Mechanics of Deformable Bodies: BASAEN, RV
Lecture 1.2 Simple Stresses τ=
𝑃𝑃
=
𝑃𝑃
𝐴𝐴𝐴𝐴 𝜋𝜋𝜋𝜋𝜋𝜋
where
𝐴𝐴𝑆𝑆 = circumference of hole multiplied by the
SHEAR STRESS – a classification of stress in which plate thickness
resisting area is parallel to the direction of the applied load.
(As =𝜋𝜋𝜋𝜋𝜋𝜋)
Types of Shear Stress
1. Direct Shear Stress 2. Indirect Shear or Induced Shear Stress
a. Single Shear
- If the stress distribution is uniform or constant 𝑃𝑃
then,
glued joint of area, As
𝑃𝑃 𝑃𝑃
cross sectional
𝜃𝜃 area, A
𝜃𝜃 = inclination of joint
𝑃𝑃 from the axis of block 𝑃𝑃= axial load of
𝑃𝑃 block
𝑃𝑃
𝑃𝑃𝑛𝑛
𝑑𝑑 𝑷𝑷 𝑷𝑷
𝑃𝑃𝑠𝑠
𝜃𝜃
𝑃𝑃 𝑃𝑃 𝑃𝑃
τ= = 𝜋𝜋
𝐴𝐴𝐴𝐴 𝑑𝑑 2
4 𝑃𝑃𝑆𝑆 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑃𝑃(𝑠𝑠𝑠𝑠𝑠𝑠2𝜃𝜃)
τ𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗 = = = :
b. Double Shear 𝐴𝐴𝑆𝑆 𝐴𝐴 2𝐴𝐴
- two identical areas resist the load 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑷𝑷(𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔)
𝑃𝑃/2 𝛕𝛕𝒋𝒋𝒋𝒋𝒋𝒋𝒋𝒋𝒋𝒋 =
𝑃𝑃 𝟐𝟐𝟐𝟐
𝑃𝑃/2
(expressed in terms of the cross sectional area
of the block A and the inclination of the joint)
𝑃𝑃 𝑃𝑃
τ= = 𝜋𝜋 2
𝐴𝐴𝐴𝐴 2 � 𝑑𝑑 � The block is under compression but shearing
4
(𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔) (sliding) may happen at the glued joint as shown.
𝑃𝑃/2
BEARING STRESS
𝑃𝑃/2 - A compressive stress (also a normal stress 𝝈𝝈) but
𝑃𝑃 its effect is external, i.e. at the contact surface
between the bodies in contact.
areas under shear - Also referred to as a “crushing or contact stress”.
c. Punching Shear
From a plate and rivet assembly,
𝑷𝑷
Punch projected area of the contact surface
(bearing area, Ab )
Plate 𝑑𝑑
𝑡𝑡
𝑡𝑡
𝑑𝑑
𝑃𝑃
1
Mechanics of Deformable Bodies: BASAEN, RV
𝑃𝑃 𝑃𝑃 determine the diameter of the smallest hole that can be
σ𝑏𝑏 = =
𝐴𝐴𝑏𝑏 𝑑𝑑𝑑𝑑 punched.
where 𝑷𝑷
Punch of diameter ′d′
𝐴𝐴𝑏𝑏 = 𝑑𝑑𝑑𝑑 (for the given plate)
Metal sheet of thickness ′t′
Note: 𝑡𝑡
Failure due to the bearing stress results in an 𝑑𝑑 compressed area, Ac
𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎, 𝐴𝐴𝑠𝑠
enlargement of the rivet hole.
