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# Lesson 06, shearing stresses

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Lesson 06, Shearing Stresses

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### Lesson 06, shearing stresses

1. 1. Lecturer; Dr. Dawood S. Atrushi Dec. 2014 – Jan. 2015 Bending- and Shearing Stresses - y1 2 ) of the beam
2. 2. Shear Stress in Beam of Rectangular Cross Section ¢  For a beam subjected to M and V with rectangular cross section having width b and height h, the shear stress τ acts parallel to shear force V. January, 2015Bending Stresses - DAT2 ular Cross Section d V with rectangular cross section e shear stress acts parallel to the the width of the beam the a small act lly as subjected the a solate a small stresses act d horizontally as urfaces are free, be vanish, i.e. 2 tion subjected the a e isolate a small ear stresses act nied horizontally as om surfaces are free, must be vanish, i.e. h/2
3. 3. The top and bottom surfaces are free, then the shear stress τmust be vanish, i.e.; January, 2015Bending Stresses - DAT3 shown the top and bottom surfaces are fre then the shear stress must be vanish, i. = 0 at y = h/2 for two equal rectangular beams o height h subjected to a concentrate load P, if no friction between the beam each beam will be in compression above i N.A., the lower longitudinal line of th upper beam will slide w.r.t. the upper lin mall act as ee, i.e. of ted ms, its the and bottom surfaces are free, stress must be vanish, i.e. y = h/2 qual rectangular beams of ubjected to a concentrated o friction between the beams, be in compression above its er longitudinal line of the ll slide w.r.t. the upper line am beam of height 2h, shear stress must exist along N.A. to
4. 4. ¢  Consider a small section of the beam subjected M and V in left face and M +dM and V+dV in right face. ¢  For the element mm1p1p, τ acts on p1p and no stress on mm1. January, 2015Bending Stresses - DAT4 onsider a small section of the beam ected M and V in left face and dM and V + dV in right face or the element mm1p1p, acts on and no stress on mm1 the beam is subjected to pure bending = constant), x acting on mp and must be equal, then = 0 on or nonuniform bending, M acts on mn and M + dM acts on , consider dA at the distance y form N.A., then on mn small section of the beam and V in left face and V + dV in right face ment mm1p1p, acts on ess on mm1 s subjected to pure bending x acting on mp and equal, then = 0 on
5. 5. ¢  For non-uniform bending, M acts on mn and M + dM acts on m1n1, consider dA at the distance y form N.A., then on mn January, 2015Bending Stresses - DAT5 orm bending, M acts on mn and M + dM acts dA at the distance y form N.A., then on mn M y = CC dA I tal horizontal force on mp is M y CC dA I (M + dM) y p1p and no stress on mm1 if the beam is subjected to pure bending (M = constant), x acting on mp and m1p1 must be equal, then = 0 on pp1 for nonuniform bending, M acts on mn and M + m1n1, consider dA at the distance y form N.A., then o M y x dA = CC dA I hence the total horizontal force on mp is nonuniform bending, M acts on mn and M + dM acts o consider dA at the distance y form N.A., then on mn M y x dA = CC dA I nce the total horizontal force on mp is M y F1 = CC dA I milarly (M + dM) y F2 = CCCCC dA I
