Chapter 3.4, Problem 3.101P

(a)

To determine

The equivalent force at end A of the beam by replacing each loading with an equivalent force couple system.

Answer to Problem 3.101P

The equivalent force and couple at end A of the beam are as below,

For the beam (a), the equivalent force $\left(R_a\right)$ and couple $\left(M_a\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{1,000\text{N}\cdot \text{m}\left(\text{clockwise}\right)}$ respectively.

For the beam (b), the equivalent force $\left(R_b\right)$ and couple $\left(M_b\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{500\text{N}\cdot \text{m}\left(\text{anti-clockwise}\right)}$ respectively.

For the beam (c), the equivalent force $\left(R_c\right)$ and couple $\left(M_c\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{500\text{N}\cdot \text{m}\left(\text{clockwise}\right)}$ respectively.

For the beam (d), the equivalent force $\left(R_d\right)$ and couple $\left(M_d\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{1,100\text{N}\cdot \text{m}\left(\text{clockwise}\right)}$ respectively.

For the beam (e), the equivalent force $\left(R_e\right)$ and couple $\left(M_e\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{1,000\text{N}\cdot \text{m}\left(\text{clockwise}\right)}$ respectively.

For the beam (f), the equivalent force $\left(R_f\right)$ and couple $\left(M_f\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{200\text{N}\cdot \text{m}\left(\text{clockwise}\right)}$ respectively.

For the beam (g), the equivalent force $\left(R_g\right)$ and couple $\left(M_g\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{2,300\text{N}\cdot \text{m}\left(\text{anti-clockwise}\right)}$ respectively.

For the beam (h), the equivalent force $\left(R_h\right)$ and couple $\left(M_h\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{650\text{N}\cdot \text{m}\left(\text{anti-clockwise}\right)}$ respectively.

Explanation of Solution

(a)

Draw the free body diagram of the beam (a) as Figure (1).

Refer Figure (1).

Calculate the equivalent force at end A $\left(R_a\right)$ using the relation:

Consider the vertical equilibrium condition.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -300-200+R_a=0\\ R_a=500\text{N}(\downarrow )\end{array}$

Calculate the moment about A $\left(M_a\right)$ using the relation:

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ M_a-400-\left(200\times 3\right)=0\\ M_a=-1000\text{N}\cdot \text{m}\left(\text{clockwise}\right)\end{array}$

Thus, for the beam (a), the equivalent force $\left(R_a\right)$ and couple $\left(M_a\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{1000\text{N}\cdot \text{m}\left(\text{clockwise}\right)}$ respectively.

(b)

Draw the free body diagram of the beam (b) as Figure (2).

Refer Figure (2).

Calculate the equivalent force at end A $\left(R_b\right)$ using the relation:

Consider the vertical equilibrium condition.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ 300+200+R_b=0\\ R_b=-500\text{N}(\uparrow )\end{array}$

Calculate the moment about A $\left(M_b\right)$ using the relation:

$\begin{array}{l}{\displaystyle \sum M_A}=M_b\\ -400+\left(300\times 3\right)=M_b\\ M_b=500\text{N}\cdot \text{m}\left(\text{anti-clockwise}\right)\end{array}$

Thus, for the beam (b), the equivalent force $\left(R_b\right)$ and couple $\left(M_b\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{500\text{N}\cdot \text{m}\left(\text{anti-clockwise}\right)}$ respectively.

(c)

Draw the free body diagram of the beam (c) as Figure (3).

Refer Figure (3).

Calculate the equivalent force at end A $\left(R_c\right)$ using the relation:

Consider the vertical equilibrium condition.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -300-200+R_c=0\\ R_c=-500\text{N}(\downarrow )\end{array}$

Calculate the moment about A $\left(M_c\right)$ using the relation:

$\begin{array}{l}{\displaystyle \sum M_A}=M_c\\ 400-\left(300\times 3\right)=M_c\\ M_c=-500\text{N}\cdot \text{m}\left(\text{clockwise}\right)\end{array}$

Thus, for the beam (c), the equivalent force $\left(R_c\right)$ and couple $\left(M_c\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{-500\text{N}\cdot \text{m}\left(\text{clockwise}\right)}$ respectively.