Solution:
a) Maximum plate thickness, t:
Shear stress:
enlarged hole 𝑃𝑃 𝑃𝑃 𝑃𝑃
�τ = = � = 𝜏𝜏𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠ℎ : = 40:
𝐴𝐴𝑠𝑠 𝜋𝜋𝜋𝜋𝜋𝜋 𝜋𝜋(2.5)𝑡𝑡
𝑃𝑃 = 100𝜋𝜋(𝑡𝑡) → 𝑒𝑒𝑒𝑒𝑒𝑒 (1)
Example 1: Compressive stress:
What force is required to punch a 20 𝑚𝑚𝑚𝑚
𝑃𝑃 𝑃𝑃 𝑃𝑃
diameter hole in a plate that is 25 𝑚𝑚𝑚𝑚 thick? The shear �σ = = 𝜋𝜋 � = 𝜎𝜎𝑤𝑤𝑐𝑐 ∶ 𝜋𝜋 = 50:
strength of the plate is 350 𝑀𝑀𝑀𝑀/𝑚𝑚2 ? 𝐴𝐴𝑐𝑐 𝑑𝑑 2 𝑑𝑑 2
4 4
𝜋𝜋 2
Solution: 𝑃𝑃 = 50 � 25 � = 245.44 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘
4
𝑷𝑷
Substituting P into eqn (1),
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃ℎ 𝑜𝑜𝑜𝑜 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ′𝑑𝑑′
245.44 = 100𝜋𝜋(𝑡𝑡):
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 ′𝑡𝑡′
𝒕𝒕 = 𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕 𝒊𝒊𝒊𝒊 (𝒂𝒂𝒂𝒂𝒂𝒂. )
𝑡𝑡
𝑑𝑑 b) Minimum hole diameter, d=? if t = 0.25 in.
𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎, 𝐴𝐴𝑠𝑠
Shear stress:
The force requirement is to completely destroy the plate by 𝑃𝑃 𝑃𝑃
punching. Thus, the rquired stress is at least equal to the �τ = = � = 40: 𝑃𝑃 = 10𝜋𝜋𝜋𝜋
𝜋𝜋𝜋𝜋𝜋𝜋 𝜋𝜋(𝑑𝑑)0.25
shear strength of the plate. 𝜏𝜏 ≥ 𝜏𝜏𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠ℎ . Computing for → 𝑒𝑒𝑒𝑒𝑒𝑒 (2)
the minimum required force P.
Compressive stress:
𝑃𝑃 𝑃𝑃
�τ = = � = �𝜏𝜏𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠ℎ � 𝑃𝑃 𝜋𝜋
𝐴𝐴𝐴𝐴 𝜋𝜋𝜋𝜋𝜋𝜋 �σ = 𝜋𝜋 � = 50: 𝑃𝑃 = 50 � 𝑑𝑑 2 � → 𝑒𝑒𝑒𝑒𝑒𝑒(3)
𝑑𝑑 2 4
𝑃𝑃 𝑁𝑁 4
2
= 350 : 𝑃𝑃 = 549 779𝑁𝑁
𝜋𝜋(20)(25)𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚2 Equating (20 to (3):
𝜋𝜋
Or, the required force is at least about 10𝜋𝜋𝜋𝜋 = 50 � 𝑑𝑑 2 �:
4
P = 550 kN (Ans.) 𝒅𝒅 = 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖. (𝑨𝑨𝑨𝑨𝑨𝑨. )
Example 2: Example 3:
A circular hole is to be punched in a plate that has The lap joint loaded as shown consists of three rivets
a shear strength of 40 𝑘𝑘𝑘𝑘𝑘𝑘. The working compressive stress of uniform diameter, d = 20mm. The thickness of the plates
for the punch is 50 𝑘𝑘𝑠𝑠𝑠𝑠. (a) Compute the maximum is 25mm and the load applied is 50 kN and the plate has a
thickness of a plate in which a hole 2.5 𝑖𝑖𝑖𝑖𝑖𝑖ℎ in diameter width of 130 mm. (a) Compute the shear stress in a rivet,
can be punched. (b) If the plate is 0.25 𝑖𝑖𝑖𝑖𝑖𝑖ℎ thickness, (b) ) the average bearing stress in each plate, and (c) the
maximum average tensile stress in a plate.
2
Mechanics of Deformable Bodies: BASAEN, RV
Neglecting stress concentration, the maximum average
𝑃𝑃 tensile stress in the plate occurs on the hole nearest the
load as shown above.
𝑃𝑃 50 000
𝜎𝜎𝑡𝑡 = =
𝑑𝑑
(𝑤𝑤 − 𝑑𝑑)𝑡𝑡 (130 − 20)25
/ 𝑡𝑡
𝝈𝝈𝒕𝒕 = 𝟏𝟏𝟏𝟏. 𝟏𝟏𝟏𝟏 𝑴𝑴𝑴𝑴𝑴𝑴 (𝑨𝑨𝑨𝑨𝑨𝑨. )
𝑤𝑤
𝑃𝑃
Example 4:
A pulley assembly as shown transmits a twisting
moment of 𝑇𝑇 = 2200 𝑁𝑁 − 𝑚𝑚, at a shaft diameter of 60mm.
