MAK4053E Homework-2

EGE DEMİRER 050170413

QUESTION-1)

a) Free Body Diagram of the system:

b) Firstly, moment and force balances are to be made (Theta symbol is used as angle

instead of “a”).

For the vehicle (FWD):

𝐹𝑧 = 0 ; 𝑊𝑓 + 𝑊𝑟 − 𝑊 ∗ cos(𝜃) − 𝑅𝑑𝑧 = 0

𝐹𝑥 = 0 ; 𝐹𝑓 − 𝑊 ∗ sin(𝜃) − 𝑅𝑥𝑟 − 𝑅𝑑𝑥 = 0

Moment balance equations from lecture notes are used to get weight on each Wheel:

𝑙2 ℎ 𝑠 𝑑
𝑊𝑓 = ∗ 𝑊 ∗ cos(𝜃) − ∗ 𝑊 ∗ sin(𝜃) − ∗ 𝑅𝑑𝑥 − ∗ 𝑅𝑑𝑧
𝐿 𝐿 𝐿 𝐿

𝑙1 ℎ 𝑠 𝑑+𝐿
𝑊𝑟 = ∗ 𝑊 ∗ cos(𝜃) + ∗ 𝑊 ∗ sin(𝜃) + ∗ 𝑅𝑑𝑥 + ∗ 𝑅𝑑𝑧
𝐿 𝐿 𝐿 𝐿
 𝑅𝑥𝑓 = 𝑊𝑓 ∗ 𝑓𝑟 , 𝑅𝑥𝑟 = 𝑊𝑟 ∗ 𝑓𝑟

For the trailer:

𝐹𝑧 = 0 ; 𝑊𝑡 + 𝑅𝑑𝑧 − 𝐺 ∗ cos(𝜃) = 0

𝐹𝑥 = 0 ; 𝑅𝑑𝑥 − 𝑅𝑥𝑡 − 𝐺 ∗ sin(𝜃) = 0

𝐹𝑟𝑜𝑚 ℎ𝑖𝑡𝑐ℎ 𝑝𝑜𝑖𝑛𝑡 𝑀ℎ = 0 ; 𝑊𝑡 ∗ 𝐿𝑡 + 𝑅𝑥𝑡 ∗ 𝑠 − 𝐺 ∗ cos(𝜃) ∗ 𝑙3 + 𝐺 ∗ sin(𝜃) ∗ (ℎ𝑡 − 𝑠) = 0

𝐹𝑟𝑜𝑚 𝑡𝑖𝑟𝑒 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑀𝑡 = 𝐺 ∗ cos(𝜃) ∗ 𝑙4 − 𝐺 ∗ sin(𝜃) ∗ ℎ𝑡 − 𝑅𝑑𝑧 ∗ 𝐿𝑡 + 𝑅𝑑𝑥 ∗ 𝑠

𝐺 ∗ sin(𝜃) ∗ (ℎ𝑡 − 𝑠) + 𝐺 ∗ cos(𝜃) ∗ 𝑙3

𝑊𝑡 =
𝐿𝑡 + 𝑓𝑟 ∗ 𝑠

𝑅𝑥𝑡 = 𝑊𝑡 ∗ 𝑓𝑟

𝑅𝑑𝑧 = 𝐺 ∗ cos(𝜃) − 𝑊𝑡

𝑅𝑑𝑥 = 𝐺 ∗ sin(𝜃) + 𝑊𝑡 ∗ 𝑓𝑟

𝐹𝑓
𝑓𝑓𝑟𝑜𝑛𝑡 = ≤ 𝜇𝑝
𝑊𝑓

𝐹𝑓 = 𝑊 ∗ sin(𝜃) + 𝑊𝑟 ∗ 𝑓𝑟 + 𝐺 ∗ sin(𝜃) + 𝑊𝑡 ∗ 𝑓𝑟

𝑙2 ℎ 𝑠 𝑑
𝑊𝑓 = ∗ 𝑊 ∗ cos(𝜃) − ∗ 𝑊 ∗ sin(𝜃) − ∗ 𝐺 ∗ sin(𝜃) + 𝑊𝑡 ∗ 𝑓𝑟 − ∗ 𝐺 ∗ cos(𝜃) − 𝑊𝑡
𝐿 𝐿 𝐿 𝐿
 c) Before calculation, it is important to remind that higher gradability is expected
from RWD vehicle for this question. Derivation of the rear Wheel drive can be seen in section
d. Gradibility plots are as follows:

d) For the vehicle (RWD):

