Engineering Mechanics of Solids by Popov Chapter 7: Shear Stresses in Beams and Related Problems

1. Chapter 7
Shear Stresses in Beams and Related Problems
Mechanics of Solids

2. Part A- Shear Stresses in Beams
• If a shear and bending moment are present at one section through a
beam, a different bending moment will exist at an adjoining section,
although the shear may remain constant.
𝑑𝑀 = 𝑉 𝑑𝑥
• Consider the shear and bending moment diagrams shown in Fig. 1
• The change in the bending moment in a distance 𝑑𝑥 is P 𝑑𝑥.
• Shear stresses on mutually perpendicular planes of an infinitesimal
element are considered equal.

3. Fig. 1: Shear and bending moment diagrams for the loading shown

4. Fig. 2: Shear flow model of an I-beam. (a) beam segment with bending stresses simulated by blocks.
(b) Shear force transmitting through a dowel. (c) for determining the force on a dowel only a change in moment
Is needed. (d) The longitudinal shear force divided by the area of the imaginary cut yields shear stress.
(e) Horizontal cut below the flange for determining the shear stress. (f) Vertical cut through the flange for
determining the shear stress.

5. Shear Flow
• Consider an elastic beam made from several continuous longitudinal
planks whose cross section is shown in Fig. 3(a).
Fig. 3: Elements for deriving shear
flow in a beam

6. • The element shown in Fig. 3(b) is subjected to a bending moment +𝑀𝐴
at end 𝐴 and to +𝑀 𝐵 at end 𝐵.
• Denoting the total force acting normal to the area 𝑓𝑔ℎ𝑗 by 𝐹𝐵,
• 𝑦 is the distance from the neutral axis to the centroid of 𝐴 𝑓𝑔ℎ𝑗.
• Consider end 𝐴 of the element in Fig. 3(c). The total force acting
normal to the area 𝑎𝑏𝑑𝑒,

7. • Since for prismatic beams, an area such as 𝑓𝑔ℎ𝑗 is equal to the area
𝑎𝑏𝑑𝑒. If the moments at 𝐴 and 𝐵 are equal, 𝐹𝐴 = 𝐹𝐵.
• If, 𝑀𝐴 ≠ 𝑀 𝐵, equilibrium of the horizontal forces may be attained only
by developing a horizontal resisting force 𝑅 in the bolt.
• If 𝑀 𝐵 > 𝑀𝐴, then 𝐹𝐵 > 𝐹𝐴 , and 𝐹𝐴 + 𝑅 = 𝐹𝐵 .
• An expression for the differential longitudinal push (or pull) 𝑑𝐹:
• The quantity 𝑑𝐹 𝑑𝑥 is designated by 𝑞 and referred to as the shear
flow.

8. • The total shear force at the section investigated is represented by 𝑉,
and the integral of 𝑦 𝑑𝐴 for determining 𝑄 extends only over the
cross-sectional area of the beam to one side of this area at which 𝑞 is
investigated.
• Illustrations of the manner of determining 𝑄 are in Fig. 4.

9. Fig. 4: Procedure for determining 𝑄

10. The Shear-Stress Formula for Beams
• A side view of an element is shown in Fig. 5(a), where the imaginary
longitudinal cut is made at a distance 𝑦1 from the neutral axis. The
cross-sectional area of the beam is shown in Fig. 5(c).
• Because of different bending moments acting on a section 𝐴 and 𝐵.
More push or pull is developed on one side of the partial area 𝑓𝑔ℎ𝑗
than on the other, this longitudinal force in a distance 𝑑𝑥 is

11. Fig. 5: Derivation of shear stresses in a beam

12. • The longitudinal shear stress and the shear stress in the plane of the
vertical section at the longitudinal cut.
• As, 𝑑𝑀 𝑑𝑥 = 𝑉 and 𝑞 = 𝑉𝑄 𝐼.
• Here, 𝑉 is the total shear force at the section, I is the moment of inertia
of the whole cross sectional area about the neutral axis.

