Determine the absolute maximum bending moment in the simply supported beam due to the wheel loads of the HS20-44 truck shown in Fig. 9.11(a). 4k16k16 k- 14 ft-f–14 ft-50 ft(a)PR = 36 k4 k16 k16 k23E14 ft–'9.33 ft4.67 ft(b)£ 2.33 ft14--22.67 ft-PR = 36 k4k16 k16 k3A,18.67 ft 2.33 ft,–14 t--11.67 t--13.33 tt-25 ft25 ft(c)FIG. 9.11

Consider all the free-body diagrams of the beam.

4k 16k 16 k - 14 ft-f–14 ft -50 ft (a) PR = 36 k 4 k 16 k 16 k 2 3 E14 ft– '9.33 ft 4.67 ft (b) £ 2.33 ft 14- -22.67 ft- PR = 36 k 4k 16 k 16 k 3 A,18.67 ft 2.33 ft, –14 t--11.67 t--13.33 tt- 25 ft 25 ft (c) FIG. 9.11

4k 16k 16 k - 14 ft-f–14 ft -50 ft (a) PR = 36 k 4 k 16 k 16 k 2 3 E14 ft– '9.33 ft 4.67 ft (b) £ 2.33 ft 14- -22.67 ft- PR = 36 k 4k 16 k 16 k 3 A,18.67 ft 2.33 ft, –14 t--11.67 t--13.33 tt- 25 ft 25 ft (c) FIG. 9.11

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

Determine the absolute maximum bending moment in the simply supported beam due to the wheel loads of the HS20-44 truck shown in Fig. 9.11(a).

4k
16k
16 k
- 14 ft-f–14 ft
-50 ft
(a)
PR = 36 k
4 k
16 k
16 k
2
3
E14 ft–
'9.33 ft
4.67 ft
(b)
£ 2.33 ft
14-
-22.67 ft-
PR = 36 k
4k
16 k
16 k
3
A,18.67 ft 2.33 ft,
–14 t--11.67 t--13.33 tt-
25 ft
25 ft
(c)
FIG. 9.11

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