6. 6. Hence the total horizontal force on mp is; January, 2015Bending Stresses - DAT6 x dA = CC dA I hence the total horizontal force on mp is M y F1 = CC dA I similarly (M + dM) y F2 = CCCCC dA I and the horizontal force on pp1 is F3 = b dx equation of equilibrium x dA = CC dA I hence the total horizontal force on mp is M y F1 = CC dA I similarly (M + dM) y F2 = CCCCC dA I and the horizontal force on pp1 is F3 = b dx equation of equilibrium 21 M y F1 = CC dA I similarly (M + dM) y F2 = CCCCC dA I and the horizontal force on pp1 is F3 = b dx equation of equilibrium F3 = F2 - F1 Similarly hence the total horizontal force on mp is M y F1 = CC dA I similarly (M + dM) y F2 = CCCCC dA I and the horizontal force on pp1 is F3 = b dx equation of equilibrium And the horizontal force on pp1 is;
7. 7. Equation of equilibrium; January, 2015Bending Stresses - DAT7 21 F3 = b dx equation of equilibrium F3 = F2 - F1 (M + dM) y M y b dx = CCCCC dA - CC dA I I dM 1 V = CC C y dA = CC y dA dx Ib I b denote Q = y dA is the first moment of the cross s above the level y (area mm1p1p) at which the shear stress act V Q = CC shear stress formula I b (M + dM) y M y b dx = CCCCC dA - CC dA I I dM 1 V = CC C y dA = CC y dA dx Ib I b denote Q = y dA is the first moment of the cro above the level y (area mm1p1p) at which the shear stress V Q = CC shear stress formula I b Denote Q = ∫y dA is the first moment of the cross section area above the level y (area mm1p1p) at which the shear stress τ, then;
8. 8. Then, the shear stress formula will be; January, 2015Bending Stresses - DAT8 denote Q = y dA is the first moment of above the level y (area mm1p1p) at which the shear stre V Q = CC shear stress formula I b for V, I, b are constants, ~ Q for a rectangular cross section h h/2 - y1 b h2 Q = b (C - y1) (y1 + CCC) = C (C 2 2 2 4 V h2 then = CC (C - y1 2 ) 2 I 4 For V, I, b are constants, τ ~ Q for a rectangular cross section dx Ib I b enote Q = y dA is the first moment of the cross section e the level y (area mm1p1p) at which the shear stress acts, then V Q = CC shear stress formula I b or V, I, b are constants, ~ Q or a rectangular cross section h h/2 - y1 b h2 Q = b (C - y1) (y1 + CCC) = C (C - y1 2 ) 2 2 2 4 V h2 2 dM 1 V = CC C y dA = CC y dA dx Ib I b denote Q = y dA is the first moment of the cross section above the level y (area mm1p1p) at which the shear stress acts, then V Q = CC shear stress formula I b for V, I, b are constants, ~ Q for a rectangular cross section h h/2 - y1 b h2 Q = b (C - y1) (y1 + CCC) = C (C - y1 2 ) 2 2 2 4 V h2 then = CC (C - y1 2 )