13. • 𝑄 is the statical moment around the neutral axis of the partial area of the
cross section to one side of the imaginary longitudinal cut, and 𝑦 is the
distance from the neutral axis to the centroid of the partial area 𝐴 𝑓𝑔ℎ𝑗. 𝑡
is the width of the imaginary longitudinal cut, usually equal to the
thickness or width of the member.
• The proper sectioning of some cross-sectional areas of beams is shown in
Figs. 6(a), (b),(d), and (e). The use of inclined plane should be avoided
unless the section is made across a small thickness.
Fig. 6: Sectioning for partial areas of cross sections for computing shear stresses

14. Warpage of Plane Sections Due to Shear
• The maximum shear stress, hence, maximum shear strain, occurs at
𝑦 = 0. and no shear strain takes place at 𝑦 = ± ℎ 2.
• This behavior warps the initially plane sections through a beam, as
shown in Fig. 7.
• However, warpage of the sections is important only for very short
members and is negligibly small for small members. This can be
substantiated by the two-dimensional finite-element studies.
Fig. 7: Shear distortions in a beam

15. Limitations of the Shear-Stress Formula
• The material is assumed to be elastic with the same elastic modulus in
tension as in compression. The theory developed applies only to
straight beams.
• Consider a section through the I beam. The vertical shear stress is zero
for element 𝑎. No shear stresses exist on the top plane of the beam.
Fig. 8: Boundary conditions are not
satisfied at the levels 2-2

16. • The top surface of the beam is a free surface. Hence, the conditions at
the boundary are satisfied.
• While, In determining the shear stresses for the I beam at levels 2-2 is
contradictory.
• The shear stresses were found to be 570 psi for the elements such as
𝑏 or 𝑐 shown in Fig. 8. this requires matching horizontal shear stresses
on the inner surfaces of the flanges.
• However, the later surfaces must be free of shear stresses as they are
free boundaries of the beam. Hence, it cannot be resolved by the
conventional methods.
• The more advanced techniques of the mathematical theory of
elasticity or three-dimensional analysis must be used to obtain an
accurate solution.

17. Shear Stresses in Beam Flanges
• For the beam shown in Fig. 9(b), positive bending moments increase
toward the reader, larger normal forces act on the near section.
• For the elements shown, 𝜏𝑡 𝑑𝑥 or 𝑞 𝑑𝑥 must aid the smaller force
acting on the partial area of the cross section.
• The magnitude of the shear stresses varies for the different vertical
cuts. If cut 𝑐−𝑐 in Fig. 9(a) is at the edge of the beam, the hatched
area of the beam’s cross section is zero.
• If the thickness of the flange is constant, and cut 𝑐−𝑐 is made
progressively closer to the web, this area increases from zero at a
linear rate.

18. Fig. 9: Shear forces in the flanges of an I beam act perpendicularly to the axis of symmetry

19. • As 𝑦 remains constant for any such area, 𝑄 also increases linearly from
zero toward the web, shear flow 𝑞 𝑐 = 𝑉𝑄 𝐼 follows the same
variation.
• If the thickness of the flange remains the same, the shear stress 𝜏 𝑐 =
𝑉𝑄 𝐼𝑡 varies similarly.
• As maybe seen from Fig. 9(b), these quantities in the plane of the
cross section act in opposite directions on the two sides.
• The variation of these shear stresses or shear flows is represented in
Fig. 9(c), it is assumed that the web has zero thickness.
• The shear stresses shown in Fig. 9(c), when integrated over the area
on which they act, are equivalent to a force.

20. • The magnitude of the horizontal force 𝐹1 for one-half of the flange,
Fig. 9(d) is equal to the average shear stress multiplied by one-half of
the whole area of the flange, i.e.,
𝐹1 =
𝜏 𝑐−𝑚𝑎𝑥
2
𝑏𝑡
2
or 𝐹1 =
𝑞 𝑐−𝑚𝑎𝑥
2
𝑏
2
• If an I beam transmits a vertical shear, these horizontal forces act in
the upper and lower flanges.
• To determine the shear flow at the juncture of the flange and the
web, cut a-a in Fig. 9(a), the whole area of the flange times 𝑦 must be
used in computing the value of 𝑄.
• Since, in finding 𝑞 𝑐−𝑚𝑎𝑥, one-half the flange area times the same 𝑦
has already been used, the sum of the two horizontal shear flows
coming in from opposite sides gives the vertical shear flow at cup a-a.

21. • As the resistance to the vertical shear 𝑉 in thin walled I beams is
developed mainly in the web, Fig. 9(d).
• The vertical shear flow splits upon reaching the lower flange. This is
represented in Fig. 9(d) by the two forces 𝐹1 that are the result of the
horizontal shear flows in the flanges.
• The shear forces that act at a section of an I beam are shown in Fig.
9(d).
• For equilibrium, the applied vertical forces must act through the
centroid of the cross-sectional area to be coincident with 𝑉. If the
forces are so applied, no torsion will occur.
• This is true for all sections having cross-sectional areas with an axis of
symmetry.

