Statics review
• Example:
a) Find reactions at supports A and C.
Hinge connection
1300 N/m B
b) Find internal forces and moment, N, V, and M at
Y=1.0 m on column AB.
C
680 N.m
Roller
3.7 m
Y
X
A
2.75 m
1
Statics review
Global FBD
Solution a) Reactions at A , C
Hinge connection
- Draw the FBD
1300 N/m B
C
680 N.m
4 unknowns: MA , Ax , Ay , Cy
Cy
3 equilibrium equations for the global FBD:
M
A
=0,
F
x
= 0,
F
y
3.7 m
=0
Not enough equations for finding unknowns → draw FBD
for substructures!
Y
X
Ay
A MA
Ax
2.75 m
2
Statics review
- Draw the FBD for members AB and BC
Global FBD
Hinge connection
1300 N/m B
By
Bx
1300 N/m
B
C
Bx
By
Note: the internal
forces must be in
opposite directions!
Ay
680 N.m
Cy
FBD for BC
FBD for AB
3.7 m
C
Cy
3.7 m
680 N.m
Y
X
Ay
A MA
Ax
2.75 m
A MA
Ax
2.75 m
3
Statics review
- Equilibrium equations for BC:
 M B = 0 ➔ -680 + 2.75 Cy = 0 ➔ Cy = 247 N
BC  Fx = 0 ➔ Bx + Cx = 0 ➔ Bx = 0
 Fy = 0 ➔ -By + Cy = 0 ➔ BY = 247 N
For moment equation, CCW is assumed positive.
FBD for BC
By
B
C
Bx
680 N.m
Cy
2.75 m
4
Statics review
- Equilibrium equations for AB:
 M A = 0 ➔ -2405(23 × 3.7 ) + MA= 0 ➔ MA= 5932 Nm
AB  Fx = 0 ➔ Ax + 2405 = 0 ➔ Ax = -2405
 Fy = 0 ➔ By + Ay = 0 ➔ AY = -247 N
For moment equation, CCW is assumed positive.
FBD for AB
1300 N/m
Bx=0
Bx=0
By=247 N
2405 N
Note: statically equivalent concentrated forces for
triangularly distributed force:
By=247 N
≡
3.7 m
2
× 3.7 𝑚
3
Ay
A MA
Ax
Ay
A MA
Ax
5
Statics review
Global FBD
Solution b) Internal forces and moment, N, V, and M at
Y=1.0 m on column AB.
Hinge connection
1300 N/m B
- Cut a section through the AB at Y=1 m, draw created
FBD and write equilibrium :
M
AD
D
=0
➔ 𝑀 + 𝑀A + Ax 1 +
 Fx = 0 ➔ V= -Ax - 1
F
y
1
× 1300
2 3.7
= 0 ➔ N =-Ay= 247
1
× 1300
3.7
1300
1
1
3
M V
D
680 N.m
Cy
× 1 = 0➔ M= -3585 Nm
1 = 0 ➔ V=2229
3.7 m
Y
X
N
1m
1 1
×
2 3.7
C
Ay
A MA
Ax
2.75 m
Y
X
Ay
A MA
Ax
6
Shear and Normal Stress
• Example:
a) Find average shear stress in the plate
b) Find the average compressive stress in the punch.
punch
Plate
Solution a)
Shear area in the plate: circumference
of the hole times the thickness of the plate:
As = 𝜋𝑑𝑡 = 𝜋 20 8 = 502.7 𝑚𝑚2
Average shear stress in the plate:
𝑃
110 × 1000
𝜏𝑎𝑣𝑒𝑟 =
=
= 219 MPa
𝐴𝑠
502.7
𝐴𝑝𝑢𝑛𝑐ℎ = 𝜋𝑑 2 /4
As = 𝜋𝑑𝑡
Solution b)
Average compressive stress in the punch:
𝑃
110 × 1000
𝜎𝑐 =
==
= 350 MPa
𝐴𝑝𝑢𝑛𝑐ℎ
𝜋𝑑 2 /4
7