Problem 4.1
Applied Loads & Reactions
Loading up that
pickup truck!
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slides.
A 1400-kg pickup truck is
loaded with two crates, each
having a mass of 350 kg.
Determine the reactions at
each of the two rear wheels
and front wheels.
Consider just the pickup truck.
Its weight is concentrated at
its center of gravity.
A
B
1.8 m
1.2 m
0.75 m
This weight is distributed over
the two axles, at RA and RB.
A
RA
B
1.8 m
1.2 m
RB
0.75 m
SUGGESTION:
Since the loads are given in
kilograms with even numbers, it
might be best to calculate with
kg and convert to Newtons at the
end.
Analyze as though a simple beam.
F = 0 = -1400 kg + RA + RB
MB = 1400(1.2 m) - RA(3.0 m)
 RA = 560 kg x 9.81m/sec2
and RB = 840 kg x 9.81m/sec2
A
RA
B
1.8 m
1.2 m
RB
0.75 m
The reactions at the wheels are:
RA = 5.49 kN
RB = 8.24 kN
A
RA
B
1.8 m
1.2 m
RB
0.75 m
Now let’s add some crates to the
bed of the truck and determine
the reactions at the wheels.
A
RA
B
1.8 m
1.2 m
RB
0.75 m
Each crate has a mass of 350 kg
and is positioned as shown.
1.7 m
C
2.8 m
D
A
RA
B
1.8 m
1.2 m
RB
0.75 m
Repeat the same calculations with
the added loads as shown.
1.7 m
C
2.8 m
D
A
RA
B
1.8 m
1.2 m
RB
0.75 m
F= 0 = -350 -350 -1400 + RA + RB
1.7 m
C
2.8 m
D
350 kg
350 kg
1400 kg
A
RA
B
1.8 m
1.2 m
RB
0.75 m
MB = 1400(1.2 m) - RA(3.0 m)
+ 350(2.05 m) + 350(3.75 m)
1.7 m
C
2.8 m
D
350 kg
2.05 m
350 kg
1400 kg
A
RA
B
1.8 m
1.2 m
RB
0.75 m
MB = 1400(1.2 m) - RA(3.0 m)
+ 350(2.05 m) + 350(3.75 m)
1,6800 + 717.5 +1,312.5 = RA(3.0 m)
3,7100 kg-m = RA(3.0 m)
3,7100 kg-m = RA
(3.0 m)
RA = 1,236.67 kg x 9.81 m/sec2
RA = 12.13 kN
F= 0 = -350 -350 -1400 + RA + RB
RB = 350 + 350 + 1400 - RA
RB = 2100 kg - 1236.67 kg
RB = 863.33 kg x 9.81 m/sec2
RB = 8.47 kN
350 kg
350 kg
1400 kg
A
RA = 12.13 kN
B
RB = 8.47 kN
Final results are shown.
Each rear wheel = 6.07 kN
Each front wheel = 4.24 kN
350 kg
350 kg
1400 kg
A
RA = 12.13 kN
B
RB = 8.47 kN