3) A simply supported beam is subjected to point load and uniformly distributed load as shown in Figure Q3. Calculate the slope and deflection at point C by using Maccaullay method. Give the answer in term of EI.
3) A simply supported beam is subjected to point load and uniformly distributed load as shown in Figure Q3. Calculate the slope and deflection at point C by using Maccaullay method. Give the answer in term of EI.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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3) A simply supported beam is subjected to point load and uniformly distributed load as shown in Figure Q3. Calculate the slope and deflection at point C by using Maccaullay method. Give the answer in term of EI.
![SZH/DCC2063/EXAMPLE SLOPE&DEFLECTION MACAULAY METHOD
EXAMPLE SLOPE AND DEFLECTION (MACAULAY METHOD)
A simply supported beam is subjected to point load and uniformly distributed load as shown in
Diagram. Calculate slope and deflection at point C by Using Macaulay Method in term of EI.
40KN
15KN/m
A
B
C
2mm
1m
7m
EMD = 0,
FAY(10) – 15(7)(1/2 (7)) - 40(8) = 0,
10 FAY- 320 + 367.5
10 FAY= 687.5 kN
FAY= 68.75 kN
Moment equation: -
Mx = El d'y = 68.75 [x] - 40 [x-2] – 15 [x - 3]2
dx,
slope equation :-
El dy =
dx
68.75 [x]? - 40 [x-2]? – 15 [x - 3] + C1
2
2
6
Deflection equation :-
El y = 68 75[x] - 40 [x-2] – 15 [x - 31* + Cx+ C2
6
24
Boundary condition:-
X = 0, y = 0, C2=0
X = 10m , y = 0, C, = ?
EI (0) = 68 75[10]³ - 40 [8]³ – 15 [71* + 10C1
6
24
0 = 11458.33 – 3413.33 – 1500.63 + 10C,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcea20e3d-92fd-4583-b0ce-fd99e602e1c5%2F768c635b-e2bf-4f21-a79f-8cd0e8f50b75%2Ff1ofstg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:SZH/DCC2063/EXAMPLE SLOPE&DEFLECTION MACAULAY METHOD
EXAMPLE SLOPE AND DEFLECTION (MACAULAY METHOD)
A simply supported beam is subjected to point load and uniformly distributed load as shown in
Diagram. Calculate slope and deflection at point C by Using Macaulay Method in term of EI.
40KN
15KN/m
A
B
C
2mm
1m
7m
EMD = 0,
FAY(10) – 15(7)(1/2 (7)) - 40(8) = 0,
10 FAY- 320 + 367.5
10 FAY= 687.5 kN
FAY= 68.75 kN
Moment equation: -
Mx = El d'y = 68.75 [x] - 40 [x-2] – 15 [x - 3]2
dx,
slope equation :-
El dy =
dx
68.75 [x]? - 40 [x-2]? – 15 [x - 3] + C1
2
2
6
Deflection equation :-
El y = 68 75[x] - 40 [x-2] – 15 [x - 31* + Cx+ C2
6
24
Boundary condition:-
X = 0, y = 0, C2=0
X = 10m , y = 0, C, = ?
EI (0) = 68 75[10]³ - 40 [8]³ – 15 [71* + 10C1
6
24
0 = 11458.33 – 3413.33 – 1500.63 + 10C,
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