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March 27, 2015

March 27, 2015

Posted by **Colour** on Sunday, June 30, 2013 at 12:19pm.

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)=

unanswered

for a≤x≤a+b, M(x)=

unanswered

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=

unanswered

at x+max=

unanswered

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:

hmin= cm unanswered

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:

M+max= N·m unanswered

M−max= N·m unanswered

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.

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% 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 % % % % % % % % % % % % % % % % % % % % % % % % %

Status: UnansweredUnsubmitted

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.)

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L =

M0 =

x1 =

x2 =

x3 =

M1 =

M2 =

M3 =

x =

M =

% Enter your plotting commands below:

area

xlabel

ylabel

Status: UnansweredUnsubmitted

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=

N·m unanswered

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)=

MPa unanswered

σ−max(A)=

MPa unanswered

σ+max(B)=

MPa unanswered

σ−max(B)=

MPa unanswered

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 ?

DADB

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)=

unanswered

For L1≤x≤L1+L2: M(x)=

unanswered

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=

MPa unanswered

xmax=

m unanswered

ymax=

m unanswered

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=

MPa unanswered

σb=

MPa unanswered

ρC=

m unanswered

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∗=

unanswered

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).

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None

Status: UnansweredUnsubmitted

- Mechanial engineering -
**simonsay**, Monday, July 1, 2013 at 12:19pmHW7_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

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