# Mechanial engineering

posted by .

HW7_1A: BENDING MOMENT DIAGRAMS AND STRESSES, PART I

We will obtain bending moment diagrams for three beams and calculate stresses inside the beams. For each beam, take the x-axis along the neutral axis of the beam, oriented from left to right, with origin at A.

HW7_1A_1 : 5.0 POINTS

For the beam above, obtain symbolic expressions in terms of P, a, b, x for the bending moment resultant, M(x):

for 0≤x≤a, M(x)=
for a≤x≤a+b, M(x)=

HW7_1A_2 : 5.0 POINTS

Obtain symbolic expressions, in terms of P, a, and b, for the maximum positive value of the bending moment, M+max, and for the coordinate, x+max, where it occurs:

M+max=
at x+max=

HW7_1A_3 : 5.0 POINTS

The beam has a constant rectangular cross section of width h and height 3h. The beam is homogeneous and composed of a linear elastic material with a failure stress of σf=180 MPa. Take a=1.8 m, b=0.4 m, and P=2.2 kN.

Calculate the numerical value for the minimum dimension (in cm), h=hmin, required for a safety factor SF=3 against failure:

HW7_1B_1 : 5.0 POINTS

The beam above has a rectangular cross section of height h=12 cm and width b=4 cm. Take L=1.2 m and P=10 kN.

Obtain numerical values in N·m, for the maximum positive and maximum negative values of the bending moment in the beam, M+max and M−max:

HW7_1B_2 : 5.0 POINTS

Let's say that a cross section where you have M+max is at location x+max, and the cross section where you have M−max is at location x−max. Use the MATLAB window below to plot the profiles of stress on sections x+max and x−max, i.e. σn(x+max,y)=σ+n(y), (Splus) and σn(x−max,y)=σ−n(y), (Smin).

A vector, y, of coordinates from the bottom (y=−0.06 m) to the top (y=0.06 m) of the cross section has already been defined for you. The vector y has 11 entries, evenly spaced, with the first entry y(1)=−0.06 at the bottom surface of the cross section, the central entry y(6)=0 at the neutral axis, and the last entry y(11)=0.06 at the top surface of the cross section.

First, enter the values of the known quantities (Mplus, Mmin, h, b) in SI units (N·m, m). Then enter an expression for I, σ+n(y), σ−n(y) (use variable names I, Splus, Smin). These expressions should be written in terms of Mplus, Mmin, h, b, and y. Note that once you have defined a variable, you can use it in successive lines of the script. (Refer to E7_1_2 for help.)

Now run the MATLAB script to plot the stress profiles. Note that in the "DO NOT EDIT" portion of the window, the script converts stress values to MPa (by dividing by 10^6) so that the values on the plots are in MPa.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
% DO NOT EDIT % % % % % % % % % % % % % % % % % % % % % % % % %
y=linspace(-0.06, 0.06, 11); %
% DO NOT EDIT % % % % % % % % % % % % % % % % % % % % % % % % %
Mplus=
Mmin=
h=
b=
I=
Splus=
Smin=
% DO NOT EDIT % % % % % % % % % % % % % % % % % % % % % % % % %
subplot(1,2,1) %
sigma_n=Splus; %
plot(sigma_n /10^6,y,'LineWidth', 2.0) %
ylabel('y') %
xlabel('\sigma_n (x^+,y)[MPa] ') %
title('stress profile at x^+') %
minsn=min(sigma_n/10^6); %
maxsn=max(sigma_n/10^6); %
axis([minsn,maxsn,-0.06,0.06]) %
dsn=(maxsn-minsn)/2; %
set(gca,'XTick',minsn:dsn:maxsn,'Ytick',-0.06:0.06:0.06) %
grid on %
subplot(1,2,2) %
sigma_n=Smin; %
plot(sigma_n/10^6,y,'LineWidth', 2.0) %
ylabel('y') %
xlabel('\sigma_n (x^-,y)[MPa] ') %
title('stress profile at x^-') %
minsn=min(sigma_n/10^6); %
maxsn=max(sigma_n/10^6); %
axis([minsn,maxsn,-0.06,0.06]) %
dsn=(maxsn-minsn)/2; %
set(gca,'XTick',minsn:dsn:maxsn,'Ytick',-0.06:0.06:0.06) %
grid on %
% DO NOT EDIT % % % % % % % % % % % % % % % % % % % % % % % % %

HW7_1C: BENDING MOMENT DIAGRAMS AND STRESSES, PART III

HW7_1C : 5.0 POINTS

For the simply supported beam in the figure, take L=1.2 m and M0=3 kN·m.

In the MATLAB window below, write MATLAB code to plot the bending moment diagram M(x). You will have to define the variables M0 and L (in N·m and m) as well as three vectors x1, x2, x3 for the x coordinate:

0≤x1≤L/3
L/3≤x2≤2L/3
2L/3≤x3≤L
Also, you will have to define three vectors M1, M2, M3 for the corresponding M(x) values.

Finally, you will have to assemble the vectors x and M from x1, x2, x3 and M1, M2, M3, and plot M vs x using the area command. (Refer to E7_2_2 for help.)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
L =
M0 =

x1 =
x2 =
x3 =

M1 =
M2 =
M3 =

x =
M =

% Enter your plotting commands below:
area
xlabel
ylabel

HW7_2: STRESSES FROM PRESCRIBED CURVATURE: WIRE AROUND PEGS

A round copper wire (E=110 GPa) of round cross section with diameter d=2 mm is bent in a figure-8 around two round pegs of diamters DA=38 cm and DB=78 cm as shown in the figure. The axial force in the wire is negligible. The wire is in continuous contact with the pegs around their outer perimeters at A and B.

