This document discusses the relationships between loads, shear forces, and bending moments in beams. It states that shear forces and bending moments are internal stress resultants that can be calculated from equations of equilibrium. Distributed loads cause shear forces to vary linearly or quadratically along the beam and bending moments to vary quadratically or cubically. Concentrated loads cause an abrupt change in shear force but no change in bending moment. Couples cause no change in shear force but an abrupt change in bending moment.

2. Free body diagram of the beam
𝜮𝑭 𝒙 = 𝟎
⇒ 𝑯 𝑨= 𝟎
𝜮𝑭 𝒚 = 𝟎
𝐑 𝑨 − 𝟒 − 𝟏. 𝟓 ∗ 𝟐 = 𝟎
𝐑 𝑨 = 𝟕 𝐤𝐍
𝛴𝑀 𝐹𝐸 = 0
- 𝑀𝐴 − 4*(1)- 1.5 *2 * (2 +1 ) =0
 𝑀𝐴 =- ( 4+ 1.5 *2 * 3)
 𝑀𝐴 = - 13 kN-m
Q4. Calculate the support reactions for the beam as shown in the figure

3. Free body diagram of the beam
𝜮𝑭 𝒙 = 𝟎
⇒ 𝑯 𝑨= 𝟎
𝜮𝑭 𝒚 = 𝟎
𝐑 𝑨 + 𝐑 𝑩 − 𝟑𝟓𝟎 −
1
2
∗ 1 ∗ 1750 = 𝟎
𝐑 𝑨 + 𝐑 𝑩 = 𝟑𝟓𝟎 + 𝟖𝟕𝟓 = 𝟏𝟐𝟐𝟓 𝐤𝐍
𝛴𝑀𝐴 = 0
−350*(1)- (
1
2
∗ 1 ∗ 1750) * (1+
2
3
) - 𝑅 𝐵 ∗ 2=0
 𝑅 𝐵 ∗ 2 = ( 350 + 875 *
5
3
)
 𝑅 𝐵 = 904.2 kN
Magnitude of UVL= Area of loading Diag
=
1
2
∗ 𝐿 ∗ 𝑞0
=
1
2
∗ 1 ∗ 1750
= 875 N
→ (1)
→ (2)
Substitute 2 in (1)
𝐑 𝑨 = 𝟏𝟐𝟐𝟓 − 𝐑 𝑩
=>𝐑 𝑨 = 𝟏𝟐𝟐𝟓 − 𝟗𝟎𝟒. 𝟐
=> 𝐑 𝑨 = 320.8 kN
Q5. Calculate the support reactions for the beam as shown in the figure

4.  Finding the reactions is usually the first step in the analysis of a beam.
 Once the reactions are known, the shear forces and bending moments can be found. If a beam is
supported in a statically determinate manner, all reactions can be found from free-body diagrams and
equations of equilibrium.
 Internal stress resultants: Shear forces and bending moments are the resultants of stresses distributed
over the cross section. Therefore, these quantities are known collectively as stress resultants. The stress
resultants in statically determinate beams can be calculated from equations of equilibrium.
 When designing a beam, we usually need to know how the shear forces and bending moments vary
throughout the length of the beam. Of special importance are the maximum and minimum values of
these quantities. Information of this kind is usually provided by graphs in which the shear force and
bending moment are plotted as ordinates and the distance x along the axis of the beam is plotted as
the abscissa. Such graphs are called shear-force and bending-moment diagrams.

6. Shear forces and bending moments, like axial forces in bars and internal torques in
shafts, are the resultants of stresses distributed over the cross section. Therefore,
these quantities are known collectively as stress resultants. The stress resultants in
statically determinate beams can be calculated from equations of equilibrium.

7. When a beam is loaded by forces or
couples, stresses and strains are created
throughout the interior of the beam. To
determine these stresses and strains, we
first must find the internal forces and
internal couples that act on cross sections
of the beam.

8. Relation between loads, shear force and bending moments:
The relationships between loads, shear forces, and bending moments in beams are quite useful when investigating the
shear forces and bending moments throughout the entire length of a beam, and they are especially helpful when
constructing shear-force and bending-moment diagrams.
Element of a beam used in deriving the relationships between loads, shear forces, and bending moments

9. q
Distributed load:
From this equation we see that the rate of change of the shear force
at any point on the axis of the beam is equal to the negative of the
intensity of the distributed load at that same point.

10. Type of distributed
Load
Shear force
V= − 𝒒 ⅆ𝒙
Variation of Shear force
No distributed load
on the segment of
beam (q=0)
V= − 𝟎 . ⅆ𝒙 = C The shear force remains constant in
that part of the beam
if the distributed load
is uniform along part
of the beam
(q= C)
V= − 𝐂 . ⅆ𝒙 = − C. 𝑥 The shear force varies linearly in that
part of the beam
If linearly varying
load acting on
segment of beam
(q=
𝒒 𝟎 𝒙
𝑳
)
V= −
𝒒 𝟎 𝒙
𝑳
. ⅆ𝒙 = −
𝒒 𝟎 𝒙 𝟐
𝟐𝑳
The shear force varies quadratic in
that part of the beam
𝑑𝑣 = −𝑞 ⅆ𝑥
V= − 𝒒 ⅆ𝒙
𝑞
𝑥
=
𝑞0
𝐿
𝒒 =
𝒒 𝟎 𝒙
𝑳

11. Distributed load:
Discarding products of differentials (because they are negligible compared to the other terms), we
obtain the following relationship:
This equation shows that the rate of change of the bending moment at any point
on the axis of a beam is equal to the shear force at that same point.

12. Type of distributed Load Shear force
V= − 𝒒 ⅆ𝒙
Variation of Shear
force
Bending Moment
M= 𝐕 ⅆ𝒙
Variation of
bending moment
No distributed load on
the segment of beam
(q=0)
𝑽= C remains constant M= 𝐂 . ⅆ𝒙 = C. 𝑥 The Bending
moment varies
linearly
if the distributed load is
uniform along part of the
beam
(q= q)
𝑽= − q. 𝑥 shear force varies
linearly
M= − q 𝑥 ⅆ𝒙 = −
𝒒𝒙 𝟐
𝟐
The Bending
moment varies
quadratically
If linearly varying load
acting on segment of
beam
(q=
𝒒 𝟎 𝒙
𝑳
)
𝑽= −
𝒒 𝟎 𝒙 𝟐
𝟐𝑳
The shear force
varies quadratic in
that part of the
beam
M= −
𝒒 𝟎 𝒙 𝟐
𝟐𝑳
ⅆ𝒙 = −
𝒒 𝟎 𝒙 𝟑
𝟔𝑳
The Bending
moment varies
cubically

13. Concentrated Loads:
Since the length dx of the element is infinitesimally
small, we see from this equation that the
increment M1 in the bending moment is also
infinitesimally small. Thus, the bending moment
does not change as we pass through the point of
application of a concentrated load.
This result means that an abrupt change in
the shear force occurs at any point where a
concentrated load acts. As we pass from
left to right through the point of load
application, the shear force decreases by
an amount equal to the magnitude of the
downward load P.

14. Load in the form of couple:
From equilibrium of the element in the vertical direction
we obtain V1= 0, which shows that the shear force does
not change at the point of application of a couple.
Equilibrium of moments about the left-hand side of the element gives
This equation shows that the bending moment decreases by M0 as we move from left to right
through the point of load application. Thus, the bending moment changes abruptly at the
point of application of a couple.