Question

For the beam shown in the given figure:

 (a) Express the internal shear (V) and moment (M) in the beam as a function of x.

 (b) Draw the shear force diagram (SFD) and bending moment diagram (BMD).

 (c) If the area moment of inertia (I) of the beam's cross section about the neutral axis is 301.3 (10^-6)m⁴, determine the absolute maximum bending stress (σ_max) in the beam. 

Answer 1

Given. s I-beam Box simple ht LOOMLM, X supported 200N 20mm A óc zomgy isomm q ext x A X fir 81 N um 20mm flox=25(4-** 9 15pm Raf Hm z100 H 25 * Wic H-> ☆ 250mm So 7) beam is in equilibrium ) EF4 :0 >> Rat Rp = (600X4) + ( xuxioo ) Rat Rp 6ooo 2) EMA ZO z) PBX8 z) XUX TOO 2 y ) +(400x482) as a Ra> [33 334 A -2 Ra= = 466-66UM aj poternal shear forcev & moment m expressed function ofs for each part as taking it that value about of that value about x-x in that plane 2504-x) - 100% >> in AC - 1V466.664 { xx (200- 3) VALE 466-664 2) adxzo z) VA = 466.664 M z) at sezus vei 1966-133.33UN voor een AC 25(67) too e) in BC z) VBC =-RB VBC E-RB 7 YBC 3-133:33U14 o from rom sea to xz8 =) Tu care -133:33 unl >>

es for bencing moment in ARO A xxxpoo-2542-x) x ]-.466464x m = AB 3 to zde xarx +(100xXx) +(25(2-x) xxx) 23 atxzo mazo 2) at xs4>> MBS 4x4x(100)xyx2 466.66484 + look -ow lz zara9 n-M = 533 • 332 22 incB Mobi-466-664xx + 1oox4 (32-2)+< x4x100 2 (x-3 a) al->(243) Mga .-466-66444 + coox4427 Uxioo x 284 MB2-733.332 11-m s at x383) Mes-466.66448 tlooxu (6) txurloc(8-5 6 22 whino =Wlz 2) abslute Maximum stoe Bendi Moment oment in the beam is at shear force 21330322MM. VACZO ding Masc z) VACS 466-664 -100425?-1001-50% – 12-5002 zo 2) 466-664-250x t12-5x² zo ?) 12.522-250x +466.66 ioc zo roots are calculated as below 297

2001 b Shear force & Bending moment diagra 100MM diagram free body . RA 466-664 curue shearforce diagram 1-133.zzuri onem curie There Scurcle Bending diagram Moment -533-332 MM -613.0129 M 250 xipz = 2502-14 X 17:5X4661664) 25 22x1 = 17:16m, Xz z 2008m - locaction of zero shear force and marimum bending Mua 100-25(4-7.09)}2008) -6466.66442.09) +(tex 2,08°) +25C-2 ore) yavast moment is IzZ608mn 2) M 2008 to (100-25(u-i Z Max IN 22 Mnał -613.012 AM Bending stress moment as Es abslute maximum May = 613.012 NAM

C) Give moment of inertia of becom crossection about Neutral axis is I 301.3xcol nu =) abslute Maximum bendin stress in the beam is Masaxt y I z? 61340123133x190 mase 301.3x18661012 ax max mac 2). o max 0-34587 m mpa Rebslute Mascimum bending stress

Answer 2

Given. s I-beam Box simple ht LOOMLM, X supported 200N 20mm A óc zomgy isomm q ext x A X fir 81 N um 20mm flox=25(4-** 9 15pm Raf Hm z100 H 25 * Wic H-> ☆ 250mm So 7) beam is in equilibrium ) EF4 :0 >> Rat Rp = (600X4) + ( xuxioo ) Rat Rp 6ooo 2) EMA ZO z) PBX8 z) XUX TOO 2 y ) +(400x482) as a Ra> [33 334 A -2 Ra= = 466-66UM aj poternal shear forcev & moment m expressed function ofs for each part as taking it that value about of that value about x-x in that plane 2504-x) - 100% >> in AC - 1V466.664 { xx (200- 3) VALE 466-664 2) adxzo z) VA = 466.664 M z) at sezus vei 1966-133.33UN voor een AC 25(67) too e) in BC z) VBC =-RB VBC E-RB 7 YBC 3-133:33U14 o from rom sea to xz8 =) Tu care -133:33 unl >>

