Spring/Summer 2021
May 7th - Week 1
Circular beam with pin connections in reality
FBD
𝑀 =0
𝐹 =0
𝑅 𝐿 − 𝐹𝑎 = 0
𝑅 +𝑅 −𝐹 =0
𝑅 =
𝐹𝑎
𝐿
𝑅 =𝐹−
Simplify
𝑅 =𝐹−𝑅
Plug 1  2
𝐹𝑎
𝐿
𝑅 =
𝐹 𝐿−𝑎
𝐿
Equations we will use to determine deflection and slope:
Juvinall-Marshek Txtbk pg 208
2. Determine deflection and slope (by direct integration) – cont.
 Local Equilibrium Equations
𝑥
𝐹
𝑅
L−a
a
𝑅
𝑺𝒉𝒆𝒂𝒓 (𝑽)
𝑩𝒆𝒏𝒅𝒊𝒏𝒈 (𝑴)
𝟎 ≤ 𝒙 ≤ 𝒂:
𝑥
𝑉=𝑅 =
𝑀
𝐹 𝐿−𝑎
𝐿
𝑀=𝑅 𝑥=
𝐹 𝐿−𝑎
𝑥
𝐿
𝑉
𝑅
𝑉 =𝑅 −𝐹
𝒂≤𝒙≤𝑳
𝑥
𝑉=
𝐹
𝑅
𝑉
𝑀
𝐹 𝐿−𝑎
−𝐹
𝐿
𝐹𝑎
𝑉=−
𝐿
𝑀 =𝑅 𝑥−𝐹 𝑥−𝑎
𝑀=
𝐹 𝐿−𝑎
𝑥−𝐹 𝑥−𝑎
𝐿
𝑀 = 𝐹𝑎 1 −
𝑥
𝐿
2. Determine deflection and slope (by direct integration) – cont.
 Shear and Bending Moment Diagrams
𝐹
A
B
a
𝑉=𝑅 =
L−a
𝐹 𝐿−𝑎
𝐿
𝑉=−
𝐹𝑎
𝐿
2. Determine deflection and slope (by direct integration) – cont.
 Direct Integration
𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡:
𝑴 𝒅𝟐 𝜹
=
𝑬𝑰 𝒅𝒙𝟐
𝑆𝑙𝑜𝑝𝑒: 𝜽 =
𝒅𝜹
𝒅𝒙
𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛: 𝜹 = 𝒙
𝐹
𝑭𝒐𝒓 𝟎 ≤ 𝒙 ≤ 𝒂:
𝑭𝒐𝒓 𝒂 ≤ 𝒙 ≤ 𝑳:
𝑅
𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡:
𝑀 𝑑 𝛿 𝐹 𝐿−𝑎
=
=
𝑥
𝐸𝐼 𝑑𝑥
𝐿 𝐸𝐼
→ 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑡𝑜 𝑔𝑒𝑡 𝑠𝑙𝑜𝑝𝑒
𝜃=
𝑑𝛿 1 𝐹 𝑥 − 𝑎
=
𝑥 +𝐶
𝑑𝑥 2 𝐿 𝐸𝐼
→ 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑡𝑜 𝑔𝑒𝑡 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
𝛿=
1𝐹 𝐿 −𝑎
𝑥 +𝐶 𝑥+𝐶
6 𝐿 𝐸𝐼
𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡:
𝑀 𝑑 𝛿 𝐹𝑎
𝑥
=
=
1−
𝐸𝐼 𝑑𝑥
𝐸𝐼
𝐿
→ 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑡𝑜 𝑔𝑒𝑡 𝑠𝑙𝑜𝑝𝑒
𝜃=
𝑑𝛿 𝐹𝑎
1 𝐹𝑎
=
𝑥−
𝑥 +𝐶
𝑑𝑥 𝐸𝐼
2 𝐸𝐼
→ 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑡𝑜 𝑔𝑒𝑡 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
