السلام عليكم انا طالب اول سنة هندسة وعندي واجب. الوجب جزاء و اول جزء تصميم وبصراحة ماني عرف كيف اصمام تفكونًً ساعدوني:
Engineering 100 / Freshman Year Seminar
Mechanical Engineering Final Design Project
I. Problem Statement
You have been asked to design and test a support system to hang a sign of known weight W (= m * g), which has a single point of attachment (as indicated schematically in Fig. 1 (L)). The support system you come up with involves a system of springs of known but possibly different spring constants, ka, kb, …, kn, where, for example, kn is the spring constant of the nth spring. The store owner has asked that the top of the sign be a distance hmax cm below the support arm. As you examine the support arm, you notice that there is a fixed center attachment point and two or four additional possible attachment points whose position along the arm is variable (Fig. 1 ®). That is, xo is a design parameter when two additional attachment points are used. This means that your system can have either three or five attachment points.
Figure 1. (L) Schematic of hanging sign; ® support arm detail; distance xo is variable.
Two additional requirements are specified by the owner:
 Your design must use more than one spring, only one of which can be vertical.
 The end attachment points cannot be more than 60 cm apart.
Table 1 outlines the design parameters. You will have access to four red springs, four blue springs, and one green spring.
Table 1. Design Parameters
Variables
Values
Mass
m = 1000 g
Distance between top of sign and support arm
20 cm <h < 26 cm
red spring stiffness
kred = 11.5 N/m
blue spring stiffness
kblue = 29 N/m
green spring stiffness
kgreen = 50 N/m
Free length of the spring
Lo = 13.5 cm
In order to design the spring support system, you need to know some basics about how force is developed in a spring (i.e., Hooke’s law spring).
II. Hooke’s Law Spring Fundamentals
Consider the following example scenario in which the left end of the spring is attached to the wall (Fig. 2).
Figure 2. Freeend of spring is displaced (by applied force F) a distance dx from its equilibrium position.
In the top sketch, the spring is in a rest or equilibrium position and has equilibrium length Lo. That is, it is neither stretched nor compressed. In the bottom sketch, an external force F has been applied to the freeend of the spring causing it to displace dx to a new equilibrium position. Thus,
dx = B B For a Hooke’s law spring, the restoring force Fs developed in a spring due to displacement is given by Fs = k * dx and is directed along the axis of the spring. In this equation, k is the spring constant and dx is the displacement of the freeend of the spring. Fs is said to be a restoring force since it is directed in the direction opposite to the displacement dx. That is, if point B moves to the right, then Fs points to the left. Alternately, if point B moves to the left, then Fs points to the right. Since, after displacement, we notice that point B
neither moves to the left nor right, we say that point Bis in equilibrium. This means that the sum of forces at point B
is equal to zero. Consider the freebody diagram (FBD) at point B(see dashed box in Fig. 2 bottom). Considering only forces applied at point B
, we see that there is the applied force F (acting to the right) and the spring restoring force Fs (acting to the left), Fig. 3.
Figure 3. Freebody diagram at point B` showing external applied force F, and spring restoring force Fs.
Equilibrium implies that the sum of forces in any direction is identically equal to zero. In the xdirection here,
F + (Fs) = 0
F = Fs
Note in the top equation, Fs is negative since it points in the negative xdirection.
For 2 springs in parallel (Fig. 4a), we would have 2 spring forces in our FBD; while for 2 springs in series (Fig. 4b), there would only be 1 spring force in the FBD. Why?
(A)(B)
Figure 4. (A) Springs in parallel; (B) springs in series.
Finally, for 2 springs in symmetrical arrangement at angle with respect to the horizontal (Fig. 5), the force balance would appear as
W + (Fs * sin ) + (Fs * sin ) = 0
W = mg = 2Fssin = 2(k * dl)sin
Note, in this example we have considered that W acts in the positive direction. Furthermore, note that the weight W and the vertical component of the spring force (Fs * sin ) have different signs since they are oppositely directed. Since dl, in the above equation, is ultimately related to the hmax requirement, then
and we have enough information to solve for the angle , and hence xo. Here, Lo is the freelength of the spring; i.e., the spring’s length if it were simply resting on a tabletop.
Figure 5. Two springs in symmetrical arrangement inclined at angle with respect to the horizontal.
III. Design Requirements

In groups of two, design a spring support system which satisfies the owner’s requirements. Your design should specify the number and type of springs used; orientation of spring(s) (i.e., , xo); and the distance h. Remember: At the center attachment position only 1 spring can be used.

Schematic and calculations to support h (distance from the top the sign to the support point) must be included. Provide the theoretical stiffness of each system. Screen shots are allowed. They must be inserted in the document.
Your design (items 1 and 2 above) is due 11/17/2015 at 1:30 pm; submit on Blackboard. This deadline is for all groups.
IIII. Testing
You will be assigned one date to test and gather the data required for the report. 
Assemble your system as designed. Load the sign (i.e., m = 1000 g) and measure the distance from the top of the sign (connection point) to the attachment point. This is the proofofconcept of your design.

Now you will be collecting mass and displacement data to generate a plot of force versus displacement. A minimum of six data sets are required. This involves taking measurements of displacement as you increase the load progressively until reaching the design load. The slope of the force versus displacement curve is the experimental stiffness of the system.