es for bencing moment in ARO A xxxpoo-2542-x) x ]-.466464x m = AB 3 to zde xarx +(100xXx) +(25(2-x) xxx) 23 atxzo mazo 2) at xs4>> MBS 4x4x(100)xyx2 466.66484 + look -ow lz zara9 n-M = 533 • 332 22 incB Mobi-466-664xx + 1oox4 (32-2)+< x4x100 2 (x-3 a) al->(243) Mga .-466-66444 + coox4427 Uxioo x 284 MB2-733.332 11-m s at x383) Mes-466.66448 tlooxu (6) txurloc(8-5 6 22 whino =Wlz 2) abslute Maximum stoe Bendi Moment oment in the beam is at shear force 21330322MM. VACZO ding Masc z) VACS 466-664 -100425?-1001-50% – 12-5002 zo 2) 466-664-250x t12-5x² zo ?) 12.522-250x +466.66 ioc zo roots are calculated as below 297

2001 b Shear force & Bending moment diagra 100MM diagram free body . RA 466-664 curue shearforce diagram 1-133.zzuri onem curie There Scurcle Bending diagram Moment -533-332 MM -613.0129 M 250 xipz = 2502-14 X 17:5X4661664) 25 22x1 = 17:16m, Xz z 2008m - locaction of zero shear force and marimum bending Mua 100-25(4-7.09)}2008) -6466.66442.09) +(tex 2,08°) +25C-2 ore) yavast moment is IzZ608mn 2) M 2008 to (100-25(u-i Z Max IN 22 Mnał -613.012 AM Bending stress moment as Es abslute maximum May = 613.012 NAM

C) Give moment of inertia of becom crossection about Neutral axis is I 301.3xcol nu =) abslute Maximum bendin stress in the beam is Masaxt y I z? 61340123133x190 mase 301.3x18661012 ax max mac 2). o max 0-34587 m mpa Rebslute Mascimum bending stress

Answer 3

fax PON! к I somm - C A laxt Ni A X isomm ums Given simple supported I-beam LOON/m; x zoment 780 zome ARB RAK im* Ox= 25(4-x) z) beam is in equilibrium a) E Fy=0 o => Rat Ro (600x4) +(Źxurloo PAT Rp 600 2) EMA ZO z) XUXCOoxy) +(100x482) PBX8 2) 90 M ws H-> z zomm q o 250mm - z) 2 2) Tre= 133.334 [RA= 3 466-66 UM a) internal sheas forcev & moment on expressed as a function of for each part as taking of that value about x-x in that plane 25C4-7xx) -100% z) in AC as lv = 466.664 - *x (200 z) Varr 466-664 2) at x zoz) VA 2466.664 M 2) at bezus) Vc 1856–13333UN VAC loox e es(47) too e) in BC z) VBC ==RB 27 YBc2 -133.33414 so o from 5.4 to 878= No Vo = -133.3344

in AC for bencing moment mas Exxx800-25408) 2.466464x that +(100xXx) +(25(4-X)xxx) Stopx 23 atxzo mazo 466.66484 2) at xs 42) MB = 1/2 X 4x(100)x 4x2 tlook - worn blott Ew cz W-mog gou a 7) 33.332 33 Mee:-466664xx + 100x4 1(3-2)+2x4x100 (x-4) al-x743 mg = -466-66444 + coox4x2 + UXIOO x 284 moe.-33.332 M-om 6 27 77 at **83) Mes-466.66 448 +100 xu (6) txuriod(8-3 2) me ONIM 2 moment a) abslute Maximum she be Bending mon in the beam is at shear force 2) Vaczo 21331332 M. Marc 2) VACS466-664 - 100X +254?-100r -50% - 12-5322 2) 466-664-250x t12-3x² zo 27 12:502 - 250x tu66.66uoc zo roots are calculated as below 987

