• Locution of centroid G, from bottom of web • Lotadion of entroid G, from botten of wtb. • Maximum Shear Stress (Tmax ): : Tman (where,(width )meb = 6 mm ) %3D Iz x (width)wb Draw the loading, shear, and moment diagram of the beam shown in the figure. Draw the diagram in proper position. details. Include signs in the magnitude for the answer box below = (5x 10 N) × (76200 mm³) (121078.714 mmt) x(6 mm ) Provide complete computation = 52. 88475 NImm ---(answer) Tman 52.885 MPa mm* There fore, Iz 1210178. 719 Vmax kN Vmax KN Qmax 76800 Tmax MPa tmax 52. 885 MPa | Se lutien * Delermine vertical Reaction af fired support : • Area of web : A- GX92 • Area of Flange: A, e 100 y- 700 m - S52 mm 100 mm 6. Sin 6. Sm 0.5 le Z Fy =0 (1+,4-) 3 kN GKN = 46 mm RA + 4 -3-6 0 RA +1-9 eo 92 mm : [RA = 5 kN] * 9, = 92 + = 3Cmm Sechon plara • Shear Forte diayram Calcalatin: Take sechon plane from the fixed support & start calculating sheat fore in each part of the beum. - VAB = RA = 5 EN • Locatiun of rentruid G of whole T-sertiun: CEN Gmm 4kN = (562 x 46 ) + (800 x 96) ( 552 + 800) RA = 5 kN Loading Diagram - Vec RA-3 - 5-3 = 2 kN J = 75.5858 mm • Location of rentroid Gy from centroidal anis z : : hi = - 9, = 75.585 - 46 = 29. 58 58 mm : hz = Y2 -9 = - Vcp = 4 KN 3kN • Bending moment diagram Caleulation : (U,5 ) SKN 96 - 75.58 58 e 20. 4142 mm %3D 2EN B. - At free end, Mp = 0 EN M • Moment af inertia (Tz ) about centroidal z-axis: GEN 1KN Iz = (Ta, + A, h,² ) + ( Ja, + Ay-h,² ) - Me = 4 kN xo.5 m = 2 EN m - Ma = (4 x1 )– €x0.5 ) = 1 KN m - MA = (4x1.5 ) -(6x1) – (3x0.5) 6 - 6 - 1.5 Shear Forte Diagróm 6x 92 12 + 552 X 29. SE582) + ( 100x + 800 x 20.41422) 12 Me =2 kN-m Iz= 1210178. 714 mm (Answer ) -- .. M=1 EN m . MA = -1.5 KN M • from the Shear forre diagraum, Ma ximum shear foxe Vman ocrurs at fixed end of tha beam Omax: Area moment of web : Pweb = A,9, = 552 x 46 = 25392 mm3 Area Moment of flange : Qmx = Relange - Qnex. = Maximum of { qweb, Qflage 3 = Qflange A2 42 = svo x 96 = 76800 mm MA = -1.5 kN m Vmax = RA Vmex = 5 kN]-- (Answer) Bending Moment Diagrum Oe = 76800 mm ---(A nswee )
• Locution of centroid G, from bottom of web • Lotadion of entroid G, from botten of wtb. • Maximum Shear Stress (Tmax ): : Tman (where,(width )meb = 6 mm ) %3D Iz x (width)wb Draw the loading, shear, and moment diagram of the beam shown in the figure. Draw the diagram in proper position. details. Include signs in the magnitude for the answer box below = (5x 10 N) × (76200 mm³) (121078.714 mmt) x(6 mm ) Provide complete computation = 52. 88475 NImm ---(answer) Tman 52.885 MPa mm* There fore, Iz 1210178. 719 Vmax kN Vmax KN Qmax 76800 Tmax MPa tmax 52. 885 MPa | Se lutien * Delermine vertical Reaction af fired support : • Area of web : A- GX92 • Area of Flange: A, e 100 y- 700 m - S52 mm 100 mm 6. Sin 6. Sm 0.5 le Z Fy =0 (1+,4-) 3 kN GKN = 46 mm RA + 4 -3-6 0 RA +1-9 eo 92 mm : [RA = 5 kN] * 9, = 92 + = 3Cmm Sechon plara • Shear Forte diayram Calcalatin: Take sechon plane from the fixed support & start calculating sheat fore in each part of the beum. - VAB = RA = 5 EN • Locatiun of rentruid G of whole T-sertiun: CEN Gmm 4kN = (562 x 46 ) + (800 x 96) ( 552 + 800) RA = 5 kN Loading Diagram - Vec RA-3 - 5-3 = 2 kN J = 75.5858 mm • Location of rentroid Gy from centroidal anis z : : hi = - 9, = 75.585 - 46 = 29. 58 58 mm : hz = Y2 -9 = - Vcp = 4 KN 3kN • Bending moment diagram Caleulation : (U,5 ) SKN 96 - 75.58 58 e 20. 4142 mm %3D 2EN B. - At free end, Mp = 0 EN M • Moment af inertia (Tz ) about centroidal z-axis: GEN 1KN Iz = (Ta, + A, h,² ) + ( Ja, + Ay-h,² ) - Me = 4 kN xo.5 m = 2 EN m - Ma = (4 x1 )– €x0.5 ) = 1 KN m - MA = (4x1.5 ) -(6x1) – (3x0.5) 6 - 6 - 1.5 Shear Forte Diagróm 6x 92 12 + 552 X 29. SE582) + ( 100x + 800 x 20.41422) 12 Me =2 kN-m Iz= 1210178. 714 mm (Answer ) -- .. M=1 EN m . MA = -1.5 KN M • from the Shear forre diagraum, Ma ximum shear foxe Vman ocrurs at fixed end of tha beam Omax: Area moment of web : Pweb = A,9, = 552 x 46 = 25392 mm3 Area Moment of flange : Qmx = Relange - Qnex. = Maximum of { qweb, Qflage 3 = Qflange A2 42 = svo x 96 = 76800 mm MA = -1.5 kN m Vmax = RA Vmex = 5 kN]-- (Answer) Bending Moment Diagrum Oe = 76800 mm ---(A nswee )
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter10: Statically Indeterminate Beams
Section: Chapter Questions
Problem 10.4.39P: A beam supporting a uniform load of intensity q throughout its length rests on pistons at points A,...