9. 9. Then; January, 2015Bending Stresses - DAT9 1 Q = b (C - y1) (y1 + CCC) = C (C - y1 2 ) 2 2 2 4 V h2 then = CC (C - y1 2 ) 2 I 4 = 0 at y1 = h/2, max occurs at y1 = 0 (N.A.) V h2 V h2 3 V 3 max = CC = CCCC = CC = C ave 8 I 8 b h3 /12 2 A 2 max is 50% larger than ave V = resultant of shear stress, V and in th direction Q = b (C - y1) (y1 + CCC) = C (C 2 2 2 4 V h2 then = CC (C - y1 2 ) 2 I 4 = 0 at y1 = h/2, max occurs at y1 = 0 (N.A.) V h2 V h2 3 V max = CC = CCCC = CC = 8 I 8 b h3 /12 2 A max is 50% larger than ave V = resultant of shear stress, V direction h/2 - y1 b h2 y1) (y1 + CCC) = C (C - y1 2 ) 2 2 4 h2 C (C - y1 2 ) I 4 = h/2, max occurs A.) V h2 3 V 3 = CCCC = CC = C ave 8 b h3 /12 2 A 2 ger than ave nt of shear stress, V and in the same h h/2 - y1 b h2 Q = b (C - y1) (y1 + CCC) = C (C - y1 2 ) 2 2 2 4 V h2 then = CC (C - y1 2 ) 2 I 4 = 0 at y1 = h/2, max occurs at y1 = 0 (N.A.) V h2 V h2 3 V 3 max = CC = CCCC = CC = C 8 I 8 b h3 /12 2 A 2 max is 50% larger than ave V = resultant of shear stress, V and h h/2 - y1 b h2 Q = b (C - y1) (y1 + CCC) = C (C - y1 2 ) 2 2 2 4 V h2 = CC (C - y1 2 ) 2 I 4 0 at y1 = h/2, max occurs = 0 (N.A.) V h2 V h2 3 V 3 max = CC = CCCC = CC = C ave 8 I 8 b h3 /12 2 A 2 is 50% larger than ave V = resultant of shear stress, V and in the sam for a rectangular cross section h h/2 - y1 b h2 Q = b (C - y1) (y1 + CCC) = C (C - y1 2 ) 2 2 2 4 V h2 then = CC (C - y1 2 ) 2 I 4 = 0 at y1 = h/2, max occurs at y1 = 0 (N.A.) V h2 V h2 3 V 3 max = CC = CCCC = CC = C ave 8 I 8 b h3 /12 2 A 2 max is 50% larger than ave V = resultant of shear stress, V and in the h h/2 - y1 b h Q = b (C - y1) (y1 + CCC) = C (C - y1 2 ) 2 2 2 4 V h2 then = CC (C - y1 2 ) 2 I 4 = 0 at y1 = h/2, max occurs at y1 = 0 (N.A.) V h2 V h2 3 V 3 max = CC = CCCC = CC = C 8 I 8 b h3 /12 2 A 2 max is 50% larger than ave V = resultant of shear stress, V and
10. 10. Example 1 A metal beam with span L = 1m, q=28kN/m, b=25mm and h=100mm. Determine bending and shear stresses at point C. January, 2015Bending Stresses - DAT10 23 ear force varies continuously along the beam, the warping of cross due to shear strains does not substantially affect the longitudinal y more experimental investigation it is quite justifiable to use the flexure formula in the case of rm bending, except the region near the concentrate load acts of ly change of the cross section (stress concentration) 5-11 al beam with span L = 1 m 8 kN/m b = 25 mm h = 100 mm
11. 11. Example 1 Solution ¢  The shear force VC and bending moment MC at the section through C are found; MC = 2.24 kNm VC = - 8.4 kN January, 2015Bending Stresses - DAT11 C and C at point C force VC and bending C at the section through C = 2.24 kN-m = - 8.4 kN nt of inertia of the section is