𝐹𝑟
𝑓𝑟𝑒𝑎𝑟 = ≤ 𝜇𝑝
𝑊𝑟

𝐹𝑟 = 𝑊 ∗ sin(𝜃) + 𝑊𝑓 ∗ 𝑓𝑟 + 𝐺 ∗ sin(𝜃) + 𝑊𝑡 ∗ 𝑓𝑟

𝑙1 ℎ 𝑠 𝑑+𝐿
𝑊𝑟 = ∗ 𝑊 ∗ cos(𝜃) + ∗ 𝑊 ∗ sin(𝜃) + ∗ 𝐺 ∗ sin(𝜃) + 𝑊𝑡 ∗ 𝑓𝑟 + ∗ 𝐺 ∗ cos(𝜃) − 𝑊𝑡
𝐿 𝐿 𝐿 𝐿
 QUESTION-2

a) Engine torque and engine power are plotted by using Matlab software:

b) For tire size selection, weight on each tire is to be known. Information about the

vehicle is given.

𝐿 = 2810 𝑚𝑚 , 𝑙1 = 1349 𝑚𝑚 , 𝑙2 = 1461 𝑚𝑚 , ℎ = 702 𝑚𝑚

Load on each axle:

𝑊𝑟 = 9601.42𝑁 , 𝑊𝑓 = 10398.58𝑁

Load is carried by two tires per axle. Loan on a tire per axle is as follows:

𝐿𝑜𝑎𝑑 𝑜𝑛 𝑜𝑛𝑒 𝑓𝑟𝑜𝑛𝑡 𝑡𝑖𝑟𝑒 = 529.95𝑘𝑔

𝐿𝑜𝑎𝑑 𝑜𝑛 𝑜𝑛𝑒 𝑟𝑒𝑎𝑟 𝑡𝑖𝑟𝑒 = 489.05𝑘𝑔

 With this information, Fig 2.13 can be used to determine appropriate load index for the
tires. Choosing one tire size for both front and real axles is more appropriate. Load on the
front tire is higher and tire pressure of a passenger car is generally 2.2 bar.

Load index is found as 90. When Fig 2.15 is inspected for tire size selection, there is
no tire size that have load index value of 90. Therefore, smallest load size after 90 is selected,
which is 91.
 Finally, tire size is selected from Fig 2.15 as 195/65 R 15 V.

V is choosen for speed symbol since maximum speed of the car is found as 216 km/h in
section “d”.
 c) When maximum speed is calculated, it is assumed that engine gives maximum

Power (111kW from onion chart) and only aerodynamic forces are taken into account as
resistance forces.

𝑤 = 1811𝑚𝑚 , ℎ = 1429𝑚𝑚 , 𝐴 = 𝑤 ∗ ℎ = 2.59𝑚2

𝑘𝑔
𝐶𝐷 = 0.32 , 𝜌𝑎𝑖𝑟 = 1.225
𝑚3

1 2
𝐴𝑒𝑟𝑜𝑑𝑦𝑛𝑎𝑚𝑖𝑐 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐹𝑎𝑒𝑟𝑜 = ∗ 1.225 ∗ 0.32 ∗ 2.59 ∗ 𝑉𝑚𝑎𝑥
2

Force that is created from the engine is as follows:

111000
𝐹𝑎𝑐𝑐 =
𝑉𝑚𝑎𝑥

111000 1 2
= ∗ 1.225 ∗ 0.32 ∗ 2.59 ∗ 𝑉𝑚𝑎𝑥 , 𝑉𝑚𝑎𝑥 = 216 𝑘𝑚/ℎ
𝑉𝑚𝑎𝑥 2

d) Vehicle is front Wheel drive. Gradability of the vehicle is as follows:

