SME
Mechanics of Materials (Solid Mechanics I) · Problem type
Bar fixed at both ends
Use this for “A bar between two walls carries 30 kN at 0.4 m from the left wall and 0.6 m from the right. Find each reaction.”
Diagram
The walls share P: RA + RB = 18 kN + 12 kN = 30 kN.
The walls hold the length: what the T part stretches, the C part squeezes back.
This is an example. Type your own values to start a new problem.
L (m)
=
a (m)
+
b (m)
RA (kN)
=
P (kN)
×
b (m)
L (m)
RB (kN)
=
P (kN)
×
a (m)
L (m)
Formulas
L
=
a
+
b
1 = 0.4 + 0.6
RA
=
P
×
b
L
18
=
30
×
0.6
1
RB
=
P
×
a
L
12
=
30
×
0.4
1
Assumptions
•
Both walls are rigid, and the bar fits between them with no force before P goes on.
•
One material and one area, so each part’s stiffness goes as 1 ÷ its length.
•
P pulls toward B: the part beside A stretches in tension, the part beside B is squeezed.
Step-by-step
WE KNOW
P = 30 kN, a = 0.4 m, b = 0.6 m
FIND
length (L), reaction at A (RA), reaction at B (RB)
Step 1 · Find length (L)
Use L = a + b
The
load
splits
the
bar
into
two
parts,
a
and
b
.
L
=
0.4
+
0.6
L
=
1
m
Step 2 · Find reaction at A (RA)
Use RA = P × b ÷ L
The
bar
can’t
change
length,
so
RA
a
=
RB
b
;
with
RA
+
RB
=
P
,
that
leaves
RA
=
Pb
L
.
RA
=
30
×
0.6
1
RA = 18 kN
Step 3 · Find reaction at B (RB)
Use RB = P × a ÷ L
The same balance from the other side: the farther wall A is, the more wall B takes.
RB
=
30
×
0.4
1
RB = 12 kN
Check
Put the numbers back in. Both sides should match.
1 = 0.4 + 0.6 ✓
18
=
30
×
0.6
1
✓
12
=
30
×
0.4
1
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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