Presentation on the analysis of beam for roof corridor-Mechanics of Materials

1. MM.DD.20XX
ADD A FOOTER
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2. ANALYSIS OF BEAM FOR
CORRIDOR ROOF
PRESENTATION
MECHANICS OF MATERIAL-II

3. MECHANICS OF MATERIAL-II
3
To calculate the Stresses,
deflections and slope present in
a beam which is substituted on a
roof top of a building.

4. ADD A FOOTER
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VISUAL VIEWS

5. INTODUCTION
5 MECHANICS OF MATERIAL-II
A beam is the structural element that primarily resists load
applied literally to the beam axis. Its mode of deflection is
primarily by bending. The load applied to the beam results in
the reaction forces at the beam support points.

6. OBJECTIVES
6 MECHANICS OF MATERIAL-II
The main objective is to find;
 Different stresses under complete
analytical calculations.
 To find the deflection and slope in a
beam.
 Normal and shear stress distributions
wherever applicable.

7. MECHANICS OF MATERIAL-II
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ASSUMPTIONS
oAll the beams are simply supported beams.
oAll the loads above the first floor are
directly transmitted to ground via columns.
oAll the active loads (student loads) are per
day based.
oTreating the entire beam as a rigid body,
determine the reaction forces.

8. MECHANICS OF MATERIAL-II
8
DIMENSIONS
 Length of beam=19×11=209 𝑖𝑛𝑐ℎ
 Thickness of beam=15 𝑖𝑛𝑐ℎ
 Height of beam=20 𝑖𝑛𝑐ℎ
 Volume of beam=(209×15×20) 𝑖𝑛𝑐ℎ3
=62700 𝑖𝑛𝑐ℎ3
 Mass of beam calculated=(0.084
lb
𝑖𝑛𝑐ℎ3 × 62700 𝑖𝑛𝑐ℎ3)=5266.8 lbm.
 Weight of beam calculated=(5266.8×32.2)=169590.96 lbf

9. MECHANICS OF MATERIAL-II
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DIMENSIONS
 Total 800 students which are to be assumed on the
corridor and 50% passing through the corridor each day
i.e. 400
 Average mass of student=60 kg=132.27 lbm
 Mass of 400 students=52910.9 lbm
 Weight of 400 students=1703732.24 lbf
 Total beams in the corridor=10
 Length of corridor=L=66×33=2178 𝑖𝑛𝑐ℎ

10. MECHANICS OF MATERIAL-II
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DIMENSIONS
 Total weight on each beam=(weight of 400 students + weight
of roof + weight of tank and machine)
=(170970.98+11022442.852+452861.022)
=1317903.38 lbf
 Weight of a beam=169590.96 lbf
 Concentrated load on beam=131903.38 lbf
 Total distributed load of beam=
131903.38
209
= 6305.75 𝑙𝑏𝑓
𝑖𝑛𝑐ℎ

11. FREE BODY DIAGRAM
11
DISTRIBUTED LOAD:
CONCENTRTIC LOAD:
MECHANICS OF MATERIAL-I

12. 12
REACTION FORCES
𝑴𝑨=0:
𝑹𝑩 (209)-169590.96(104.5)-1317903.38(104.5)
𝑹𝑩=743747.17 lbf. …………… (i)
𝑭𝒚=0:
𝑹𝑩 + 𝑹𝑨-1317903.38-169590.96=0
𝑹𝑨=743747.17 lbf. …………… (ii)
MECHANICS OF MATERIAL-II

13. 13

14. 14 MECHANICS OF MATERIAL-II
NOW FROM SINGULARITY FUNCTION TABLE;
Through this table we can calculate the values
𝑴 𝒙 = 𝑹𝑨 < 𝒙 − 𝟎 >𝟏 −
𝟏
𝟐
(𝒘𝒐) < 𝒙 − 𝟎 >𝟐− 𝒘 < 𝒙 − 𝟏𝟎𝟒. 𝟓 >𝟏
𝑴 𝒙 =743747.17 < 𝒙 − 𝟎 >𝟐 −
𝟏
𝟐
𝟔𝟑𝟎𝟓. 𝟕𝟓 × < 𝒙 − 𝟎 >𝟐− 𝟏𝟔𝟗𝟓𝟗𝟎. 𝟗𝟔 < 𝒙 − 𝟏𝟎𝟒. 𝟓𝟕 >
𝐀𝐬, 𝐄𝐈.
𝒅𝟐
𝒚
𝒅𝒙𝟐 = 𝑴 𝒙
𝐄𝐈.
𝒅𝟐𝒚
𝒅𝒙𝟐 =743747.17 < 𝒙 − 𝟎 >𝟐 −
𝟏
𝟐
𝟔𝟑𝟎𝟓. 𝟕𝟓 × < 𝒙 − 𝟎 >𝟐− 𝟏𝟔𝟗𝟓𝟗𝟎. 𝟗𝟔 < 𝒙 − 𝟏𝟎𝟒. 𝟓𝟕 >
Integrating both sides;
𝑬𝑰. 𝜽 = 𝑬𝑰.
𝒅𝒚
𝒅𝒙
= 𝟕𝟒𝟑𝟕𝟒𝟕. 𝟏𝟕
𝒙𝟐
𝟐
− 𝟔𝟑𝟎𝟓. 𝟕𝟓
𝒙𝟑
𝟔
− 𝟏𝟔𝟗𝟓𝟗𝟎. 𝟗𝟔
<𝒙−𝟏𝟎𝟒.𝟓>𝟐
𝟐
+ 𝒄𝟏 …………… (A)