250 I xiz: 250 2502-(4 X 17-584661664) 25 Xz z 2008m Xu 2008 m 27 Xic170416m, locaction of zero shear force and maximum bending moment is Leme (100-25(4-2008) 25(4+2.08)$2008) -6466-664x2.og +(00420037) +(25(4-7.68) 42,00 2008 2) M Mraz Z E 22 Muar -613.012 AM es absleete maximum Bending stress moment Mu Max · 613.012 NM

Answer 4

Answer 5

fax PON! к I somm - C A laxt Ni A X isomm ums Given simple supported I-beam LOON/m; x zoment 780 zome ARB RAK im* Ox= 25(4-x) z) beam is in equilibrium a) E Fy=0 o => Rat Ro (600x4) +(Źxurloo PAT Rp 600 2) EMA ZO z) XUXCOoxy) +(100x482) PBX8 2) 90 M ws H-> z zomm q o 250mm - z) 2 2) Tre= 133.334 [RA= 3 466-66 UM a) internal sheas forcev & moment on expressed as a function of for each part as taking of that value about x-x in that plane 25C4-7xx) -100% z) in AC as lv = 466.664 - *x (200 z) Varr 466-664 2) at x zoz) VA 2466.664 M 2) at bezus) Vc 1856–13333UN VAC loox e es(47) too e) in BC z) VBC ==RB 27 YBc2 -133.33414 so o from 5.4 to 878= No Vo = -133.3344

in AC for bencing moment mas Exxx800-25408) 2.466464x that +(100xXx) +(25(4-X)xxx) Stopx 23 atxzo mazo 466.66484 2) at xs 42) MB = 1/2 X 4x(100)x 4x2 tlook - worn blott Ew cz W-mog gou a 7) 33.332 33 Mee:-466664xx + 100x4 1(3-2)+2x4x100 (x-4) al-x743 mg = -466-66444 + coox4x2 + UXIOO x 284 moe.-33.332 M-om 6 27 77 at **83) Mes-466.66 448 +100 xu (6) txuriod(8-3 2) me ONIM 2 moment a) abslute Maximum she be Bending mon in the beam is at shear force 2) Vaczo 21331332 M. Marc 2) VACS466-664 - 100X +254?-100r -50% - 12-5322 2) 466-664-250x t12-3x² zo 27 12:502 - 250x tu66.66uoc zo roots are calculated as below 987

250 I xiz: 250 2502-(4 X 17-584661664) 25 Xz z 2008m Xu 2008 m 27 Xic170416m, locaction of zero shear force and maximum bending moment is Leme (100-25(4-2008) 25(4+2.08)$2008) -6466-664x2.og +(00420037) +(25(4-7.68) 42,00 2008 2) M Mraz Z E 22 Muar -613.012 AM es absleete maximum Bending stress moment Mu Max · 613.012 NM

200 mily ioon bo free body diagrari IC RAN gy $199.99h shearforie diagram 1-133-3zuri onM!! Bending diagram Moment -333-332.mom -6130012 MM

Answer 6

fax PON! к I somm - C A laxt Ni A X isomm ums Given simple supported I-beam LOON/m; x zoment 780 zome ARB RAK im* Ox= 25(4-x) z) beam is in equilibrium a) E Fy=0 o => Rat Ro (600x4) +(Źxurloo PAT Rp 600 2) EMA ZO z) XUXCOoxy) +(100x482) PBX8 2) 90 M ws H-> z zomm q o 250mm - z) 2 2) Tre= 133.334 [RA= 3 466-66 UM a) internal sheas forcev & moment on expressed as a function of for each part as taking of that value about x-x in that plane 25C4-7xx) -100% z) in AC as lv = 466.664 - *x (200 z) Varr 466-664 2) at x zoz) VA 2466.664 M 2) at bezus) Vc 1856–13333UN VAC loox e es(47) too e) in BC z) VBC ==RB 27 YBc2 -133.33414 so o from 5.4 to 878= No Vo = -133.3344