12. 12. The moment of inertia of the section is; January, 2015Bending Stresses - DAT12 VC = - 8.4 kN the moment of inertia of the section is b h3 1 I = CC = C x 25 x 1003 = 2,083 x 10 12 12 normal stress at C is M y 2.24 x 106 N-mm x 25 mm C = - CC = - CCCCCCCCCCC = I 2,083 x 103 mm4 shear stress at C, calculate QC first AC = 25 x 25 = 625 mm2 yC = 37.5 mm QC = AC yC = 23,400 mm3 kN-m kN ia of the section is 1 = C x 25 x 1003 = 2,083 x 103 mm4 12 is 2.24 x 106 N-mm x 25 mm = - CCCCCCCCCCC = - 26.9 MPa 2,083 x 103 mm4 calculate QC first The normal stress at C is; VC = - 8.4 kN the moment of inertia of the section is b h3 1 I = CC = C x 25 x 1003 = 2,083 x 103 m 12 12 normal stress at C is M y 2.24 x 106 N-mm x 25 mm C = - CC = - CCCCCCCCCCC = - 2 I 2,083 x 103 mm4 shear stress at C, calculate QC first AC = 25 x 25 = 625 mm2 yC = 37.5 mm MC = 2.24 kN-m VC = - 8.4 kN e moment of inertia of the section is b h3 1 I = CC = C x 25 x 1003 = 2,083 x 103 mm4 12 12 ormal stress at C is M y 2.24 x 106 N-mm x 25 mm C = - CC = - CCCCCCCCCCC = - 26.9 MPa I 2,083 x 103 mm4 hear stress at C, calculate QC first
13. 13. Shear stress at C, calculate Qc first; January, 2015Bending Stresses - DAT13 I 2,083 x 103 mm4 shear stress at C, calculate QC first AC = 25 x 25 = 625 mm2 yC = 37.5 mm QC = AC yC = 23,400 mm3 VC QC 8,400 x 23,400 C = CCC = CCCCCCC = 3.8 MPa I b 2,083 x 103 x 25 the stress element at point C is shown Example 5-12 a wood beam AB supporting two concentrated loads P b = 100 mm h = 150 mm M y 2.24 x 106 N-mm x 25 mm - CC = - CCCCCCCCCCC = - 26.9 MPa I 2,083 x 103 mm4 s at C, calculate QC first 25 x 25 = 625 mm2 yC = 37.5 mm = AC yC = 23,400 mm3 VC QC 8,400 x 23,400 = CCC = CCCCCCC = 3.8 MPa I b 2,083 x 103 x 25 element at point C is shown 2 beam AB supporting two loads P 0 mm h = 150 mm M y 2.24 x 106 N-mm x 25 mm C = - CC = - CCCCCCCCCCC = - 26.9 MP I 2,083 x 103 mm4 shear stress at C, calculate QC first AC = 25 x 25 = 625 mm2 yC = 37.5 mm QC = AC yC = 23,400 mm3 VC QC 8,400 x 23,400 C = CCC = CCCCCCC = 3.8 MPa I b 2,083 x 103 x 25 the stress element at point C is shown Example 5-12 a wood beam AB supporting two concentrated loads P b = 100 mm h = 150 mm M y 2.24 x 106 N-mm x 25 mm C = - CC = - CCCCCCCCCCC = - 26.9 MP I 2,083 x 103 mm4 shear stress at C, calculate QC first AC = 25 x 25 = 625 mm2 yC = 37.5 mm QC = AC yC = 23,400 mm3 VC QC 8,400 x 23,400 C = CCC = CCCCCCC = 3.8 MPa I b 2,083 x 103 x 25 the stress element at point C is shown Example 5-12 a wood beam AB supporting two concentrated loads P b = 100 mm h = 150 mm h3 1 C = C x 25 x 1003 = 2,083 x 103 mm4 2 12 t C is y 2.24 x 106 N-mm x 25 mm C = - CCCCCCCCCCC = - 26.9 MPa 2,083 x 103 mm4 C, calculate QC first 25 = 625 mm2 yC = 37.5 mm AC yC = 23,400 mm3 C QC 8,400 x 23,400 The stress element at point C is shown 3.8 MPA 3.8 MPA 26.9 MPA26.9 MPA
14. 14. Example 2 A wood beam AB supporting two concentrated loads P. Cross sectional dimentions; b = 100mm, h = 150mm, a = 0.5 m. Given σallow = 11MPa, τallow = 1.2 Mpa, determine Pmax. January, 2015Bending Stresses - DAT14 ate QC first m2 yC = 37.5 mm 3,400 mm3 8,400 x 23,400 CCCCCCC = 3.8 MPa 2,083 x 103 x 25 C is shown pporting two 50 mm a = 0.5 m allow = 11 MPa allow = 1.2 MPa determine Pmax the maximum shear force and bending moment are Vmax = P Mmax = P a the section modulus and area are b h2 S = CC A = b h 6
