Thursday

July 24, 2014

July 24, 2014

Posted by **AMK** on Sunday, June 23, 2013 at 4:47pm.

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=

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: BENDING MOMENT DIAGRAMS AND STRESSES, PART II

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.

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).

1

None

Status: UnansweredUnsubmitted

- Strength of materials -
**MathMate**, Sunday, June 23, 2013 at 6:34pmYou have posted a number of questions without your input, and most of all, without the accompanying diagrams, nor the descriptions to replace the diagrams.

If you believe that we can imagine what the loading is like, you are mistaken.

I suggest you concentrate on one particular problem, try your best, and if you still need help, post the question with a description of the diagram.

For example, for the first question, all you need is to say

Beam: simply supported at both ends

Length, L= ? ft

Point load a=? ft from the left

Point load b=? ft from first point load,

etc.

- MIT 2.01X -
**simonsay**, Thursday, June 27, 2013 at 6:27pmHW7_1A: BENDING MOMENT DIAGRAMS AND STRESSES

Length, L= a+b

Point load a=a

Point load b=P

- MIT 2.01X -
**MAN**, Friday, June 28, 2013 at 4:59pmWait I am confused?

- MIT 2.01X -
**simonsay**, Saturday, June 29, 2013 at 8:38amHW7_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

- MIT 2.01X -
**Marcus**, Monday, July 1, 2013 at 1:39pmHW7_1B_1 : Mmax+=4000

- MIT 2.01X -
**Just Help**, Monday, July 1, 2013 at 3:58pmHow about 7.1B1 Mmax-? 7.1B2?

- MIT 2.01X -
**Just Help**, Monday, July 1, 2013 at 4:03pmAnybody 7.1.C, 7.2.1,7.2.2, 7.3.1, 7.3.2, 7.4.1 ?

- MIT 2.01X -
**MAN**, Monday, July 1, 2013 at 5:04pm7-1B.2, 7-1C, 7-2-1, 7-2-2, 7-3-1, 7-3-2, 7-3-3, 7-4-1,7-4x?

- MIT 2.01X -
**Just Help**, Monday, July 1, 2013 at 5:15pm7.1B1 Mmax-?

- MIT 2.01X -
**Just Help**, Monday, July 1, 2013 at 5:18pm7.1B1 Mmax-?

- MIT 2.01X -
**Just Help**, Monday, July 1, 2013 at 5:22pmThe due date is almost there, please anyone?

- MIT 2.01X -
**simonsay**, Monday, July 1, 2013 at 5:23pm7.1B1 Mmax- =-8000

- MIT 2.01X -
**Just Help**, Monday, July 1, 2013 at 5:28pmThanks, 7.1B2 (Matlab), 7.1.C (Matlab), 7.2.1,7.2.2, 7.3.1, 7.3.2, 7.4.1

- MIT 2.01X -
**simonsay**, Monday, July 1, 2013 at 8:22pmHW 7_2_2

576

-576

281.33

-281.33

HW 7_3_1

-P*x

- MIT 2.01X -
**MAN**, Monday, July 1, 2013 at 8:49pmHelp with the rest?

- MIT 2.01X -
**simonsay**, Monday, July 1, 2013 at 9:02pmHW7_3_2

part 2 0.5

part 3 0.05

- MIT 2.01X -
**MAN**, Monday, July 1, 2013 at 9:53pmHelp with the rest of all the questions???? PLEASE I AM BEGGING YOU

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