𝛿=
1 𝐹𝑎
1 𝐹𝑎
𝑥 −
𝑥 +𝐶 𝑥+𝐶
2 𝐸𝐼
6 𝐸𝐼
a
L−a
𝑅
2. Determine deflection and slope (by direct integration) – cont.
 Boundary and Continuity Conditions
Use Boundary Conditions and Continuity Conditions to solve for constants: 𝐶 , 𝐶 , 𝐶 , 𝐶
Boundary Conditions:
**No deflection can occur at the supports/pins, therefore, 𝜹 = 𝟎 when 𝒙 = 𝟎 and 𝒙 = 𝑳
Plugging equation 𝜹 𝟎 = 𝟎 into
and 𝜹 𝑳 = 𝟎 into
BC 1: 𝜹 𝟎 = 𝟎
BC 2: 𝜹 𝑳 = 𝟎
𝛿=
1𝐹 𝐿 −𝑎
𝑥 +𝐶 𝑥+𝐶
6 𝐿 𝐸𝐼
𝛿=
1 𝐹𝑎
1 𝐹𝑎
𝑥 −
𝑥 +𝐶 𝑥+𝐶
2 𝐸𝐼
6 𝐸𝐼
𝛿=
1𝐹 𝐿 − 𝑎
(𝑂) + 𝐶 0 + 𝐶 = 0
6 𝐿 𝐸𝐼
𝛿=
1 𝐹𝑎
1 𝐹𝑎
(𝐿) −
𝐿
2 𝐸𝐼
6 𝐸𝐼
𝐶 =0
+𝐶 𝐿 +𝐶 =0
2. Determine deflection and slope (by direct integration) – cont.
 Boundary and Continuity Conditions
Continuity Conditions:
Slope and deflection are continuous at 𝒙 = 𝒂.
We can therefore equate equations
and
;
and
Slope (equating Eq 1 and 3, and subbing in x=a)
1 𝐹 (𝑎) − 𝑎
𝐹𝑎
1 𝐹𝑎
(𝑎) +𝐶 =
𝑥−
(𝑎) +𝐶
2
𝐿 𝐸𝐼
𝐸𝐼
2 𝐸𝐼
Deflection (equating Eq. 2 and 4 and subbing in x=a)
1𝐹 𝐿 − 𝑎
1 𝐹𝑎
(𝑎) + 𝐶 (𝑎) =
𝑎
6 𝐿 𝐸𝐼
2 𝐸𝐼
Solve Eqs
−
1 𝐹𝑎
𝑎
6 𝐸𝐼
+ 𝐶 (𝑎) + 𝐶
simultaneously to determine 𝐶 , 𝐶 and 𝐶
2. Determine deflection and slope (by direct integration) – cont.
 Boundary and Continuity Conditions
Solving yields the following solutions:
Deflection
𝟎≤𝒙≤𝒂
𝛿=−
𝐹 𝐿−𝑎 𝑥
𝐿 −𝑥 − 𝐿−𝑎
6 𝐸𝐼 𝐿
=
𝐹 𝐿−𝑎 𝑥
(𝑥 + 𝑎 − 2𝐿𝑎)
6 𝐸𝐼 𝐿
𝒂≤𝒙≤𝑳
𝛿=−
𝐹𝑎 𝐿 − 𝑥
𝐿 − 𝐿−𝑥
6𝐸𝐼 𝐿
=
𝐹𝑎 𝐿 − 𝑥
(𝑥 − 2𝐿𝑥 + 𝑎 )
6 𝐸𝐼 𝐿
−𝑎
Slope
𝐹 𝐿−𝑎
6 𝐸𝐼 𝐿
𝟎≤𝒙≤𝒂
𝜃=
𝒂≤𝒙≤𝑳
𝜃=−
𝑎 − 2𝐿𝑎 + 3𝑥
𝐹𝑎
2𝐿 − 6𝐿𝑥 + 𝑎 + 3𝑥
6 𝐸𝐼 𝐿
Note that solutions to the general
equations for slope and deflection
of simply supported beams are well
known and already exist!
Visit page 851 in your textbook for
a helpful table!