Solution: The key used has a length, Lk of 70mm and a thickness, t
a) Shear stress in a rivet of 10mm. (a) What is the key width, b, necessary if the
working shear stress is 60MPa? (b) Determine the bearing
𝑑𝑑 = 20𝑚𝑚𝑚𝑚 stress of the key.
𝑃𝑃
𝑃𝑃 Solution:
(a) Width, b = ?
The total load is equally shared among the three rivets
and thus, each rivet gets one-third of the total load.
𝑃𝑃 50 000
𝜏𝜏𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 𝜋𝜋 = 𝜋𝜋
𝑛𝑛 𝑑𝑑 2
4 3 � 202 �
4
𝝉𝝉𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓 = 𝟓𝟓𝟓𝟓. 𝟎𝟎𝟎𝟎 𝑴𝑴𝑴𝑴𝑴𝑴 (𝒂𝒂𝒂𝒂𝒂𝒂. )
b) Bearing stress in each plate
𝑃𝑃
𝑃𝑃 𝑇𝑇
𝐹𝐹
The three holes on the plate also equally share the + 𝑑𝑑
applied load. The bearing stress in the plate is then
𝑃𝑃 50000 𝐹𝐹
𝜎𝜎𝑏𝑏𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = =
𝑛𝑛(𝑑𝑑𝑑𝑑) 3(20)(25)
𝝈𝝈𝒃𝒃𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑 = 𝟑𝟑𝟑𝟑. 𝟑𝟑𝟑𝟑 𝑴𝑴𝑴𝑴𝑴𝑴 (𝑨𝑨𝑨𝑨𝑨𝑨. )
[Σ𝑀𝑀+ = 0]
c) Maximum average tensile stress in the plate 𝑑𝑑 2𝑇𝑇 2(2200 000)
The maximum tensile stress occurs at the location of 𝑇𝑇 = 𝐹𝐹 � � : 𝐹𝐹 = = :
the hole 2 𝑑𝑑 60
𝐹𝐹 = 73 333.3𝑁𝑁
𝑃𝑃
Shear stress in the key
𝐹𝐹 73 333.3
𝑤𝑤 𝑡𝑡 �𝜏𝜏𝑘𝑘 = � ⇒ 𝜏𝜏𝑤𝑤 : = 60
(𝐿𝐿𝑘𝑘 𝑏𝑏) 70(𝑏𝑏)
𝑑𝑑
3
Mechanics of Deformable Bodies: BASAEN, RV
𝒃𝒃 = 𝟏𝟏𝟏𝟏. 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒 (𝑨𝑨𝑨𝑨𝑨𝑨. ) Required rod diameter at B:
From
(b) Bearing stress in the key, 𝜎𝜎𝑏𝑏 = ?
𝑃𝑃 31 180
�𝜎𝜎𝑡𝑡 = 𝜋𝜋 � ⇒ 𝜎𝜎𝑤𝑤 𝑡𝑡 : 𝜋𝜋 = 100:
2
𝑑𝑑 𝑑𝑑 2
4 𝑟𝑟𝑟𝑟𝑟𝑟 4 𝑟𝑟𝑟𝑟𝑟𝑟
𝒅𝒅𝒓𝒓𝒓𝒓𝒓𝒓 = 𝟏𝟏𝟏𝟏. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 ≈ 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐(𝑨𝑨𝑨𝑨𝑨𝑨. )
𝐹𝐹 b) Shear stress at pin D:
Pin at D is under double shear
𝑅𝑅𝐷𝐷 52 990
𝜏𝜏𝑝𝑝𝑝𝑝𝑝𝑝 = 𝜋𝜋 = 𝜋𝜋
2 � 𝑑𝑑 2 � 2 � 202 �
𝐹𝐹 𝐹𝐹 73 333.3 4 4
𝜎𝜎𝑏𝑏 = = =
𝐴𝐴𝑏𝑏 � 𝑡𝑡 � (𝐿𝐿 ) �10� (70) 𝝉𝝉𝒑𝒑𝒑𝒑𝒑𝒑 = 𝟖𝟖𝟖𝟖. 𝟐𝟐𝟐𝟐 𝑴𝑴𝑴𝑴𝑴𝑴 (𝑨𝑨𝑨𝑨𝑨𝑨. )
2 𝑘𝑘
2
𝝈𝝈𝒃𝒃 = 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟓𝟓𝟓𝟓 𝑴𝑴𝑴𝑴𝑴𝑴 (𝑨𝑨𝑨𝑨𝑨𝑨. ) c) Thickness of bar BDC
