Question

A mild steel flat 150mm wide and 20mm thick of length 5000mm
carries a tensite load of 300 KN, if E-2-10' N/mm' and 1/m = 0.25.
Calculate the change in length, width and volume of the bar.​

Answer 1

Answer:

Explanation:

Remember 1 GN/m2 = 1 kN/mm2

∴ E = 200 kN/mm2

We know E = Fl

Adl ∴ dl = F.l

A.E

Therefore, for portion of 700 mm,

the extension dl1 = 55000×700×7×4

200000×22×56×56 = 5

64

mm = 0.0178 mm

Now dl2 for 1400 mm length of 35 mm dia,

dl2 = 55000×1400×7×4

200000×22×35×35 = 0.4 mm

Total extension = dl1 +dl2 = 0.0178+0.4 = 0.4178 mm

Compound bars: When two or more materials (members) are rigidly fixed together so that they

share the same load and extend or compress by same amount, the two members form compound

bar. Let us say that in Fig. 2.7 we have to find stress in each material and amount of compression.

l

P Material A

of ES

Material B

of EB

Figure 2.8

Let the outer tube of material A has outside

dia as d1 and inside dia as d2 and inner tube of

material B has outside dia as d3 and inside dia

as d4. Both ends are joined rigidly to make compound bar of length l.