15. 15. Example 2 Solution The maximum shear force and bending moment are Vmax = P Mmax = Pa January, 2015Bending Stresses - DAT15 The section modulus and are are; a = 0.5 m allow = 11 MPa al determine Pmax the maximum shear force and bending momen Vmax = P Mmax = P a the section modulus and area are b h2 S = CC A = b h 6 maximum normal and shear stresses are
16. 16. Maximum normal and shear stresses are; January, 2015Bending Stresses - DAT16 b h2 S = CC A = b h 6 maximum normal and shear stresses are Mmax 6 P a 3 Vmax 3 P max = CC = CC max = CCC = CCC S b h2 2 A 2 b h allow b h2 11 x 100 x 1502 Pbending = CCCC = CCCCCCC = 8,250 N = 8.25 kN 6 a 6 x 500 2 allow b h 2 x 1.2 x 100 x 150 Pshear = CCCC = CCCCCCCC = 12,000 N = 12 kN 3 3 Pmax = 8.25 kN 8-9 Shear Stresses in Beam of Circular Cross Section V Q r4 b h2 S = CC A = b h 6 maximum normal and shear stresses are Mmax 6 P a 3 Vmax 3 P max = CC = CC max = CCC = CCC S b h2 2 A 2 b h allow b h2 11 x 100 x 1502 Pbending = CCCC = CCCCCCC = 8,250 N = 8.25 kN 6 a 6 x 500 2 allow b h 2 x 1.2 x 100 x 150 Pshear = CCCC = CCCCCCCC = 12,000 N = 12 kN 3 3 Pmax = 8.25 kN 8-9 Shear Stresses in Beam of Circular Cross Section V Q r4 = CC I = CC for solid section ce and bending moment are Mmax = P a d area are A = b h shear stresses are 6 P a 3 Vmax 3 P = CC max = CCC = CCC b h2 2 A 2 b h b h2 11 x 100 x 1502 CC = CCCCCCC = 8,250 N = 8.25 kN 6 x 500 h 2 x 1.2 x 100 x 150 C = CCCCCCCC = 12,000 N = 12 kN 3 Vmax = P Mmax = P a the section modulus and area are b h2 S = CC A = b h 6 maximum normal and shear stresses are Mmax 6 P a 3 Vmax 3 P max = CC = CC max = CCC = CCC S b h2 2 A 2 b h allow b h2 11 x 100 x 1502 Pbending = CCCC = CCCCCCC = 8,250 N = 8.25 kN 6 a 6 x 500 2 allow b h 2 x 1.2 x 100 x 150 Pshear = CCCC = CCCCCCCC = 12,000 N = 12 kN 3 3 Pmax = 8.25 kN Vmax = P Mmax = P a the section modulus and area are b h2 S = CC A = b h 6 maximum normal and shear stresses are Mmax 6 P a 3 Vmax max = CC = CC max = CCC = S b h2 2 A allow b h2 11 x 100 x 1502 Pbending = CCCC = CCCCCCC = 8,250 N 6 a 6 x 500 2 allow b h 2 x 1.2 x 100 x 150 Pshear = CCCC = CCCCCCCC = 12,000 3 3 Pmax = 8.25 kN
17. 17. Shear Stresses in Beam of Circular Cross Section The shear stress at the neutral axis; January, 2015Bending Stresses - DAT17 Pmax = 8.25 kN 8-9 Shear Stresses in Beam of Circular Cross Section V Q r4 = CC I = CC for solid section I b 4 the shear stress at the neutral axis r2 4 r 2 r3 Q = A y = (CC) (CC) = CC 2 3 3 hear = CCCC = CCCCCCCC = 12,000 N = 1 3 3 max = 8.25 kN Stresses in Beam of Circular Cross Section V Q r4 = CC I = CC for solid section I b 4 ar stress at the neutral axis r2 4 r 2 r3 = A y = (CC) (CC) = CC b = 2 r Pshear = CCCC = CCCCCCCC = 1 3 3 Pmax = 8.25 kN 8-9 Shear Stresses in Beam of Circular Cross Section V Q r4 = CC I = CC for solid section I b 4 the shear stress at the neutral axis r2 4 r 2 r3 Q = A y = (CC) (CC) = CC 2 3 3 25 Q r C I = CC for solid section 4 at the neutral axis r2 4 r 2 r3 y = (CC) (CC) = CC b = 2 r 2 3 3