Example 5: Using the largest force acting at bar BDC, the
A bell crank lever is pinned to point D with a diameter maximum bearing stress is then related to the
of 20mm and is loaded by end forces as shown. (a) working bearing stress. Thus,
Calculate the diameter of a rod (not shown) to be connected 𝑅𝑅𝐷𝐷 52 990
�𝜎𝜎𝑏𝑏𝐷𝐷 = � ⇒ 𝜎𝜎𝑤𝑤 𝑏𝑏 : = 80
to point B if the tensile working stress is 100 MPa. (b) What 𝑑𝑑𝑝𝑝𝑝𝑝𝑝𝑝 (𝑡𝑡) 20(𝑡𝑡)
is the shear stress required for the pin at D? (c) What is the
thickness required for the crank BDC if the working 𝒕𝒕 = 𝟑𝟑𝟑𝟑. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 (𝑨𝑨𝑨𝑨𝑨𝑨. )
bearing stress is 80 MPa.
Example 6:
𝑃𝑃
A joint is made by gluing two plywood gussets of
𝑃𝑃 thickness t to wood boards. The tensile working stresses are
𝐵𝐵
1200 psi for the plywood and 700 psi for the boards. The
working shear stress for the glue is 50 psi. Determine the
200𝑚𝑚𝑚𝑚 dimensions b and t so that the joint will be as strong as the
boards.
240𝑚𝑚𝑚𝑚
𝐶𝐶
𝑃𝑃
𝐷𝐷
60𝑜𝑜
30𝑘𝑘𝑘𝑘 𝑡𝑡
30𝑘𝑘𝑘𝑘
𝑏𝑏 1.75"
Solution: 𝑃𝑃
a) 𝑑𝑑𝑟𝑟𝑟𝑟𝑟𝑟 =? 5.75"
Force P:
[Σ𝑀𝑀𝐷𝐷 = 0] Solution:
𝑃𝑃(200) = 30𝑠𝑠𝑠𝑠𝑠𝑠60𝑜𝑜 (240): 𝑃𝑃 = 31.18𝑘𝑘𝑘𝑘 Stress analysis for the joint:
Reaction force at D:
[Σ𝐹𝐹𝑥𝑥 = 0]
𝑅𝑅𝐷𝐷 𝑥𝑥 = 30𝑐𝑐𝑐𝑐𝑐𝑐60𝑜𝑜 + 31.18: 𝑷𝑷 Tension on
plywood gussets
𝑅𝑅𝐷𝐷 𝑥𝑥 = 46.18 𝑘𝑘𝑘𝑘 𝑃𝑃
𝜎𝜎𝑝𝑝𝑝𝑝 =
[Σ𝐹𝐹𝑦𝑦 = 0] 2(5.75𝑡𝑡)
𝑅𝑅𝐷𝐷 𝑦𝑦 = 30𝑠𝑠𝑠𝑠𝑠𝑠60𝑜𝑜 : 𝑅𝑅𝐷𝐷 𝑦𝑦 = 25.98 𝑘𝑘𝑘𝑘 𝑷𝑷
Shear on the
glued joints
∴ 𝑅𝑅𝐷𝐷 = √46.182 + 25.982 𝜏𝜏𝑔𝑔𝑔𝑔 =
𝑃𝑃
: (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒)
2(5.75𝑏𝑏)
𝑅𝑅𝐷𝐷 = 52.99 𝑘𝑘𝑘𝑘
𝑷𝑷 Tension on
wood boards
𝑃𝑃
𝜎𝜎𝑤𝑤𝑤𝑤 =
(5.75)(1.75))
4
Mechanics of Deformable Bodies: BASAEN, RV
Tension on wood boards Thus, the maximum permisssible load P is only P =3681.6
𝑃𝑃 lb. The maximum torque or twisting moment that the
�𝜎𝜎𝑤𝑤𝑤𝑤 = � ⇒ 𝜎𝜎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑤𝑤𝑤𝑤
(5.75)(1.75)) coupling can transmit is:
𝑃𝑃
= 700: 𝑷𝑷
(5.75)(1.75))
𝑻𝑻 𝑷𝑷 𝑹𝑹
𝑃𝑃 = 7043.75 𝑙𝑙𝑙𝑙
+
( maximum permissible load for the boards) 𝑷𝑷
Tension on plywood gussets 𝑷𝑷
Using the above maximum load,
𝑃𝑃 [Σ𝑀𝑀+ = 0]
�𝜎𝜎𝑝𝑝𝑝𝑝 = � ⇒ 𝜎𝜎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑤𝑤 𝑝𝑝𝑝𝑝
2(5.75𝑡𝑡) 𝑇𝑇 = 𝑃𝑃𝑃𝑃𝑃𝑃: 𝑛𝑛 = 𝑛𝑛𝑛𝑛. 𝑜𝑜𝑜𝑜 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 = 4:
7043.75 𝑇𝑇 = 3681.6(3.5)4:
= 1200:
2(5.75𝑡𝑡) 𝑇𝑇 = 51 542.4 𝑖𝑖𝑖𝑖 𝑙𝑙𝑙𝑙:
Or,
𝒕𝒕 = 𝟎𝟎. 𝟓𝟓𝟓𝟓 𝒊𝒊𝒊𝒊. (𝑨𝑨𝑨𝑨𝑨𝑨. ) 𝑻𝑻 = 𝟓𝟓𝟓𝟓. 𝟓𝟓𝟓𝟓𝟓𝟓 𝒌𝒌𝒌𝒌𝒌𝒌 − 𝒊𝒊𝒊𝒊 (𝑨𝑨𝑨𝑨𝑨𝑨. )
Shear on the glued joints Example 8:
𝑃𝑃 7043.75 The figure shows a roof truss and the detail of the
[𝜏𝜏𝑔𝑔𝑔𝑔 = ] ⇒ 𝜏𝜏𝑤𝑤 𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗 : = 50: connection at joint 𝐵𝐵. Members 𝐵𝐵𝐵𝐵 𝑎𝑎𝑎𝑎𝑎𝑎 𝐵𝐵𝐵𝐵 are angle
2(5.75𝑏𝑏) 2(5.75𝑏𝑏)
sections with the thickness shown in the figure. The
working stresses are 70 𝑀𝑀𝑀𝑀𝑀𝑀 for shear in the rivets and
𝒃𝒃 = 𝟏𝟏𝟏𝟏. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐. (𝑨𝑨𝑨𝑨𝑨𝑨. )
140 𝑀𝑀𝑀𝑀𝑀𝑀 for bearing stress due to the rivets. How many
19 𝑚𝑚𝑚𝑚 diameter rivets are required to fasten the following
Example 7: members to the gusset plate: (a) 𝐵𝐵𝐵𝐵; and (b) 𝐵𝐵𝐵𝐵?
The halves of the coupling are held together by
four 5/8" diameter bolts. The working stresses are 12 𝑘𝑘𝑘𝑘𝑘𝑘 𝐷𝐷
for shear in the bolts and 15 𝑘𝑘𝑘𝑘𝑘𝑘 for bearing in the coupling.
Find the largest torque “𝑇𝑇” that can be safely transmitted by
𝐹𝐹
the coupling. Assume that the forces in the bolts have equal 𝐵𝐵
6𝑚𝑚
magnitudes. The bolts are located along a bolt circle of 3.5”
radius and the flanges of the coupling are 0.5” thick.
𝐻𝐻
𝐴𝐴
4𝑚𝑚 𝐶𝐶 4𝑚𝑚 𝐸𝐸 4𝑚𝑚 𝐺𝐺 4𝑚𝑚
𝑻𝑻
𝑷𝑷 96𝑘𝑘𝑘𝑘 96𝑘𝑘𝑘𝑘
200𝑘𝑘𝑘𝑘
𝑻𝑻 𝑷𝑷 𝑹𝑹
+
𝑷𝑷
𝑷𝑷
𝑻𝑻
Solution:
Considering shearing of bolts:
𝑃𝑃 𝑃𝑃
𝜏𝜏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 = 𝜋𝜋 ⇒ 𝜏𝜏𝑤𝑤 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 : ⇒ 12 000
� 𝑑𝑑 2 � 𝜋𝜋 5 2
4 � � � �
4 8
𝑃𝑃 = 3681.6 𝑙𝑙𝑙𝑙.