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Again Integrating both sides;
𝑬𝑰. 𝒚 = 𝟕𝟒𝟑𝟕𝟒𝟕. 𝟏𝟕
𝒙𝟑
𝟔
− 𝟔𝟑𝟎𝟓. 𝟕𝟓
𝒙𝟒
𝟐𝟒
− 𝟏𝟔𝟗𝟓𝟗𝟎. 𝟗𝟔
<𝒙−𝟏𝟎𝟒.𝟓>𝟑
𝟔
+ 𝒄𝟏𝒙 + 𝒄𝟐 …………… (B)
When x=0; y=0 then
𝒄𝟐 = 𝟎
When 𝒙 = 𝟐𝟎𝟗 inch then 𝒚 = 𝟎; because in simply supported beam at supports deflection is zero
Eq (B) becomes;
𝟎 = 𝟕𝟒𝟑𝟕𝟒𝟕. 𝟏𝟕
(𝟐𝟎𝟗)𝟑
𝟔
− 𝟔𝟑𝟎𝟓. 𝟕𝟓
𝟐𝟎𝟗 𝟒
𝟐𝟒
− 𝟏𝟔𝟗𝟓𝟗𝟎. 𝟗𝟔
(𝟏𝟎𝟒.𝟓)𝟑
𝟔
+ 𝒄𝟏(𝟐𝟎𝟗)
𝟎 = 𝟏. 𝟏𝟑𝟏𝟔 × 𝟏𝟎𝟏𝟐 − 𝟓. 𝟎𝟏𝟑 × 𝟏𝟎𝟏𝟏 − 𝟑. 𝟐𝟐𝟓 × 𝟏𝟎𝟏𝟎 + 𝒄𝟏(𝟐𝟎𝟗)
𝒄𝟏 𝟐𝟎𝟗 = 𝟓. 𝟎𝟏𝟑 × 𝟏𝟎𝟏𝟏
+ 𝟑. 𝟐𝟐𝟓 × 𝟏𝟎𝟏𝟎
− 𝟏. 𝟏𝟑𝟏𝟔 × 𝟏𝟎𝟏𝟐
𝒄𝟏 = −𝟐. 𝟖𝟔𝟓 × 𝟏𝟎𝟗

16. MECHANICS OF MATERIAL-II
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Then eq (B) becomes,
𝑬𝑰. 𝒚 = 𝟕𝟒𝟑𝟕𝟒𝟕. 𝟏𝟕
𝒙𝟑
𝟔
− 𝟔𝟑𝟎𝟓. 𝟕𝟓
𝒙𝟒
𝟐𝟒
− 𝟏𝟔𝟗𝟓𝟗𝟎. 𝟗𝟔
< 𝒙 − 𝟏𝟎𝟒. 𝟓 >𝟑
𝟔
− 𝟐. 𝟖𝟔 × 𝟏𝟎𝟗𝒙
𝒚𝒎𝒂𝒙 𝒘𝒉𝒆𝒏 𝒙 = 𝟏𝟎𝟒. 𝟓
𝑬𝑰. 𝒚 = 𝟕𝟒𝟑𝟕𝟒𝟕. 𝟏𝟕
(𝟏𝟎𝟒. 𝟓)𝟑
𝟔
− 𝟔𝟑𝟎𝟓. 𝟕𝟓
𝟏𝟎𝟒. 𝟓 𝟒
𝟐𝟒
− 𝟏𝟔𝟗𝟓𝟗𝟎. 𝟗𝟔
< 𝟏𝟎𝟒. 𝟓 − 𝟏𝟎𝟒. 𝟓 >𝟑
𝟔
− 𝟐. 𝟖𝟔 × 𝟏𝟎𝟗
(𝟏𝟎𝟒. 𝟓)
𝑬𝑰. 𝒚 = 𝟏. 𝟒𝟏𝟒𝟔 × 𝟏𝟎𝟏𝟏 − 𝟑. 𝟏𝟑𝟑𝟐 × 𝟏𝟎𝟏𝟎 − 𝟐. 𝟖𝟔 × 𝟏𝟎𝟒. 𝟓
𝑬𝑰. 𝒚𝒎𝒂𝒙 = −𝟏. 𝟖𝟖𝟕𝟒𝟐 × 𝟏𝟎𝟏𝟏
𝒚𝒎𝒂𝒙 =
𝟏
𝑬𝑰
(−𝟏. 𝟖𝟖𝟕𝟒𝟐 × 𝟏𝟎𝟏𝟏)
𝑬𝑰. 𝜽 = 𝑬𝑰.
𝒅𝒚
𝒅𝒙
= 𝟕𝟒𝟑𝟕𝟒𝟕. 𝟏𝟕
𝒙𝟐
𝟐
− 𝟔𝟑𝟎𝟓. 𝟕𝟓
𝒙𝟑
𝟔
− 𝟏𝟔𝟗𝟓𝟗𝟎. 𝟗𝟔
< 𝒙 − 𝟏𝟎𝟒. 𝟓 >𝟐
𝟐
− 𝟐. 𝟖𝟔𝟓 × 𝟏𝟎𝟗
maximum slope at x=0 or x=209
𝜽𝒎𝒂𝒙
−𝟐. 𝟖𝟔𝟓 × 𝟏𝟎𝟗
𝑬𝑰