A
𝑅
a
𝑅
𝑅
a
=
𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜 𝐹 𝑎𝑙𝑜𝑛𝑒
a
𝑅
𝐿
𝐿
𝑇𝑜𝑡𝑎𝑙 𝐵𝑒𝑎𝑚 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝛿
A
A
B
a
𝐹
𝐹
𝐹
𝐹
𝐿
+
𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜 𝐹 𝑎𝑙𝑜𝑛𝑒
…using Method of Superposition to determine deflection of simply supported beam
𝐹
A
𝑅
𝐹
A
a
a
𝑅
𝐿
𝐿
Applying equations from Problem 1 (also found in textbook)
𝟎 ≤ 𝒙 ≤ 𝒂𝟏
𝛿 =
𝐹 𝐿−𝑎 𝑥
(𝑥 + 𝑎 − 2𝐿𝑎 )
6 𝐸𝐼 𝐿
𝒂𝟏 ≤ 𝒙 ≤ 𝑳
𝛿 =
𝐹 𝑎 𝐿−𝑥
(𝑥 − 2𝐿𝑥 + 𝑎 )
6 𝐸𝐼 𝐿
𝟎 ≤ 𝒙 ≤ 𝒂𝟐
𝛿 =
𝐹 𝐿−𝑎 𝑥
(𝑥 + 𝑎 − 2𝐿𝑎 )
6 𝐸𝐼 𝐿
𝒂𝟐 ≤ 𝒙 ≤ 𝑳
𝛿 =
𝐹 𝑎 𝐿−𝑥
(𝑥 − 2𝐿𝑥 + 𝑎 )
6 𝐸𝐼 𝐿
…using Method of Superposition to determine deflection of simply supported beam
→ 𝛿
=𝛿 +𝛿
𝒂𝟏 ≤ 𝒙 ≤ 𝒂 𝟐 → 𝛿
=𝛿 +𝛿
→ 𝛿
=𝛿 +𝛿
𝟎 ≤ 𝒙 ≤ 𝒂𝟏
𝒂𝟐 ≤ 𝒙 ≤ 𝑳
𝟎 ≤ 𝒙 ≤ 𝒂𝟏
𝛿
=𝛿 +𝛿 =
𝐹 𝐿−𝑎 𝑥
𝑥 + 𝑎 − 2𝐿𝑎
6 𝐸𝐼 𝐿
=𝛿 +𝛿 =
𝐹 𝑎 𝐿−𝑥
6 𝐸𝐼 𝐿
𝑥 − 2𝐿𝑥 + 𝑎
+
𝐹 𝐿−𝑎 𝑥
(𝑥 + 𝑎 − 2𝐿𝑎 )
6 𝐸𝐼 𝐿
=𝛿 +𝛿 =
𝐹 𝑎 𝐿−𝑥
6 𝐸𝐼 𝐿
𝑥 − 2𝐿𝑥 + 𝑎
+
𝐹 𝑎 𝐿−𝑥
(𝑥 − 2𝐿𝑥 + 𝑎 )
6 𝐸𝐼 𝐿
+
𝐹 𝐿−𝑎 𝑥
𝑥 + 𝑎 − 2𝐿𝑎
6 𝐸𝐼 𝐿
𝒂𝟏 ≤ 𝒙 ≤ 𝒂 𝟐
𝛿
𝒂𝟐 ≤ 𝒙 ≤ 𝑳
𝛿
3 in
3 in
4 in
𝑀 =0
𝐹 =0
𝑅 −
𝑅 = 4𝑝
𝑀 +𝑀−
𝑝𝑑𝑥 = 0
𝑅 = 80 𝑙𝑏𝑓
𝑀 = −𝑀 +
𝑥𝑝𝑑𝑥 = 0
1
𝑥 𝑝
2
𝑀 = 160 𝑙𝑏𝑓 ∗ 𝑖𝑛
𝑝
𝑥
3 in
3 in
M
𝑀
4 in
𝑅
𝑺𝒉𝒆𝒂𝒓 (𝑽)
𝑩𝒆𝒏𝒅𝒊𝒏𝒈 (𝑴)
𝟎 ≤ 𝒙 ≤ 𝟑 𝒊𝒏:
𝑉= 𝑅
𝑀
𝑀 = −𝑀 + 𝑅 𝑥
𝑀
𝑅
𝑉
𝟑 ≤ 𝒙 ≤ 𝟕 𝒊𝒏:
𝑉=𝑅 −
𝑝
𝑝𝑑𝑠
𝑉 = 𝑅 − 𝑝(𝑥 − 3)
𝑀
𝑀 = −𝑀 + 𝑅 𝑥 − ∫
𝑥 − 𝑠 𝑝𝑑𝑠
𝑀 = −𝑀 + 𝑅 𝑥 − 𝑝 𝑥𝑠 − 𝑠
𝑀
𝑅
𝑉
𝑀 = −𝑀 + 𝑅 𝑥 − 𝑝
𝑥 − 3𝑥 +
𝑝
𝑥
3 in
3 in
M
𝑀
4 in
𝑅
𝑺𝒉𝒆𝒂𝒓 (𝑽)
𝑩𝒆𝒏𝒅𝒊𝒏𝒈 (𝑴)
𝟕 ≤ 𝒙 ≤ 𝟏𝟎 𝒊𝒏:
𝑉=𝑅 −
𝑝𝑑𝑠
𝑀 = −𝑀 + 𝑅 𝑥 − ∫
𝑥 − 𝑠 𝑝𝑑𝑠
𝑝
𝑉 = 𝑅 − 4𝑝
𝑀
𝑀 = −𝑀 + 𝑅 𝑥 − 𝑝 𝑥𝑠 − 𝑠
𝑀
𝑅
𝑉
𝑀 = −𝑀 + 𝑅 𝑥 − 𝑝 4𝑥 − 20
3. Equations needed to find deflection and slope by direct integration
𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡:
𝑴 𝒅𝟐 𝜹
=
𝑬𝑰 𝒅𝒙𝟐
𝑆𝑙𝑜𝑝𝑒: 𝜽 =
𝑭𝒐𝒓 𝟎 ≤ 𝒙 ≤ 𝟑 𝒊𝒏:
𝒅𝜹
𝒅𝒙
𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛: 𝜹 = 𝒙
𝑝
𝑥
3 in
𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡:
𝑑𝛿
1
1
= (−𝑀 𝑥 + 𝑅 𝑥 ) + 𝐶
𝑑𝑥 𝐸𝐼
2
Plugging equations 𝜹 𝟎 = 𝟎 and 𝜽 𝟎 = 𝟎 into
𝜃=
+𝐶 𝑥+𝐶
and
𝐶 =𝐶 =0
→ 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑡𝑜 𝑔𝑒𝑡 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