22. Shear Center
• Consider a beam having the cross section of a channel, Fig. 10(a).
• The walls of this channel are assumed to be sufficiently thin that the
computations may be based on center line dimension.
Fig. 10: Deriving location of shear center for a channel

23. • Assuming that this channel resists a vertical shear, the bending
moments will vary from one section through the beam to another.
• By taking an arbitrary vertical cut as 𝑐−𝑐 in Fig. 10(a), 𝑞 and 𝜏 maybe
found. The variation of 𝑞 and 𝜏 is parabolic along the web, Fig. 10(b).
• The average shear stress 𝜏 𝑎 2 multiplied by the areas of flange gives a
force 𝐹1 = 𝜏 𝑎 2 𝑏𝑡, and the sum of the vertical shear stresses over
the area of the web is the shear 𝑉 = −ℎ 2
+ℎ 2
𝜏𝑡 𝑑𝑦.
• These shear forces acting in the plane of the cross section is shown in
fig. 10(c) and indicate that a force 𝑉 and a couple 𝐹1ℎ are developed at
the section through the channel.
• To prevent twisting and thus maintaining the applicability of the
initially assumed bending-stress distribution, the externally force must
be applied to balance the internal couple 𝐹1ℎ.

24. • Consider the segment of a cantilever beam of negligible weight, as
shown in Fig. 10(d), to which a vertical force 𝑃 is applied parallel to
the web at a distance 𝑒 from the web’s center line.
• To maintain this applied force in equilibrium, an equal and opposite
shear force 𝑉 must be developed in the web.
• To cause no twisting of the channel, couple 𝑃𝑒 must equal couple 𝐹1ℎ.
• At the same section through the channel, bending moment 𝑃𝐿 is
resisted by the usual flexural stresses.
• Distance 𝑒, locating the plane in which force 𝑃 must be applied so as
to cause no twist in the channel.
𝑒 =
𝐹1ℎ
𝑃
=
1 2 𝜏 𝑎 𝑏𝑡ℎ
𝑃
=
𝑏𝑡ℎ
2𝑃
𝑉𝑄
𝐼𝑡
=
𝑏𝑡ℎ
2𝑃
𝑉𝑏𝑡 ℎ 2
𝐼𝑡
=
𝑏2ℎ2 𝑡
4𝐼

25. • Distance 𝑒 is a property of a section and is measured outward from the
center of the web to the applied force.
• We can locate the plane in which the horizontal forces must be applied
so as to cause no twist in the channel. For the channel considered, by
virtue of symmetry, This plane coincides with the neutral plane of the
former case.
• The intersection of these two mutually perpendicular planes with the
plane of the cross section locates a point that is called the shear
center. It is designated by the letter 𝑆 in Fig. 10(c).
• The shear center for any cross section lies on a longitudinal line parallel
to the axis of the beam. Any transverse force applied through the
shear center causes no torsion of the beam.

26. • When a member of any cross-sectional area is twisted, the twist takes
place around the shear center, which is fixed. For this reason, the
shear center is also called the center of twist.
• For cross-sectional areas having one axis of symmetry, the shear
center always lie on the axis of symmetry an coincides with the
centroid of the cross-sectional area.
• The shear force for a symmetrical angle is located at the intersection
of the center line of its legs, as shown in Figs. 11(a) and (b).
• The shear flow at every section, as 𝑐 − 𝑐, is directed along the center
line of a leg. These shear flows yield two identical forces, 𝐹1 in the
legs.
• The vertical components of the forces equal the vertical shear applied
through 𝑆.

27. • An analogous situation for any angle or T section, as shown in Figs.
12(a) and (b).
Fig. 11: Shear stress for an equal Fig. 12: Shear center for the sections
leg angle is at 𝑆 shown is at 𝑆
• The location of the shear center for various members is particularly
important in aircraft applications.
• If force 𝑃 is applied outside shear center 𝑆, as shown in Fig. 13, two
equal but opposite forces 𝑃 can be introduced at 𝑆.

28. Fig. 13: Torsion-bending of a channel
• In addition to the stresses caused by 𝑃 applied at 𝑆, the torsional
stresses caused by the torque equal to 𝑃𝑑 must be considered.

29. Superposition of Shear Stresses
Combined Direct and Torsional Shear Stresses
• The direct shear stresses are determined using the procedures of Part
A, the shear stresses caused by torques susceptible to the methods of
analysis of torsion is used. This must be determined for the same
elementary area.
• Dividing the vector sum of these forces by the initial area, combined
shear stress is obtained.
• Generally, the maximum torsional shear stresses as well as the
maximum direct shear stress for beams occur at the boundaries of
cross section and are collinear, so algebraic sum can be done.