17. MECHANICS OF MATERIAL-II
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SHEAR FORCE
DIAGRAM
𝑽 𝒙 = 𝑹𝑨 < 𝒙 − 𝟎 >𝟎 − (𝒘𝒐) < 𝒙 − 𝟎 >𝟏− 𝒘 < 𝒙 − 𝟏𝟎𝟒. 𝟓 >𝟎
𝟎 < 𝒙 < 𝟐𝟎𝟗

18. MECHANICS OF MATERIAL-II
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BENDING MOMENT
DIAGRAM
𝑴 𝒙 = 𝑹𝑨 < 𝒙 − 𝟎 >𝟏 −
𝟏
𝟐
𝒘𝒐 < 𝒙 − 𝟎 >𝟐− 𝒘 < 𝒙 − 𝟏𝟎𝟒. 𝟓 >𝟏
𝟎 < 𝒙 < 𝟐𝟎𝟗

19. 19
This is the cross-section of the beam;
 As seen in the cross-section the beam is made up of
concrete and three steel rods of diameter 5/8 inch.
 Concrete beams subjected to bending moments are reinforced by steel rods.
The steel rods carry the entire tensile load below the neutral surface. The
upper part of the concrete beam carries the compressive load. In the
transformed section, the cross sectional area of the steel, As, is replaced by
the equivalent area n As where
n = Es/Ec

20. MECHANICS OF MATERIAL-II
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Diameter of rod=5/8 inch
Modulus of Elasticity of Steel=29x106
psi
Modulus of Elasticity of Concrete=3.6x106
psi
n = Es/Ec
n=29×106
3.6×106=8.06
Area of rod=
𝜋
4
(
5
8
)2=0.3068 𝑖𝑛𝑐ℎ2
No. of rod=3
Total Area=3x0.3068 𝑖𝑛𝑐ℎ2=0.9204 𝑖𝑛𝑐ℎ2
nAs=8.06x0.9204=7.418 𝑖𝑛𝑐ℎ2
First moment of Area about neutral axis;
(bx)x/2 – n.As(d-x) = 0
(15)𝑥2/2 – 7.418(18-x) = 0
so, x=3.75, x=-4.79(neglect)
And , I=𝑏𝑥3/3+n.As(d-x) = 0 => I=1769.99 𝑖𝑛𝑐ℎ4

