Determine the reactions at the wheels at D, E, and F, and the force on the hitch H that is mounted on the car, for the extreme positions A and B of the load.. Forces and Moments in Beams
Trang 11-14 Section 1
Example 4
A load of 7 kN may be placed anywhere within A and B in the trailer of negligible weight Determine the reactions at the wheels at D, E, and F, and the force on the hitch H that is mounted on the car, for the extreme positions A and B of the load The mass of the car is 1500 kg, and its weight is acting at
C (see Figure 1.2.16)
Solution The scalar method is best here.
Forces and Moments in Beams
Beams are common structural members whose main function is to resist bending The geometric changes and safety aspects of beams are analyzed by first assuming that they are rigid The preceding sections enable one to determine (1) the external (supporting) reactions acting on a statically determinate beam, and (2) the internal forces and moments at any cross section in a beam
FIGURE 1.2.16 Analysis of a car with trailer.
Put the load at position A first
For the trailer alone, with y as the vertical axis
∑M F = 7(1) – H y (3) = 0, H y = 2.33 kN
On the car
H y = 2.33 kN ↓ Ans.
∑F y = 2.33 – 7 + F y = 0, F y = 4.67 kN ↑ Ans.
For the car alone
∑M E = –2.33(1.2) – D y(4) + 14.72(1.8) = 0
D y = 5.93 kN ↑ Ans.
∑F y = 5.93 + E y – 14.72 – 2.33 = 0
E y = 11.12 kN ↑ Ans.
Put the load at position B next
For the trailer alone
∑M F = 0.8(7) – H y (3) = 0, H y = –1.87 kN
On the car
H y = 1.87 kN ↓ Ans.
∑F y = –1.87 – 7 + E y = 0
E y = 8.87 kN ↑ Ans.
For the car alone
∑M E = –(1.87)(1.2) – D y(4) + 14.72(1.8) = 0
D y = 7.19 kN ↑ Ans.
∑F y = 7.19 + E y – 14.72 – (–1.87) = 0
E y = 5.66 kN ↑ Ans.
FO =100 N j
M O
∑ =0
MO+MC+(rOA×WA)+(rOB×WB)=0
MO+(100 20 50 0 6 0 3 0 4 60 0 2 40 0i− j+ k) N m m N ⋅ + −( i j+ k) × −( j)+ −(i j) m N × −( j)=
MO+100 20 50 36 24 40 0 N m N m N m N m N m N m ⋅ i− ⋅ j+ ⋅ k− ⋅ k+ ⋅ i− ⋅ k=
MO = − + +( 124 20 26i j k) N m⋅