18. 18. January, 2015Bending Stresses - DAT18 V (2 r3 / 3) 4 V 4 V max = CCCCCC = CCC = CC ( r4 / 4) (2 r) 3 r2 3 A for a hollow circular cross section 2 I = C (r2 4 - r1 4 ) Q = C (r2 3 - r1 3 ) 4 3 b = 2 (r2 - r1) then the maximum shear stress at N.A. is V Q 4 V r2 2 + r2r1 + r1 2 max = CC = CC (CCCCCC) I b 3 A r2 2 + r1 2 V (2 r3 / 3) 4 V 4 V 4 CCCCCC = CCC = CC = C ave ( r4 / 4) (2 r) 3 r2 3 A 3 rcular cross section 4 V 4 V 4 C = CCC = CC = C ave r) 3 r2 3 A 3 ction 2 Q = C (r2 3 - r1 3 ) 3 ss at N.A. is 2 2
19. 19. For a hollow circular cross section; January, 2015Bending Stresses - DAT19 for a hollow circular cross section 2 I = C (r2 4 - r1 4 ) Q = C (r2 3 - r1 3 ) 4 3 b = 2 (r2 - r1) then the maximum shear stress at N.A. is V Q 4 V r2 2 + r2r1 + r1 2 max = CC = CC (CCCCCC) I b 3 A r2 2 + r1 2 where A = (r2 2 - r1 2 ) Example 5-13 circular cross section 2 C (r2 4 - r1 4 ) Q = C (r2 3 - r1 3 ) 4 3 2 (r2 - r1) mum shear stress at N.A. is V Q 4 V r2 2 + r2r1 + r1 2 CC = CC (CCCCCC) I b 3 A r2 2 + r1 2 A = (r2 2 - r1 2 ) Then the maximum shear stress at N.A. is; then the maximum shear stress at N.A. V Q 4 V r2 2 + max = CC = CC (CCC I b 3 A r2 2 where A = (r2 2 - r1 2 ) Example 5-13 a vertical pole of a circular tube d2 = 100 mm d1 = 80 mm P = 6,675 (a) determine the max in the pole (b) for same P and same max, calcula ( r / 4) (2 r) 3 r 3 A for a hollow circular cross section 2 I = C (r2 4 - r1 4 ) Q = C (r2 3 - r1 3 ) 4 3 b = 2 (r2 - r1) then the maximum shear stress at N.A. is V Q 4 V r2 2 + r2r1 + r1 2 max = CC = CC (CCCCCC) I b 3 A r2 2 + r1 2 where A = (r2 2 - r1 2 ) Example 5-13 V (2 r3 / 3) 4 V 4 V max = CCCCCC = CCC = CC ( r4 / 4) (2 r) 3 r2 3 A for a hollow circular cross section 2 I = C (r2 4 - r1 4 ) Q = C (r2 3 - r1 3 ) 4 3 b = 2 (r2 - r1) then the maximum shear stress at N.A. is V Q 4 V r2 2 + r2r1 + r1 2 max = CC = CC (CCCCCC) I b 3 A r2 2 + r1 2 where A = (r 2 - r 2 )
20. 20. Example 3 A vertical pole of a circular tube d2 =100mm, d1=80mm, P=6,675 N. (a) determine the σmax in the pole (b) for same P and same τmax, calculate d0 of a solid circular pole January, 2015Bending Stresses - DAT20 A = (r2 - r1 ) le of a circular tube m d1 = 80 mm P = 6,675 N e the max in the pole e P and same max, calculate d0 rcular pole imum shear stress of a circular
21. 21. Shear Stress in the Webs of Beams with Flanges January, 2015Bending Stresses - DAT21 circular pole has a diameter approximately 5/8 that of ress in the Webs of Beams with Flanges am of wide-flange shape shear force V, shear stress complicated he shear force is carried by shear web he shear stress at ef, the same in the case in rectangular beam, i.e. For a beam of wide- flange shape subjected to shear force V, shear stress is much more complicated. Most of the shear force is carried by shear stresses in the web