𝑃𝑃𝐵𝐵𝐵𝐵
𝑃𝑃𝐵𝐵𝐵𝐵
Considering bearing in the coupling flanges:
𝑃𝑃 𝑃𝑃
𝜎𝜎𝑏𝑏 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 = ⇒ 𝜎𝜎𝑤𝑤 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 : ⇒ 15 000
(𝑑𝑑𝑑𝑑) 5
�� � 0.5�
8
𝑃𝑃 = 4687.5 𝑙𝑙𝑙𝑙
5
Mechanics of Deformable Bodies: BASAEN, RV
Solution:
Forces:
Due to symmetry of loading, RA = RH = 196 kN.
Joint C is a special joint, from which 𝑃𝑃𝐵𝐵𝐵𝐵 = 96 𝑘𝑘𝑘𝑘.
The force on member BE can be more easily solved
using the Method of sections --
[Σ𝑀𝑀𝐴𝐴 = 0]
𝐵𝐵
𝑷𝑷𝑩𝑩𝑩𝑩
3𝑚𝑚 3 5
𝐴𝐴 4
4𝑚𝑚 𝐶𝐶 4𝑚𝑚
3 96𝑘𝑘𝑘𝑘
𝑃𝑃𝐵𝐵𝐵𝐵 � � 8 = 96(4): 𝑃𝑃𝐵𝐵𝐵𝐵 = 80 𝑘𝑘𝑘𝑘
5
a) Number of rivets for BC:
• Based on shear stress in rivets:
𝑃𝑃𝐵𝐵𝐵𝐵 96 000
𝜏𝜏𝐵𝐵𝐵𝐵 = 𝜋𝜋 ⇒ 𝜏𝜏𝑤𝑤 : 𝜋𝜋 = 70
𝑛𝑛 𝑑𝑑 2 𝑛𝑛 (19)2
4 4
𝑛𝑛 = 4.8 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 → 5 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
• Based on bearing stress in plate:
𝑃𝑃𝐵𝐵𝐵𝐵 96 000
𝜎𝜎𝐵𝐵𝐵𝐵 = ⇒ 𝜎𝜎𝑏𝑏 𝑤𝑤 : = 140
𝑛𝑛𝑛𝑛𝑡𝑡𝑝𝑝 𝑛𝑛(19)(6)
The thickness tp necessary should be the smallest
in the joint, i.e. tp =6mm
𝑛𝑛 = 6.01 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 → 7 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
Therefore, connecting BC to the plate requires 7
rivets.
𝒏𝒏𝑩𝑩𝑩𝑩 = 𝟕𝟕 𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓 (𝑨𝑨𝑨𝑨𝑨𝑨. )
b) Number of rivets for BE:
c) Based on shear stress in rivets:
𝑃𝑃𝐵𝐵𝐵𝐵 80 000
𝜏𝜏𝐵𝐵𝐵𝐵 = 𝜋𝜋 ⇒ 𝜏𝜏𝑤𝑤 : 𝜋𝜋 = 70:
𝑛𝑛 𝑑𝑑 2 𝑛𝑛 (19)2
4 4
𝑛𝑛 = 4.03 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 → 5 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
d) Based on bearing stress in plate;
𝑃𝑃𝐵𝐵𝐵𝐵 96 000
𝜎𝜎𝐵𝐵𝐵𝐵 = ⇒ 𝜎𝜎𝑏𝑏 𝑤𝑤 : = 140:
𝑛𝑛𝑛𝑛𝑡𝑡𝑝𝑝 𝑛𝑛(19)(13)
The thickness tp necessary should be the
smallest, tp =13mm
𝑛𝑛 = 2.3 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 → 3 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
Therefore, connecting BE to the plate requires 5
rivets.
𝒏𝒏𝑩𝑩𝑩𝑩 = 𝟓𝟓 𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓 (𝑨𝑨𝑨𝑨𝑨𝑨. )