in AC for bencing moment mas Exxx800-25408) 2.466464x that +(100xXx) +(25(4-X)xxx) Stopx 23 atxzo mazo 466.66484 2) at xs 42) MB = 1/2 X 4x(100)x 4x2 tlook - worn blott Ew cz W-mog gou a 7) 33.332 33 Mee:-466664xx + 100x4 1(3-2)+2x4x100 (x-4) al-x743 mg = -466-66444 + coox4x2 + UXIOO x 284 moe.-33.332 M-om 6 27 77 at **83) Mes-466.66 448 +100 xu (6) txuriod(8-3 2) me ONIM 2 moment a) abslute Maximum she be Bending mon in the beam is at shear force 2) Vaczo 21331332 M. Marc 2) VACS466-664 - 100X +254?-100r -50% - 12-5322 2) 466-664-250x t12-3x² zo 27 12:502 - 250x tu66.66uoc zo roots are calculated as below 987

250 I xiz: 250 2502-(4 X 17-584661664) 25 Xz z 2008m Xu 2008 m 27 Xic170416m, locaction of zero shear force and maximum bending moment is Leme (100-25(4-2008) 25(4+2.08)$2008) -6466-664x2.og +(00420037) +(25(4-7.68) 42,00 2008 2) M Mraz Z E 22 Muar -613.012 AM es absleete maximum Bending stress moment Mu Max · 613.012 NM

Answer 7

fax PON! к I somm - C A laxt Ni A X isomm ums Given simple supported I-beam LOON/m; x zoment 780 zome ARB RAK im* Ox= 25(4-x) z) beam is in equilibrium a) E Fy=0 o => Rat Ro (600x4) +(Źxurloo PAT Rp 600 2) EMA ZO z) XUXCOoxy) +(100x482) PBX8 2) 90 M ws H-> z zomm q o 250mm - z) 2 2) Tre= 133.334 [RA= 3 466-66 UM a) internal sheas forcev & moment on expressed as a function of for each part as taking of that value about x-x in that plane 25C4-7xx) -100% z) in AC as lv = 466.664 - *x (200 z) Varr 466-664 2) at x zoz) VA 2466.664 M 2) at bezus) Vc 1856–13333UN VAC loox e es(47) too e) in BC z) VBC ==RB 27 YBc2 -133.33414 so o from 5.4 to 878= No Vo = -133.3344

in AC for bencing moment mas Exxx800-25408) 2.466464x that +(100xXx) +(25(4-X)xxx) Stopx 23 atxzo mazo 466.66484 2) at xs 42) MB = 1/2 X 4x(100)x 4x2 tlook - worn blott Ew cz W-mog gou a 7) 33.332 33 Mee:-466664xx + 100x4 1(3-2)+2x4x100 (x-4) al-x743 mg = -466-66444 + coox4x2 + UXIOO x 284 moe.-33.332 M-om 6 27 77 at **83) Mes-466.66 448 +100 xu (6) txuriod(8-3 2) me ONIM 2 moment a) abslute Maximum she be Bending mon in the beam is at shear force 2) Vaczo 21331332 M. Marc 2) VACS466-664 - 100X +254?-100r -50% - 12-5322 2) 466-664-250x t12-3x² zo 27 12:502 - 250x tu66.66uoc zo roots are calculated as below 987

250 I xiz: 250 2502-(4 X 17-584661664) 25 Xz z 2008m Xu 2008 m 27 Xic170416m, locaction of zero shear force and maximum bending moment is Leme (100-25(4-2008) 25(4+2.08)$2008) -6466-664x2.og +(00420037) +(25(4-7.68) 42,00 2008 2) M Mraz Z E 22 Muar -613.012 AM es absleete maximum Bending stress moment Mu Max · 613.012 NM

200 mily ioon bo free body diagrari IC RAN gy $199.99h shearforie diagram 1-133-3zuri onM!! Bending diagram Moment -333-332.mom -6130012 MM

c) Give moment of inertia of beam crosseckan about aleutral axis is I = 3016 3 x col nu i' >) absleite maximum bendir bean is bending stress in the & max max I 22 maisc 6134012x60x170 301-38186 2) maxi 0-34587 m Max mpa 2 subslute Mascimum bending stress