22. 22. mly distributed across t valid with b = t of vided the area n the January, 2015Bending Stresses - DAT22 Consider the shear stress at ef, the same assumption as in the case in rectangular beam, i.e. τ parallel to y-axis and uniformly distributed across t. stresses in the web consider the shear st assumption as in the case // y axis and uniform V Q = CC is still I b the first moment Q the shaded area is di into two parts, i.e. upper flange and the is much more complicated most of the shear force is carried stresses in the web consider the shear stress at ef, assumption as in the case in rectangular // y axis and uniformly distributed V Q = CC is still valid with b I b the first moment Q of the shaded area is divided subjected to shear force V, shear stress is much more complicated most of the shear force is carried by shear stresses in the web consider the shear stress at ef, the same assumption as in the case in rectangular beam, i.e. // y axis and uniformly distributed across t V Q = CC is still valid with b = t I b the first moment Q of
23. 23. mption as in the case in rectangular beam, i.e. / y axis and uniformly distributed across t V Q = CC is still valid with b = t I b he first moment Q of shaded area is divided two parts, i.e. the er flange and the area ween bc and ef in the January, 2015Bending Stresses - DAT23 Then the first moment of A1 and A2 w.r.t. N.A. is; A1 = b (C - C) A2 = t (C - y1) 2 2 2 then the first moment of A1 and A2 w.r.t. N.A. is h1 h/2 - h1/2 h1/2 - y1 Q = A1 (C + CCCC) + A2 (y1 + CCCC) 2 2 2 b t = C (h2 - h1 2 ) + C (h1 2 - 4 y1 2 ) 8 8 V Q V b t = CC = CC [C (h2 - h1 2 ) + C (h1 2 - 4 y1 2 )] I b 8 I t 8 8 b h3 (b - t) h1 3 1 where I = CC - CCCC = C (b h3 - b h1 3 + t h1 12 12 12 maximum shear stress in the web occurs at N.A., y1 = 0 2 2 2 then the first moment of A1 and A2 w.r.t. N.A. is h1 h/2 - h1/2 h1/2 - y1 Q = A1 (C + CCCC) + A2 (y1 + CCCC) 2 2 2 b t = C (h2 - h1 2 ) + C (h1 2 - 4 y1 2 ) 8 8 V Q V b t = CC = CC [C (h2 - h1 2 ) + C (h1 2 - 4 y1 2 )] I b 8 I t 8 8 b h3 (b - t) h1 3 1 where I = CC - CCCC = C (b h3 - b h1 3 + t h1 12 12 12 maximum shear stress in the web occurs at N.A., y1 = 0 V then the first moment of A1 and A2 w.r.t. N.A. is h1 h/2 - h1/2 h1/2 - y1 Q = A1 (C + CCCC) + A2 (y1 + CCCC) 2 2 2 b t = C (h2 - h1 2 ) + C (h1 2 - 4 y1 2 ) 8 8 V Q V b t = CC = CC [C (h2 - h1 2 ) + C (h1 2 - 4 y1 2 )] I b 8 I t 8 8 b h3 (b - t) h1 3 1 where I = CC - CCCC = C (b h3 - b h1 3 + t h1 3 ) 12 12 12 maximum shear stress in the web occurs at N.A., y1 = 0 V 2 2 2 h h1 h1 A1 = b (C - C) A2 = t (C - y1) 2 2 2 he first moment of A1 and A2 w.r.t. N.A. is h1 h/2 - h1/2 h1/2 - y1 Q = A1 (C + CCCC) + A2 (y1 + CCCC) 2 2 2 b t = C (h2 - h1 2 ) + C (h1 2 - 4 y1 2 ) 8 8 V Q V b t = CC = CC [C (h2 - h1 2 ) + C (h1 2 - 4 y1 2 )] I b 8 I t 8 8 b h3 (b - t) h1 3 1 I = CC - CCCC = C (b h3 - b h1 3 + t h1 3 ) 12 12 12 h1 ) A2 = t (C - y1) 2 A1 and A2 w.r.t. N.A. is /2 - h1/2 h1/2 - y1 CCCC) + A2 (y1 + CCCC) 2 2 t + C (h1 2 - 4 y1 2 ) 8 b t C [C (h2 - h1 2 ) + C (h1 2 - 4 y1 2 )] t 8 8 3