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Provide an explanation (in words) of how the problem was solved, the step-by-step process that you think was used, the strategy, the formula used, etc.
Transcribed Image Text:• Area of web : A, GX92 = 552 mm
• Maximum Shear Stress (Toman ):
(where, (width )aeb
- 6 mm)
Tman =
Iz x (width)wrb
Draw the loading, shear, and moment diagram of
the beam shown in the figure. Draw the diagram in
proper position.
details. Include signs in the magnitude for the
(5x 10 N) × (76z00 mm³)
(121017 8. 714 mmt ) x(6 mm )
Provide complete computation
= 52. 88475 NImm
answer box below
--(Answes)
...
Tmax 52.885 MPa
mm*
There fore,
1210178.719
mm 9
Iz
Vmax
kN
Vmax
5
KN
Qmax
mm3
76800
mm 3
Tmax
MPa
tmax
52. 885
MPa
|
So lution:
* Deler mine vertical Reaction af
fired support:
100 mm
• Area of Flange : A, e 10o oY- 700 mm
6. Sin
o. 5m
0.5
le
-(1+,d-)
Z Fy =0
* RA + 1 -3 –6 =0
RA +1-9 eo
:RA =5 EN
:-
: 9, = 22
3 kN
GKN
= 46 mm
• Lo tatfon af centrojd G. frum bottom of wtb.
: 9, = 92 + = 96 mm.
92 mm
%3D
Sechon plara
• Shear Forte diayram Calcalatin:
Take sechon plane from tho fixed
support & start cal culatiny sheal
forre in earh part of the beam.
-> VAR = R, = 5 EN
• Locatiun of centruid G of whole T-sertjun:
CEN
4kN
Gmm
= (552 x 46 ) + (800 x 96)
(552 + 800)
RA = 5 kN
Loading Piagram
- Vnc = Ra - 3 = 5 -3 = 2 kN
5 = 75.5858 mm
- Vep = 4 kN
• Location of rentroid 6, frum centroidl axis z :
-: hi = ý - 9, = 75.5858 - 46 = 29. 58 58 mm
20. 4142 mm
3KN
5 kN
Bending moment diagram
Calculat
.: h2
Y2 -9 = 96 - 75.58 58
%3D
: O, 5 )
8.
- At free end, Mp =0 EN M
• Moment af inertia (Tz) about centroial z-axis :
GEN
1KN
(Ta, + A, h,) + (Ia, + A,-h,² )
?) + (
- Me = 4 kN x o.5 m = 2 EN m
Iz =
3
Shear Forte Diogram
- Me = 4 x 1 )- 6x0.5 ) = 1 kN m
2.
+ 552 X 29. 5858
100x8
+ 800 x 20.4 142 )
12
- MA = (4x1.5 ) -(6x1) - (3x0.5)
Me =2 kN-m
6 - 6 - 1-5
Iz =
1210178 , 714 mm
(Answer )
--- ....
Me =1 EN m
. MA = -1.5 KN-M
• from the Shear fore diagram,
Ma ximum shear fore Vmax occurs
at fixed end of tha beam
Qmax :
Area moment of web : Qweb = A,9, = 552 x 46
Area Moment of flange : Rmox = Qelame
= 25392 mm
= A2 92 = SUo X96 = 76800 mm
: Qmax.
Vman = RA
= 5 kN--- Answer)
= Maximum of { Q webo QAlonge }
Qflange
MA = -1.5 kN M
::
Vmex
Bending Moment Diagrim
Qmox = 76800 mm
---(Answee )
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