Wight 5 N High=10in 10 3.5 20 7 Width=1 in 40 16 20 9.9 10 Depth=5in 4 2 Calculation: Consider a block of rubber of length (L.), height (h) and thickness(t). If a force (F) is applied downwards at a the supper surface of the slab will be deflected by (d). Arca under shear stress L.t (m) Shear stress - F/L.t (N/m) When force F(N)-mass(Kg)*acceleration (m /sec") Neglecting any displacement due to bending which is very small the whole of the displacement due to shearing. The shear strain is equal to the angle expressed in radians. Please write detailed The solution because is small tan Modulus of Rigidity -G G. shear stress shear strain Lt-d Not the modulus of rigidity is some times expressed as G or N

Mechanics of Materials (MindTap Course List)
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Chapter5: Stresses In Beams (basic Topics)
Section: Chapter Questions
Problem 5.5.29P: A steel post (E=30×106) having thickness t = 1/8 in. and height L = 72 in. support a stop sign (see...
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Modulus of rigidity
Wight Deflection
5 N
High=10in
10
3.5
20
7
Width=1 in
40
16
20
9.9
10
6
Depth=5in
4
Calculation:
Consider a block of rubber of length (L), height (h) and thickness(t).
If a force (F) is applied downwards at a the supper surface of the slab will be deflected by (d).
Area under shear stress = L. t (m)
Shear stress F/L.t (N/m)
When force F(N)Fmass(Kg)*acceleration (m /sec)
Neglecting any displacement due to bending which is very small the whole of the
displacement due to shearing. The shear strain is equal to the angle expressed in radians,
Please write detailed The
solution
tan =
because e is small
tan d=
Modulus of Rigidity =G
shear stress
G
shear strain
G=LL F.
L-t-d
Not the modulus of rigidity is some times expressed as G or N
Mass on
Hanger M
(Kg)
Deformation
Force(N)
Modulus of
Rigidity
Shear
Shear
d (mm)
stress
strain
Transcribed Image Text:Wight Deflection 5 N High=10in 10 3.5 20 7 Width=1 in 40 16 20 9.9 10 6 Depth=5in 4 Calculation: Consider a block of rubber of length (L), height (h) and thickness(t). If a force (F) is applied downwards at a the supper surface of the slab will be deflected by (d). Area under shear stress = L. t (m) Shear stress F/L.t (N/m) When force F(N)Fmass(Kg)*acceleration (m /sec) Neglecting any displacement due to bending which is very small the whole of the displacement due to shearing. The shear strain is equal to the angle expressed in radians, Please write detailed The solution tan = because e is small tan d= Modulus of Rigidity =G shear stress G shear strain G=LL F. L-t-d Not the modulus of rigidity is some times expressed as G or N Mass on Hanger M (Kg) Deformation Force(N) Modulus of Rigidity Shear Shear d (mm) stress strain
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