Worksheet 4 - Axial design and materials PDF

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Worksheet 4 - Axial design and materials...

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MONASH UNIVERSITY

ENG1001 LIGHTER, FASTER, STRONGER

WORKSHEET 4 – MECHANICAL PROPERTIES OF MATERIALS AND AXIAL MEMBER DESIGN

Name: Student ID: Team No.:

You need to watch the videos for Week 4 for completing the following tasks

1. What are the equations (or inequalities) we need to check for design when a member experiences (a) Axial Tension

(b) Axial Compression (c) What is the maximum load a member in tension or compression can carry?

2. A steel column of square cross-section is subject to an axial compression P. Calculate the maximum value of P assuming the end supports are pin supports and the column length is (a) L = 2500 mm; (b) L = 400 mm. Assume E = 200000 MPa (N/mm2), yield stress = 300 MPa (Ans: 4.21kN, 120kN) P

Step 1: Calculate the second moment of area, I, for the column (I=b4/12).

Step 2: Calculate the maximum squash load (Pc) of the column. 20mm 20mm

Step 3: (a) For L = 2500 mm, calculate the maximum buckling load (the Euler’s load PE) and compare with Pc. Draw your conclusion: the maximum load the column can carry, P =

Step 4: (b) For Le = 400 mm, calculate the maximum buckling load (the Euler’s load PE) and compare with Pc. Draw your conclusion: the maximum load P =

Step 5: Draw the corresponding failure envelope, i.e plot both the Euler buckle and Squash load lines.

3. The same section in Q2 now has fully fixed support ends. How does this effect the maximum P value when (a) L = 2500 mm; (b) L = 400 mm. Plot these points above and draw the corresponding failure envelope. How do the two envelopes compare? (Ans: 16.84kN, 120kN).

P (L=400mm) =

P(L=2500mm) =

30kN

4. Using a steel wire for one member (which on steel circular rod for the other member, calcula radius required of these two members for the s Assume that both members are pinned to the g the load is an ultimate limit state load. Take the material properties of steel as yield stre Young’s modulus = 200,000 MPa. (Ans: 7.4mm, 23.6mm) Note: As all joints are pinned, you may treat this structure as a truss.

A

4m

B

C

4m Step 1: Draw a free-body diagram which enables you to calculate the forces in these 2 members and use the method of joints to calculate the member forces.

Force in AB is

(Tension/Compression)

Force in AC is

(Tension/Compression)

Step 2: For the member in tension, state the design condition for yielding and hence calculate the minimum radius required.

The minimum radius of the tension member is

Step 3: For the member in compression, (a) state the design condition for yielding and hence calculate the minimum radius required.

(b)State the design condition for buckling and hence calculate the minimum radius required. Note: For a circular cross-section 𝐼 = 𝜋𝑟 4 /4

(c) Conclusion: from (a) and (b), the minimum radius required is

5. The structure ABC shown is subject to a vertical design load of 250 kN at mid-span of AB. The joint at B is an internal pin. Ignore self weight. (a) Calculate the reactions at A and C, and the force in BC. Indicate whether member BC is in tension or compression. (Ans: 125kN, 1250kN, 166.6kN, -166.6kN) (b) Choose a suitable minimum weight square hollow section (SHS) from the tables attached for member BC. Young’s modulus, E = 200,000 N/mm2. Note: Grade C350 steel means a yield stress of 350MPa

250 kN

Note: The total number of reactions at the supports of this structure is 4. Therefore we need more than 1 free-body diagram to solve this problem. (You may view this as an ‘internal pin’ problem)

(a) Step 1: Draw the reaction forces at the supports. Draw 2 free-body diagrams, one for AB and another for BC, both cutting through B.

Step 2: Use the free-body diagram for AB, take moments about B to solve for Ay.

Step 3: Consider the whole structure now and solve for Cy by considering ∑Fy = 0

Step 4: Use the free-body diagram for BC and the value for Cy and take moments at B to solve for Cx.

Step 5: Continue to use the whole structure and solve for Ax using ∑ 𝐹𝑥 = 0

Note: If an element has pins at both ends (be they pin supports and/or internal pins) and has no load acting in-between, like member BC, it is called a two-force member. These elements have two forces acting on them, one at each end. If only two forces act on a body that is in equilibrium, then they must be equal in magnitude, co-linear and opposite in direction, i.e. they must act along the longitudinal axis of the member. Step 6: Calculate the resultant of the reaction forces at support C and confirm its orientation lines up with the member BC longitudinal axis. (Ans: 208.3 kN)

Force in BC is

(tension/compression)

(b) Step 1: Check the design condition for buckling and compare the minimum required second moment of area, I, from those listed in the SHS tables.

Choose section (write down both A and I of this section) Step 2: Check the design condition for yielding and compare the minimum required cross-sectional area, A, from those listed in the SHS tables

Choose section Conclusion: from Step 1 & 2: Choose section

6. Which of the rods is under the highest normal stress? Calculate the stresses and tick the answer Tick

x σ=

kPa

σ=

kPa

σ=

kPa

7. Calculate the engineering stress on a bar 28cm long and having a cross-section of 9.5mm x 4.5mm

that is subjected to a load of 4200kg.

σ = 982

MPa

8. A specimen of aluminium having an elastic modulus of 69GPa and a rectangular cross section 10mm x 12.7 mm is pulled in tension with 35,500 N force, producing only elastic deformation. Calculate the resulting strain

ε = 0.0041 or 0.41%

9. A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa and an original diameter of 3.8 mm will experience only elastic deformation when a tensile load of 2000N is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.42 mm. Work to two significant figures.

l = 250 mm 10. Using the stress-strain curve below, estimate or calculate the yield stress, elastic modulus, ultimate tensile strength, uniform elongation, 0.2% proof stress, ductility. Note that you may need to use other points to those shown.

(190, 0.015)

(205, 0.020) (205, 0.022)

(180, 0.026)

(180, 0.026) (155, 0.002) (150, 0.001)

(75, 0.0005) Area = 4.5

E=

150

GPa

σy =

150

U. E. = 0.022

σUTS =

σ0.2% PS = 160 MPa

Ductility = 0.0248

Toughness =

4.5 x 106

205

MPa

J/m3

MPa

11. A sample of porous ceramic in the form of a cylinder of cross-sectional area 2 cm2 is loaded on its upper end by a mass of 30 kg. Careful measurement reveals that the length decreased by 0.007 percent. What is the Young's modulus of the porous ceramic?

E=

GPa

12. A metal bar of length 6 cm, width 2 mm and thickness 1.5 mm is tested in tension. The following force versus extension curve is produced. Use this data to calculate the elastic modulus and the 0.2% proof stress.

E=

GPa

σ0.2% PS =

MPa

13. A cylindrical rod 380 mm long, having a diameter of 10.0 mm is to be subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than 0.9 mm when the applied load is 24,500N, which of the four metals or alloys listed below are possible candidates? Justify your choice(s). Material Alluminium alloy Brass Copper Steel Alloy

Modulus (GPa) 70 100 135 207

Yield Strength (MPa) 255 345 250 450

Tensile Strength (MPa) 420 420 290 550

Required properties: σy > MPa E> GPa Possible material(s):...