1. Problem 5-b a
24 kN
30 kN
The beam AB supports two concentrated loads and rests on soil which exerts a linearly distributed upward load as shown. Determine (a) the distance a for which wA = 20 kN/m, (b) the corresponding value wB.
0.3 m A B wA wB
1.8 m

2. Solving Problems on Your Own a
Problem 5-b
Solving Problems on Your Own a
24 kN
30 kN
0.3 m
The beam AB supports two concentrated loads and rests on soil which exerts a linearly distributed upward load as shown. Determine (a) the distance a for which wA = 20 kN/m, (b) the corresponding value wB. A B wA wB
1.8 m
1. Replace the distributed load by a single equivalent force.
The magnitude of this force is equal to the area under the
distributed load curve and its line of action passes through
the centroid of the area.
2. When possible, complex distributed loads should be
divided into common shape areas.

3. RII = (1.8 m)(wB kN/m) = 0.9 wB kN
Problem 5-b Solution a
24 kN
30 kN
0.3 m
Replace the distributed
load by a pair of
equivalent forces. C A B
20 kN/m wB
0.6 m
0.6 m RI
RII 1 2
We have
RI = (1.8 m)(20 kN/m) = 18 kN 1 2
RII = (1.8 m)(wB kN/m) = 0.9 wB kN

4. SFy = 0: -24 kN + 18 kN + (0.9 wB) kN - 30 kN= 0 or wB = 40 kN/m
Problem 5-b Solution a
24 kN
30 kN
0.3 m C A B wB
0.6 m
0.6 m
RI = 18 kN
RII = 0.9 wB kN
(a) +
SMC = 0: (1.2 - a)m x 24 kN m x 18 kN
- 0.3m x 30 kN = 0
or a = m
(b) +
SFy = 0: -24 kN + 18 kN + (0.9 wB) kN - 30 kN= 0
or wB = 40 kN/m