HW7_2_1 : 5.0 POINTS

Obtain the numerical value, in N⋅m, for the maximum magnitude of bending moment in the wire:

Mmax=

HW7_2_2 : 20.0 POINTS

Obtain the numerical value, in MPa, for the maximum tensile and compressive stress (σ+max,σ−max) in the wire at sections A and B.

σ+max(A)=
σ−max(A)=
σ+max(B)=
σ−max(B)=

HW7_2_3 : 5.0 POINTS

If you want to decrease the stress in the wire should you increase or decrease the diameter of the wire?

increasedecrease
If you want to decrease the stress in the wire, should you make both pegs of diameter DA or DB ?

HW7_3 MOMENTS, CURVATURE, STRESS AT A POINT

The beam in the figure is homogeneous, with modulus E=61.1 GPa, and has round cross section with diameter d=100 mm. The supports at A and B exert only vertical reactions. A load P=12 kN is applied as indicated, at a distance L1=0.5 m from the left support A. The distance between the supports ( at A and B) is L2=2.5 m.

Take the x-axis (from left to right) along the neutral axis, with the origin at the left end of the beam (where the load P is applied)

HW7_3_1 : 8.0 POINTS

Obtain symbolic expressions, in terms of x, P, L1, L2, (written as L_1and L_2), for the bending moment resultant:

For 0≤x≤L1: M(x)=
For L1≤x≤L1+L2: M(x)=

HW7_3_2 : 12.0 POINTS

Obtain numerical values for the maximum tensile stress (σmax, in MPa ) in the beam and for the coordinates (xmax,ymax, in m) where σmax occurs:

σmax=
xmax=
ymax=

HW7_3_3 : 10.0 POINTS

Consider the beam cross section at C, the mid-span between A and B, as indicated.

On section C, obtain numerical values for the stress at points a (σa, in MPa) and b (σb, in MPa), and for the radius of curvature (ρC, in m) at the neutral axis.

σa=
σb=
ρC=

HW7_4: OPTIMIZATION OF SUPPORT PLACEMENT - THE TIRE SWING

You have to position the supports for a tire swing at the local playground. The top beam is of length 2L, and it supports three identical tire swings of weight W as indicated in the figure. The supporting legs will be placed symmetrically at a distance b (b > 2/3L) from the central swing.

HW7_4_1 : 30.0 POINTS

Obtain a symbolic expression, in terms of L, for the optimal distance b∗, which will make the maximum magnitude of the bending moment resultant in the top beam as small as possible.

b∗=

HW7_4_X : 0.0 POINTS

CHALLENGE QUESTION

In the blank MATLAB window below, write a MATLAB script to plot the max positive and max negative moments in the beam, as a function of b, providing a graphic solution to the determination of b∗ (use E7_3 as a reference).

1
None

• Mechanial engineering -

HW7_1A_1
P*(1-a/(a+b))*x
-(a/(a+b))*P*x+P*a

HW7_1A_2
-(a/(a+b))*P*a+P*a
a

HW7_1A_3
2

## Similar Questions

1. ### Physics/Mechanics

We will obtain bending moment diagrams for three beams and calculate stresses inside the beams. For each beam, take the x-axis along the neutral axis of the beam, oriented from left to right, with origin at A. HW7_1A_1 : 5.0 POINTS …
2. ### MIT 2.01X

Posted by MAN on Tuesday, June 18, 2013 at 9:24pm. We will obtain bending moment diagrams for three beams and calculate stresses inside the beams. For each beam, take the x-axis along the neutral axis of the beam, oriented from left …
3. ### MIT 2.01X

We will obtain bending moment diagrams for three beams and calculate stresses inside the beams. For each beam, take the x-axis along the neutral axis of the beam, oriented from left to right, with origin at A. HW7_1A_1 : 5.0 POINTS …
4. ### mitx 2.01x

We will obtain bending moment diagrams for three beams and calculate stresses inside the beams. For each beam, take the -axis along the neutral axis of the beam, oriented from left to right, with origin at .
5. ### Materials Science

A 3-point bending test is performed on a simply supported beam of length L=4m and of rectangular cross-section, with width b and height h. For a central load of 10kN, draw the bending moment and shear force diagrams for the beam. Then …
6. ### Materials Science

A 4-point bending test is performed on the same beam, with symmetric loads of 5kN separated by 2m. Draw the bending moment and shear force diagrams for the beam. Then answer the following questions about the shear force V and bending …
7. ### Materials Science

A hollow cylinder with outer radius r0 and inner radius ri is loaded by a uniformly distributed load q across its length L. The beam is supported by a roller at its left end and a pin L5 from its right end. What is the maximum bending …
8. ### mechanics

A 3m long balance beam is simply supported with an overhang at each end of a=0.5m. A gymnast weighing 50kg is standing a distance c to the right of support A and a distance b to the left of support B. How does the maximum bending moment …
9. ### Calculus applications

The bending moment (M) along a beam is M = WLx/2 -Wx2/2 kNm where x is the distance of a beam length L from the left hand end. W is the weight per unit length. (a) Shear force is calculated as the differential of bending moment. Find …
10. ### calculus

The bending moment (BM) along a beam BM = 30x -5x2 kNm where x is the distance in metres from the left hand end. Find the position and value of the maximum bending moment.

More Similar Questions