30. Stresses in Closely Coiled Helical Springs
• Any one coil of such a spring is assumed to lie in a plane that is nearly
perpendicular to the axis of the spring, adjoining coils must be close
together.
• A section taken perpendicular to the axis of the spring’s rod becomes
nearly vertical.
• For an equilibrium, shear force 𝑉 = 𝐹 and a torque 𝑇 = 𝐹 𝑟 are
required at any section through the rod, Fig. 14(b). 𝑟 is the distance
from the axis of the spring to the centroid of the rod’s cross-sectional
area.
• A shear acts at every section of the rod, yet no bending moment nor a
change in it appears to occur, because the rod is curved.

31. Fig. 14: Closely coiled helical spring Fig. 15
• Such an element of the rod viewed from the top is shown in Fig. 15.
• At both ends of the element, the torques are equal to 𝐹 𝑟 acting in the
direction shown. The component of these vectors toward the axis of
the spring 𝑂, resolved at the point of intersection of the vectors,

32. 2𝐹 𝑟 𝑑Φ 2 = 𝐹 𝑟 𝑑Φ
• This vector component opposes the couple developed by the vertical
shears 𝑉 = 𝐹, which are 𝑟 𝑑Φ apart.
• The nominal direct shear stress for any point on the cross section is 𝜏 =
𝐹 𝐴. Superposition of this nominal direct and the torsional shear stress
at 𝐸 gives the maximum combined shear stress.
• Thus, since 𝑇 = 𝐹 𝑟, 𝑑 = 2𝑐, and 𝐽 = 𝜋𝑑4 32
𝜏 𝑚𝑎𝑥 =
𝐹
𝐴
+
𝑇𝑐
𝐽
=
𝑇𝑐
𝐽
𝐹𝐽
𝐴𝑇𝑐
+ 1 =
16𝐹 𝑟
𝜋𝑑3
𝑑
4 𝑟
+ 1
• As 𝑑 becomes numerically comparable to 𝑟, the length of the inside
fibers of the coil differs greatly from the length of the outside fib, and
the assumptions of strain used in the torsion formula are not applicable.

33. • The spring problem has been solved exactly by the methods of the
mathematical theory of elasticity, depending on a parameter 𝑚 =
2 𝑟 𝑑, Spring index.
𝜏 𝑚𝑎𝑥 = 𝐾
16𝐹 𝑟
𝜋𝑑3
Fig. 16: Stress-concentration factors for
helical round-wire compression
or tension springs

34. • 𝐾 can be represented as a stress-concentration factor for closely
coiled helical springs made from circular rods.
• For heavy spring, the spring index is small; hence, the stress-
concentration factor 𝐾 becomes important.
• For good quality spring steel, working shear stresses range from 200
to 700 MPa (30 to 100 ksi).

35. Deflection of Closely Coiled Helical Spring
• The diameter of the wire is assumed small in comparison with the
radius of the coil.
• The effect of direct shear on the deflection of the spring is ignored.
• Consider a helical spring such as shown in Fig. 17. A typical element
𝐴𝐵 of this spring is subjected throughout its length to a torque 𝑇 =
𝐹 𝑟.
• With sufficient accuracy, the amount of rotation,
𝑑Φ = 𝑇 𝑑𝑥 𝐽𝐺
• From figure,
𝐶𝐷 = 𝐵𝐶 𝑑Φ

36. Fig. 17: Deriving deflection for a helical spring
• By finding the vertical increment 𝐸𝐷 of the deflection of force 𝐹 due to
an element of spring 𝐴𝐵 and summing such increments for all elements
of the spring, the deflection of the whole spring is obtained.
• From similar triangles 𝐶𝐷𝐸 and 𝐶𝐵𝐻,
𝐸𝐷
𝐶𝐷
=
𝐻𝐵
𝐵𝐶

37. • However, 𝐶𝐷 = 𝐵𝐶 𝑑Φ, 𝐻𝐵 = 𝑟, and 𝐸𝐷 may be denoted by 𝑑Δ, as it
represents an infinitesimal vertical deflection of the spring due to
rotation of an element 𝐴𝐵. Thus, 𝑑Δ = 𝑟 𝑑Φ and
Δ = 𝑑Δ = 𝑟 𝑑Φ = 0
𝐿
𝑟
𝑇 𝑑𝑥
𝐽𝐺
=
𝑇𝐿 𝑟
𝐽𝐺
• However, 𝑇 = 𝐹 𝑟, and for a closely coiled spring, the length 𝐿 of the
wire may be taken with sufficient accuracy as 2𝜋 𝑟𝑁, where 𝑁 is the
number of live or active coils of the spring.
Δ =
2𝜋𝐹 𝑟3 𝑁
𝐽𝐺
=
64𝐹 𝑟3 𝑁
𝐺𝑑4
• The spring constant,
𝐾 =
𝐹
Δ
=
𝐺𝑑4
64 𝑟3 𝑁
𝑁
𝑚
or
𝑙𝑏
𝑖𝑛