𝑙2 𝑙1 1461 1349
∗ 𝜇 −
tan(𝜃𝐹𝑊𝐷 ) ≤ 𝐿
∗ 𝜇 𝑝 − 𝐿 ∗ 𝑓𝑟
= 2810
𝑝 2810 ∗ 0.015
ℎ 702
1 + 𝐿 ∗ (𝜇𝑝 + 𝑓𝑟 ) 1 + 2810 ∗ (0.8 + 0.015)

𝐺𝑟𝑎𝑑𝑖𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 34%
 e) It is appropriate to start from gear ratio of the axle drive. Question states that 4th

gear is direct drive and slip is neglected.

𝑟𝑑𝑦𝑛 = 0.290𝑚 (𝑓𝑟𝑜𝑚 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑 𝑡𝑖𝑟𝑒)

𝑉𝑚𝑎𝑥 = 60.25 𝑚/𝑠

𝑟𝑎𝑑
𝑛𝑒1 = 6200 𝑟𝑝𝑚 = 649.3 (𝑓𝑟𝑜𝑚 𝑜𝑛𝑖𝑜𝑛 𝑐ℎ𝑎𝑟𝑡)
𝑠

𝑛𝑒1 ∗ 𝑟 ∗ (1 − 𝑖)
𝜉𝑛 =
𝑉𝑚𝑎𝑥 ∗ 𝜉𝑎𝑥

𝑛𝑒1 ∗ 𝑟 ∗ 649.3 ∗ 0.29

𝜉4 = 1, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝜉𝑎𝑥 = = = 3.125
𝑉𝑚𝑎𝑥 60.25

𝑊 ∗ (tan(𝜃𝑚𝑎𝑥 ) + 𝑓𝑟 ) ∗ 𝑟𝑑𝑦𝑛
𝜉1 =
𝑀𝑚𝑎𝑥 ∗ 𝜉𝑎𝑥 ∗ 𝜂𝑡 ∗ √1 + tan(𝜃max )2

𝑀𝑚𝑎𝑥 : 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑒𝑛𝑔𝑖𝑛𝑒 𝑡𝑜𝑟𝑞𝑢𝑒

tan(𝜃𝑚𝑎𝑥 ) : 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑐𝑙𝑖𝑚𝑏 𝑔𝑟𝑎𝑑𝑒 (0.34 𝑓𝑟𝑜𝑚 𝑒))

𝜂𝑡 : 𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑠𝑠𝑖𝑜𝑛 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦

20000 ∗ (0.34 + 0.015) ∗ 0.29

𝜉1 = = 3.45
190 ∗ 3.125 ∗ 0.95 ∗ √1 + 0.342

𝑛𝑔−1
𝜉𝑛 3 𝜉4 3 1
𝐾𝑔 = √ =√ =√ = 0.636
𝜉1 𝜉1 3.88

Now, other gear ratios can be found:

𝜉1 = 3.45

𝜉2 = 𝜉1 ∗ 𝐾𝑔 = 3.45 ∗ 0.636 = 2.194

𝜉3 = 𝜉2 ∗ 𝐾𝑔 = 2.194 ∗ 0.636 = 1.395

𝜉4 = 1
f) Gear ratio of the 5th gear is selected as follows:

𝑛𝑒 ∗ 𝑟 ∗ (1 − 𝑖)
𝑉=
𝜉0

𝑘𝑚
𝑉 = 110 = 30.55 𝑚/𝑠
ℎ

𝑟𝑎𝑑
𝑛𝑒 = 2600 𝑟𝑝𝑚 = 272.27 (𝑠𝑝𝑒𝑒𝑑 𝑎𝑡 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛)
𝑠

𝜉0 : 𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑅𝑎𝑡𝑖𝑜

𝑛𝑒 ∗ 𝑟
𝜉0 = = 2.58
𝑉

𝜉0
𝜉5 = = 0.827
𝜉𝑎𝑥