraph 9.3. (Fig.9.4). The shear center C is located on the smmetr ais G, to the distances h s respectivel h i with respect to the median line of the superior respectivel inferior flange (Fig.9.8). Fig.9.8 For straight bending with respect to G ais, which is the neutral ais, the corresponding shear force that produces the straight shearing is. As we have seen in paragraph 9.3., is decomposed in s and i, which give a parabolic distribution for in each flange (Fig. 9.8). from in the cross section web is null (the static moment S 0). The maimum values of in each flange are: S b b b t t t 4 8 b 8 The resultants R and R are fractile from, respectivel s R and i R

10 b t 3 R b t 3 R b t 3 i n the above relations and are the moments of inertia of each flange, with respect to the neutral ais G. The torsion moment with respect to the shear center C, is null: ( M t ) 0 : R hs + R hi 0 C But: hs + hi h hs h hi Replacing h s and R and R we get: ( h hi ) + hi 0 But: w hd s ( + ) h 0 h + i Replacing in the above equation, we finall find the shear center position, through the distances: And: hi h hs h The calculation steps used to determinate the position of the shear center C We eplain each step which should be followed in order to find the position of C, on the cross section with webs from paragraph 9.3. (Fig. 9.9). a. We position the shear center C on the smmetr ais (in our eample G ais) b. n C we appl a shear force perpendicular to the smmetr ais ( in our case) c. From this shear force we establish the direction of the shear stresses flow in all narrow rectangles and we represent the diagrams of the shear stresses, with parabolic distribution in the cross section flange, respectivel with linear variation in the cross section webs, but having different orientation and signs. d. We compute all the significant values of the shear stresses from these diagrams. For our cross section we appl Juravski s formula

11 from :. As in all the values of the ratio remain b S constant, we ma note this ratio with k, so k cst., and all the significant values of will be function this constant : f ( k) Fig.9.9 e. We compute the resultants R i corresponding to each diagram i represented on the narrow rectangle of thickness t i. We appl the formula: Ri A t i i, where A i is the area of i diagram. f. We write the condition that the torsion moment with respect to the shear center C, is null. For our cross section (Fig. 9.9): ( t ) M C 0 : R η R R3 d η + R3 d 0 η R + R g. From this condition we find the position of the shear center C given b this distance η. f the cross section has no smmetr ais, the same steps must be followed, but separatel on each ais, first from with G the neutral ais and then from with G the neutral ais. Finall we shall obtain coordinates for the shear center C: η and ζ.! All the calculations must be made in the principal sstem of ais G.

12 9.5 APPLCATON TO BAXAL BENDNG WTH SHEARNG For the simple supported beam with the static scheme and the cross section from the figure bellow calculate: a. diagrams of stresses b. the strength verification at bending ( 00 / ). n the critical sections at bending the neutral ais and the normal stress diagram σ will be represented c. in section B (B left) the shear stress diagram with the resultant values at the levels - and - The diagrams of stresses:

13 The critical sections at bending are: - section E, where M ma 6.56kNm and M 3.75kNm, so a section of biaial bending - section B, where M 0 and M ma 495kNm, so a section of uniaial bending With the second moments of area 9cm 4 and 38975cm 4, the maimum normal stresses and the neutral ais are: - in section E: + ( ) < ( ) The neutral ais equation:

14 - in section B: (495 0 ) < 00 n section B the shear forces: ma -.5kN and ma 65kN

Chapter 8 BIAXIAL BENDING

Chapter 8 BIAXIAL BENDING Chapter 8 BAXAL BENDN 8.1 DEFNTON A cross section is subjected to biaial (oblique) bending if the normal (direct) stresses from section are reduced to two bending moments and. enerall oblique bending is