21. 21
𝑬𝑰. 𝒚𝒎𝒂𝒙 = −𝟏. 𝟖𝟖𝟕𝟒𝟐 × 𝟏𝟎𝟏𝟏
𝒚𝒎𝒂𝒙 =
𝟏
𝑬𝑰
(−𝟏. 𝟖𝟖𝟕𝟒𝟐 × 𝟏𝟎𝟏𝟏
)
For Low Strength concrete:
𝐸𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 = 3.6 × 106𝑝𝑠𝑖 𝑎𝑛𝑑 𝐼 = 1769.99 𝑖𝑛𝑐ℎ4
𝒚𝒎𝒂𝒙 =
𝟏
(𝟑. 𝟔 × 𝟏𝟎𝟔(𝟏𝟕𝟔𝟗. 𝟗𝟗)
(−𝟏. 𝟖𝟖𝟕𝟒𝟐 × 𝟏𝟎𝟏𝟏)
𝒚𝒎𝒂𝒙 = −𝟐𝟗. 𝟔𝟐 𝒊𝒏𝒄𝒉 (Deflection downward and slope is clockwise)
For maximum slope;
𝑬𝑰. 𝜽 = 𝟕𝟒𝟑𝟕𝟒𝟕. 𝟏𝟕
𝒙𝟐
𝟐
− 𝟔𝟑𝟎𝟓. 𝟕𝟓
𝒙𝟑
𝟔
− 𝟏𝟔𝟗𝟓𝟗𝟎. 𝟗𝟔 < 𝒙 − 𝟏𝟎𝟒. 𝟓 >𝟐 −𝟐. 𝟖𝟔𝟓 × 𝟏𝟎𝟗
𝜽𝒎𝒂𝒙 𝒘𝒉𝒆𝒏 𝒙 = 𝟎 𝒐𝒓 𝒘𝒉𝒆𝒏 𝒙 = 𝟐. 𝟎𝟗 𝒊𝒏𝒄𝒉
𝐹𝑜𝑟 𝑥 = 0;
𝜽𝒎𝒂𝒙 =
−𝟐.𝟖𝟔𝟓×𝟏𝟎𝟗
(𝟑.𝟔×𝟏𝟎𝟔)(𝟏𝟕𝟔𝟗.𝟗𝟗)
=≫ 𝜽𝒎𝒂𝒙 = −0.449 (Deflection downward and slope is clockwise)

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Then from the table(Appendix-D);
For Simply Supported Beam:
𝑦𝑚𝑎𝑥 =
−5𝑤𝐿4
384𝐸𝐼
𝑦𝑚𝑎𝑥 =
−5(6305.75)(209)4
(384)(3.6 × 106)(1769.99)
𝑦𝑚𝑎𝑥 = −24.58 inch
% error=
29..62−24.58
29.62
× 100 = 17%
𝜃𝑚𝑎𝑥 =
−𝑤𝐿3
244𝐸𝐼
=
−(6305.75)(209)3
24 × (3.6 × 106)(1769.99)
𝜃𝑚𝑎𝑥 =-0.376

23. 23
 Now to calculate the respective stresses in concrete
(compressive) and in steel rods (tensile)
𝜎𝑐 = −𝑀𝑐/𝐼
𝑐 = 𝑥 = 3.75
𝜎𝑐 = − 43291396.05 x3.75/1769.99
𝜎𝑐 = −91.719 𝑘𝑠𝑖(𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛)
And,
𝜎𝑠 = 𝑛𝑀𝑐/𝐼
𝑐 = (𝑑 − 𝑥) = (18 − 3.75) = 14.25 𝑎𝑛𝑑 𝑛 = 8.06
𝜎𝑠 = 8.06x 43291396.05 x14.25/1769.99
𝜎𝑠 = 2809.187 𝑘𝑠𝑖(𝑇𝑒𝑛𝑠𝑖𝑙𝑒)
MECHANICS OF MATERIAL-II
MECHANICS OF MATERIAL-I

24. MECHANICS OF MATERIAL-II
24
% error=
0.449−0.376
0.449
× 100 = 16.25%
As Area under shear force diagram give moment So, moment maximum from
bending moment diagram.
𝑀 = 43291396.05 𝑙𝑏. 𝑖𝑛𝑐ℎ
As, 𝜏𝑥𝑦 = 0, 𝜎𝑠 = 2809.187 𝑘𝑠𝑖 and 𝜎𝑐 = −91.719 𝑘𝑠𝑖
The principal stresses 𝜎𝑎 and 𝜎𝑏 are;
𝜎𝑠= 𝜎𝑎 = 2809.187 𝑘𝑠𝑖
𝜎𝑐= 𝜎𝑏 = −91.719 𝑘𝑠𝑖
then for factor of safety we use Torsional criterial for brittle material.
𝜎𝑠
𝜎𝑈𝑇
=
𝜎𝑐
𝜎𝑈𝐶
= 𝐹. 𝑆
2809
58
−
−91
6.0
= 𝐹. 𝑆
𝐹. 𝑆 = 62 > 1
Ultimate Tensile Stress for steel = 𝜎𝑈𝑇=58 ksi
Ultimate Tensile Stress for concrete = 𝜎𝑈𝑇=6.0 ksi

25. MECHANICS OF MATERIAL-II
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CONCLUSION
From the above calculations it is concluded that the loads acting on
the corridor per day are within the allowable compressive and tensile
strength so the building is safe for the future unless any problems like
Earthquake acted on the beam.

26. THANK YOU!