1
1
1
− 𝑀 𝑥 + 𝑅 𝑥
𝐸𝐼
2
6
4 in
Boundary Conditions:
𝜹 = 𝟎 and 𝜽 = 𝟎 when 𝒙 = 𝟎
→ 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑡𝑜 𝑔𝑒𝑡 𝑠𝑙𝑜𝑝𝑒
𝛿=
M
𝑅
𝑀 𝑑 𝛿
1
=
= (−𝑀 + 𝑅 𝑥)
𝐸𝐼 𝑑𝑥
𝐸𝐼
𝜃=
3 in
𝑀
1
1
(−𝑀 𝑥 + 𝑅 𝑥 )
𝐸𝐼
2
𝛿=
1
1
1
− 𝑀 𝑥 + 𝑅 𝑥
𝐸𝐼
2
6
:
3. Equations needed to find deflection and slope by direct integration – cont.
𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡:
𝑴 𝒅𝟐 𝜹
=
𝑬𝑰 𝒅𝒙𝟐
𝑆𝑙𝑜𝑝𝑒: 𝜽 =
𝑭𝒐𝒓 𝟑 ≤ 𝒙 ≤ 𝟕 𝒊𝒏:
𝒅𝜹
𝒅𝒙
Continuity Conditions:
Slope and deflection are continuous at 𝒙 = 𝟑 in
Use to determine constants 𝐶 and 𝐶 .
𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡:
𝑀 𝑑 𝛿
1
1
9
=
=
−𝑀 + 𝑅 𝑥 − 𝑝 𝑥 − 3𝑥 +
𝐸𝐼 𝑑𝑥
𝐸𝐼
2
2
→ 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑡𝑜 𝑔𝑒𝑡 𝑠𝑙𝑜𝑝𝑒
𝜃=
𝑑𝛿
1
1
1
3
9
=
−𝑀 𝑥 + 𝑅 𝑥 − 𝑝 𝑥 − 𝑥 + 𝑥
𝑑𝑥 𝐸𝐼
2
6
2
2
+𝐶
→ 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑡𝑜 𝑔𝑒𝑡 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
𝛿=
1
1
1
1
1
9
− 𝑀 𝑥 + 𝑅 𝑥 −𝑝
𝑥 − 𝑥 + 𝑥
𝐸𝐼
2
6
24
2
4
𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛: 𝜹 = 𝒙
+𝐶 𝑥+𝐶
3. Equations needed to find deflection and slope by direct integration – cont.
𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡:
𝑭𝒐𝒓 𝟕 ≤ 𝒙 ≤ 𝟏𝟎 𝒊𝒏:
𝑴 𝒅𝟐 𝜹
=
𝑬𝑰 𝒅𝒙𝟐
𝑆𝑙𝑜𝑝𝑒: 𝜽 =
𝒅𝜹
𝒅𝒙
𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛: 𝜹 = 𝒙
Use the same
process!
𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡:
𝑀 𝑑 𝛿
=
=⋯
𝐸𝐼 𝑑𝑥
→ 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑡𝑜 𝑔𝑒𝑡 𝑠𝑙𝑜𝑝𝑒
𝜃=
𝑑𝛿
=⋯
𝑑𝑥
→ 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑡𝑜 𝑔𝑒𝑡 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
𝛿=⋯
This can be done by hand - but becomes tedious quickly.
In applications, we typically use point loads (not distributed
loads) to represent roller bearings supporting a shaft unless
absolutely necessary. This makes it much easier to solve!
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