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19991022

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TABLE OF CONTENT:
Introduction ……………………………………………………………………………
Aim…………………………………………………………………………………………
Apparatus………………………………………………………………………………..
Procedure………………………………………………………………………………..
Non symmetrical……………………………………………………………………….
Symmetrical………………………………………………………………………………
Pictures…………………………………………………………………………………….
Conclusion and discussion…………………………………………………………..
References…………………………………………………………………………………
1
INTRODUCTION:
Shear force is the internal force on a member when the force is not applied at the axis.
Shearing force is the force divided by the cross-sectional area.
Bending moment is the force trying to rotate member. Moment is the perpendicular
distance from the force to the axis multiplied by the force.
AIM:
 To determine the support reactions for point-loaded beam.
APPARATUS:





Beam to carry a load
Two load cells to test the shearing force and bending moments
Dial gauge for measurements
Tape measure to measure the distance
The weights to test compression and tensile strength of the beam
PROCEDURE:
 The length of the beam was measured.
 One of the load cells was then set up at the quarter span of to
the left of the beam.
 The second load cell was also set upped at same place as
first load but at the right of the beam and the knife-edge
was then locked.
 The two hangers was positioned equidistant from the mid of
the beam.
 Both lock knife-edges were locked and the load cell
indicators was adjusted to zero.
 Loads was applied to the hangers in a systematic manner,
the beam was tapped 3 times gently and readings of load
cells was considered.
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RESULTS AND CALCULATIONS
FOR NON-SYMMETRICAL:
For 2N:
∆ = (𝑅1 +𝑅2 ) – (𝑊1 +𝑊2 )
%= 100∆/ (𝑊1 + 𝑊2 )
100∆
= (1, 1+1, 01) - (2+2)
% =𝑤
1 +𝑤2
= -1, 89 N
100×1,89
=
2+2
= 47, 25%
FOR 4N: ∆= (𝑅1 +𝑅2 )-(𝑊1 +𝑊2 )
100∆
%=
= (1, 89 +1, 81)-(4+4)
𝑤1 +𝑤2
=
=-4, 3N
100×4,3
4+4
=53, 75%
100∆
FOR 6N: ∆= (𝑅1 +𝑅2 )-(𝑊1 +𝑊2 )
% =𝑤
1 +𝑤2
= (3+2, 92)-(6+6)
=
= -6, 08N
100×6,08
6+6
= 50, 67%
FOR 8N: ∆= (𝑅1 +𝑅2 )-(𝑊1 +𝑊2 )
100∆
%=
𝑤1 +𝑤2
100×8,01
= (4+3, 99)-(8+8)
=
=-8, 01N
= 50, 06%
FOR 10N: ∆ = (𝑅1 +𝑅2 )-(𝑊1 +𝑊2 )
%=
8+8
100∆
𝑤1 +𝑤2
100×10,05
= (5+4, 95)-(10+10)
=
= -10, 05N
= 52, 25%
3
10+10
𝑅1 (N)
𝑊1 (N)
𝑊2 (N)
𝑅2 (N)
1,1
1,89
3
4
5
2
4
6
8
10
2
4
6
8
10
1,01
1,81
2,92
3,99
4,95
𝑅1 +
𝑅2 (N)
2,11
3,7
5,92
7,99
9,95
𝑅2 N
20
( 𝑅1 + 𝑅2 )
40
∆(N)
%
-1,89
-4,3
-6,08
-8,01
-10,05
47,25
53,75
50,67
50,06
50,25
FOR SYMMETRICAL:
Calculations:
∑𝑀𝑅1 𝐶𝐿𝑂𝐶𝐾𝑊𝐼𝑆𝐸 = ∑𝑀𝑅1𝐴𝑁𝑇𝐼𝐶𝐿𝑂𝐶𝐾𝑊𝐼𝑆𝐸
20X0, 169 +20X0, 507 =𝑅2 X0, 676
0,676𝑅2
0,676
13,52
=0,676
𝑅2 =20N
∑V↑ = ∑V↓
20 + 𝑅2 = 20 +20
𝑅2 = 20
𝑅1 N
20
𝑊1 N
20
𝑊2 N
20
4
∆𝑁
0
%
0
PICTURES:
WEIGHT HOOCKER
TAPE MESUARE
BEAM
5
CONLUSION AND DISCUSSION:
During the conduction of the experiment we found out that the shearing force of
symmetrical differs from that of un-symmetrical. In un-symmetrical forces was not in a
state of equilibrium that’s why the shear force diagram of symmetrical closes and that
of un-symmetrical not closes.
The bending moment diagram of symmetrical differs from that of un-symmetrical, the
bending moment diagram of symmetrical didn’t cross the axis particularly x-axis while
that of un-symmetrical crossed the x-axis. Reasons in this observation was the unequal
of reactions and weights causes the areas to be unequal.
REFERNCES:
 Lab technician: Miss Nopote
 Lab assistance: Mr. Pakathi
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