24. 24. January, 2015Bending Stresses - DAT24 ¢  Maximum shear stress in the web occurs at N.A., y1 = 0 I b 8 I t 8 8 b h3 (b - t) h1 3 1 where I = CC - CCCC = C 12 12 12 maximum shear stress in the web occurs at N.A., V max = CC (b h2 - b h1 2 + t h1 2 ) 8 I t minimum shear stress occurs where the web mee h1/2 V b min = CC (h2 - h1 2 ) = CC = CC [C (h2 - h1 2 ) + C (h1 2 - 4 y1 2 )] I b 8 I t 8 8 b h3 (b - t) h1 3 1 where I = CC - CCCC = C (b h3 - b h1 3 + t h1 3 ) 12 12 12 maximum shear stress in the web occurs at N.A., y1 = 0 V max = CC (b h2 - b h1 2 + t h1 2 ) 8 I t minimum shear stress occurs where the web meets the flange, y1 = h1/2 V b min = CC (h2 - h1 2 ) 8 I t the maximum stress in the web is from 10% to 60% greater than the I b 8 I t 8 8 b h3 (b - t) h1 3 1 where I = CC - CCCC = C (b h3 12 12 12 maximum shear stress in the web occurs at N.A., y1 V max = CC (b h2 - b h1 2 + t h1 2 ) 8 I t minimum shear stress occurs where the web meets th h1/2 V b min = CC (h2 - h1 2 ) 8 I t ¢  Minimum shear stress occurs where the web meets the flange, y = ±h1/2; Where
25. 25. with b = t f d e a e Shear stress distribution on the web; January, 2015Bending Stresses - DAT25
26. 26. January, 2015Bending Stresses - DAT26 ¢  The shear force carried by the web consists two parts, a rectangle of area h1 τmin and a parabolic segment of area 2/3 h1 (τmax - τmin) 28 the maximum stress in the web is from 10% to minimum stress the shear force carried by the web consists two pa h1 min and a parabolic segment of area b h1 ( max Vweb = h1 min + b h1 ( max - min) t h1 = CC (2 max + min) 3 Vweb = 90% ~ 98% of total V for design work, the approximation to calculate V <= total shear force ¢  For design work, the approximation to calculate τmax is t h1 = CC (2 max + min) 3 Vweb = 90% ~ 98% of total V for design work, the approximation to calculate V <= total shear force max = CC t h1 <= web area
27. 27. Example 4 January, 2015Bending Stresses - DAT27 a beam of wide-flange shape wit and h1 = 290 mm, vertical shear forc determine max, min and total the moment of inertia of the cros A beam of wide-flange shape with b = 165 mm, t = 7.5mm, h = 320 mm, and h1 = 290 mm, vertical shear force V = 45 kN. Determine τmax, τmin and total shear force in the web.
28. 28. V CC = 20.7 MPa h1 a T-shaped cross section = 24 mm h = 200 mm (top of the web) and max 6 x 24 x 12 + 200 x 24 x 100 CCCCCCCCCCC = 75.77 mm Example 5 A beam having a T- shaped cross section b=100mm, t=24mm, h=200mm, V = 45kN. Determine τmin (top of the web) and τmax. January, 2015Bending Stresses - DAT28
29. 29. Thank You January, 2015Bending Stresses - DAT29