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.
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?

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

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_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 magnitude of the maximum bending moment 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).

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

HW7_1B_1: Mmax 4000

HW7_1B_1: Mmin -8000

HW7_2_3_1: Decrease

HW7_2_3_2: Db

HW7_3_1

PART 2
-P*x+P*(x-L_1)*(L_1+L_2)/L_2

7-1B.2, 7-1C, 7-2-1, 7-2-2, 7-3-1, 7-3-2, 7-3-3, 7-4-1,7-4x?

HW 7_2_2


576
-576
281.33
-281.33

HW 7_3_1
-p*x

HW 7_3_1

-P*x (P capital)

HW7_1B_2, HW7_1C, HW7_2_1, HW7_3_2 , HW7_3_3, HW7_4_1, HW7_4_1x???

HW7_3_2

part 2 0.5

part 3 0.05