Question

Please could you explain how you would go about solving the following questions? Thanks in advance!

Self-tut Question A simply supported beam AD is loaded as shown in the Figure below. Note: E = 200 GPa, Iz = 1.5.10 m M=10 Nm P=100 N q=30 N/m A D L= 3 m Use the superposition method and Appendix H from the Textbook to determine: 1. The angle of rotation at point A , rad 2. The angle of rotation at point Do, rad 3. The deflection at point B, m 4. The deflection at point C, m

Answer 1

Given Information :

• Length of the beam :
  L = 3 m
  
• Point load :
  P= 100 N
  
• Uniformly Distributed Load :
  w = 30 N/m
  
• Bending Moment :
  M 10 N-m
  
• Young's Modulus :
  E = 200 G Pa
  
• Area Moment of Inertia :
  1. = 1.5 * 10-6 m 4
  

Notations : Let the reactions at the simple supports be :
R
and
Pu
respectively

Solution :

Calculating the reaction using the static equilibrium equations :

ΣF, = 0

Ra + Rd 100 + 30 *

Ra + Rd 130

Taking Moments about support at A

ΣΜ, = 0

Rd*3+10 – 30 * = - 100 *1 = 0 2

R*3 + 10 – 115 = 0

R*3= 105

Rd 35 N

From the above equation :

R, = 95 N

Using Macaulay's Method to formulate the equations for slope and deflections across the beam.

Since the UDL is upto the end support, we needn't extend it to the other end support in order to apply the macaulay's method.

Writing the equation for bending moment by using the discontinuity equations :

2 dạy EI * 24 Ra *T - W* -|(100 * (x- - 1)| – 1013 d.12 2

The limits within which the respective discontinuous equation can be applied is also specified in the above equation :

Integrating the equation :

22 dy EI * dic R* * W* — 6 - (100 + C1 10 (2 - 1) 2 - 10 *x 2 No

Integrating the equation again

12 El» =R *-***0*x+0+ (10 (100* (2 - 1)3 6 10 * 2

Writing the Boundary Conditions :

T
:
y 0

r=3
:
y 0

Apply the above boundary conditions in the respective Discontinuity function to find the values of the constants of integration :

First Boundary Condition :

T
:
y 0

03 04 EI *0= Ra* 6 W*+C1*0+ C 24

C2 = 0

Second Boundary Condition

r=3
:
y 0

33 34 EI *0= Ra* (3-1) +C1*3+0 - (100 * 6 W*- 24 - 10* 6

33 34 32. 0 = 95 * - 30 } - 24 (3 - 1). + Cl +3+0 - (100 * - - 10 * 6 6 2

855 405 90 0 800 + C1*3+0 - 6 2 4 2

0 = +C1 * 3+ 147.916

C1 = -49.30

Substitute the values of constants of integration into the deflection equation :

7 EI*y = 23 Ra* 6 - W*- 24 - 49.30 * 3 10 1-1-1 (100 * 10 * (:2 - 1) 6 2

Substitute the values of constants of integration into the slope equation :

12 R*-w* 2 EI *2= (x - 1) ( - 1410 (100 * - 49.30 10 - 10 *x 6 2

Now Calculating the following :

(a) Angle of rotation at point A

Ans :

The point A is at  
T

Substitute the same in the slope equation :

T EI + 8 = Ra - มล - - 49.30 6

02 03 EI*= Ra* 20 * - 49.30 2. 6

EI * = -49.30

в — 49.30 200 * 109 * 1.5* 10-б

a = -1.643 * 107 rad

(b) Angle of rotation at point D

Ans :

The point D is at  
r=3

Substitute the same in the slope equation :

12 R*-w* 2 EI *2= (x - 1) ( - 1410 (100 * - 49.30 10 - 10 *x 6 2

33 32 EI *2= = 95 * - 2 - 30 * - 49.30 -|(10 (100 * (3 – 1)? 2 - 10 * 3 6

EI * = 13.2

в 13.2 200 * 109 * 1.5* 10-6

0 = 4.4 * 10-5 rad

(c) Deflection at point B:

Ans:

The point B is at  
=

7 EI*y = 23 Ra* 6 - W*- 24 - 49.30 * 3 10 1-1-1 (100 * 10 * (:2 - 1) 6 2

Using the relevant discontinuity function :

$EI*y=R_a*\frac{x^3}{6}-w*\frac{x^4}{24} -49.30*x$

$EI*y=95*\frac{1^3}{6}-30*\frac{1^4}{24} -49.30*1$

$y=\frac{-34.7166}{200*10^9*1.5*10^{-6}}$

Yo = -1.157 * 10-4 m

(d) Deflection at point C:

Ans:

The point C is at  $x=2$

7 EI*y = 23 Ra* 6 - W*- 24 - 49.30 * 3 10 1-1-1 (100 * 10 * (:2 - 1) 6 2

Using the relevant discontinuity function :

23 24 EI *y = 95* -30 * 24 - 49.30 * 2- (100 (2-1) 6 6

$EI*y=-8.6$

$y=\frac{-8.6}{200*10^9*1.5*10^{-6}}$