Two cables, each having a length i. of approximately 40 m, support a loaded -container of weight W (see figure). The cables, which have an effective cross-sectional area A = 48.0 mm 2 and effective modulus of elasticity E = 160 GPa. are identical except that one cable is longer than the other when they are hanging separately and unloaded. The difference in lengths is d = 100 mm. The cables are made of steel having an elastoplastic stress-strain diagram with a r = 500 MPa. Assume that the weight ft' is initially zero and is slowly increased by the addition of material to the container. (a) Determine the weight W y that lirsl produces yielding of the shorter cable. Also, determine the corresponding elongation 5 of the shorter cable. (b) Determine the weight W p that produces yielding of both cables. Also, determine the elongation S p of the shorter cable when the weight W just reaches the value W p . (c) Construct a load-displacement diagram showing the weight W as ordinate and the elongation & of the shorter cable as abscissa. Hint: The load displacement diagram is not a single straight line in the region 0 ^ W ^ W Y ,
Two cables, each having a length i. of approximately 40 m, support a loaded -container of weight W (see figure). The cables, which have an effective cross-sectional area A = 48.0 mm 2 and effective modulus of elasticity E = 160 GPa. are identical except that one cable is longer than the other when they are hanging separately and unloaded. The difference in lengths is d = 100 mm. The cables are made of steel having an elastoplastic stress-strain diagram with a r = 500 MPa. Assume that the weight ft' is initially zero and is slowly increased by the addition of material to the container. (a) Determine the weight W y that lirsl produces yielding of the shorter cable. Also, determine the corresponding elongation 5 of the shorter cable. (b) Determine the weight W p that produces yielding of both cables. Also, determine the elongation S p of the shorter cable when the weight W just reaches the value W p . (c) Construct a load-displacement diagram showing the weight W as ordinate and the elongation & of the shorter cable as abscissa. Hint: The load displacement diagram is not a single straight line in the region 0 ^ W ^ W Y ,
Two cables, each having a length i. of approximately 40 m, support a loaded -container of weight W (see figure). The cables, which have an effective cross-sectional area A = 48.0 mm2 and effective modulus of elasticity E = 160 GPa. are identical except that one cable is longer than the other when they are hanging separately and unloaded. The difference in lengths is d = 100 mm. The cables are made of steel having an elastoplastic stress-strain diagram with a r= 500 MPa. Assume that the weight ft' is initially zero and is slowly increased by the addition of material to the container.
(a) Determine the weight Wythat lirsl produces yielding of the shorter cable. Also, determine the corresponding elongation 5 of the shorter cable.
(b) Determine the weight Wpthat produces yielding of both cables. Also, determine the elongation Spof the shorter cable when the weight W just reaches the value Wp.
(c) Construct a load-displacement diagram showing the weight W as ordinate and the elongation & of the shorter cable as abscissa. Hint: The load displacement diagram is not a single straight line in the region 0 ^ W ^ WY,
8. In the following check to see if the set S is a vector subspace of the corresponding Rn. If
it is not, explain why not. If it is, then find a basis and the dimension.
X1
(a) S
=
X2
{[2], n ≤ n } c
X1 X2
CR²
X1
(b) S
X2
=
X3
X4
x1 + x2 x3 = 0
2) Suppose that two unequal masses m₁ and m₂ are moving with initial velocities V₁ and V₂,
respectively. The masses hit each other and have a coefficient of restitution e. After the impact,
mass 1 and 2 head to their respective gaps at angles a and ẞ, respectively. Derive expressions
for each of the angles in terms of the initial velocities and the coefficient of restitution.
m1
m2
8
m1
↑
บา
m2
ñ
В
The fallowing question is from a reeds book on applied heat i am studying. Although the answer is provided, im struggling to understand the whole answer and the formulas and the steps theyre using. Also where some ov the values such as Hg and Hf come from in part i for example. Please explain step per step in detail thanks
In an NH, refrigerator, the ammonia leaves the evaporatorand enters the cornpressor as dry saturated vapour at 2.68 bar,it leaves the compressor and enters the condenser at 8.57 bar with50" of superheat. it is condensed at constant pressure and leavesthe condenser as saturated liquid. If the rate of flow of the refrigerantthrough the circuit is 0.45 kglmin calculate
(i) the compressorpower,
(ii) the heat rejected to the condenser cooling water in kJ/s,an (iii) the refrigerating effect in kJ/s.
From tables page 12, NH,:2.68 bar, hg= 1430.58.57 bar, hf = 275.1 h supht 50" = 1597.2Mass flow of refrigerant--- - - 0.0075 kgls 60Enthalpy gain per kg of refrigerant in…
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