(d)

Draw the free body diagram of the beam (d) as Figure (4).

Refer Figure (4).

Calculate the equivalent force at end A $\left(R_d\right)$ using the relation:

Consider the vertical equilibrium condition.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -500=R_d\\ R_d=-500\text{N}(\downarrow )\end{array}$

Calculate the moment about A $\left(M_d\right)$ using the relation:

$\begin{array}{l}{\displaystyle \sum M_A}=M_d\\ 400-\left(500\times 3\right)=M_d\\ M_d=-1100\text{N}\cdot \text{m}\left(\text{clockwise}\right)\end{array}$

Thus, for the beam (d), the equivalent force $\left(R_d\right)$ and couple $\left(M_d\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{-1100\text{N}\cdot \text{m}\left(\text{clockwise}\right)}$ respectively.

(e)

Draw the free body diagram of the beam (e) as Figure (5).

Refer Figure (5).

Calculate the equivalent force at end A $\left(R_e\right)$ using the relation:

Consider the vertical equilibrium condition.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ 300-800=R_e\\ R_e=-500\text{N}(\downarrow )\end{array}$

Calculate the moment about A $\left(M_e\right)$ using the relation:

$\begin{array}{l}{\displaystyle \sum M_A}=M_e\\ 400+1000-\left(800\times 3\right)=M_e\\ M_e=-1000\text{N}\cdot \text{m}\left(\text{clockwise}\right)\end{array}$

Thus, for the beam (e), the equivalent force $\left(R_e\right)$ and couple $\left(M_e\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{1000\text{N}\cdot \text{m}\left(\text{clockwise}\right)}$ respectively.

(f)

Draw the free body diagram of the beam (f) as Figure (6).

Refer Figure (6).

Calculate the equivalent force at end A $\left(R_f\right)$ using the relation:

Consider the vertical equilibrium condition.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -300-200=R_f\\ R_f=-500\text{N}(\downarrow )\end{array}$

Calculate the moment about A $\left(M_f\right)$ using the relation:

$\begin{array}{l}{\displaystyle \sum M_A}=M_f\\ 400-\left(200\times 3\right)=M_f\\ M_f=-200\text{N}\cdot \text{m}\left(\text{clockwise}\right)\end{array}$

Thus, for the beam (f), the equivalent force $\left(R_f\right)$ and couple $\left(M_f\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{200\text{N}\cdot \text{m}\left(\text{clockwise}\right)}$ respectively.

(g)

Draw the free body diagram of the beam (g) as Figure (7).

Refer Figure (7).

Calculate the equivalent force at end A $\left(R_g\right)$ using the relation:

Consider the vertical equilibrium condition.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -800+300=R_g\\ R_g=-500\text{N}(\downarrow )\end{array}$

Calculate the moment about A $\left(M_g\right)$ using the relation:

$\begin{array}{l}{\displaystyle \sum M_A}=M_g\\ 400+1000+\left(300\times 3\right)=M_g\\ M_g=230\text{N}\cdot \text{m}\left(\text{anti-clockwise}\right)\end{array}$

Thus, for the beam (g), the equivalent force $\left(R_g\right)$ and couple $\left(M_g\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{230\text{N}\cdot \text{m}\left(\text{anti-clockwise}\right)}$ respectively.

(h)

Draw the free body diagram of the beam (h) as Figure 8.

Refer Figure (8).

Calculate the equivalent force at end A $\left(R_h\right)$ using the relation:

Consider the vertical equilibrium condition.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -250-250=R_h\\ R_h=-500\text{N}(\downarrow )\end{array}$

Calculate the moment about A $\left(M_h\right)$ using the relation:

$\begin{array}{l}{\displaystyle \sum M_A}=M_h\\ 400+1000-\left(250\times 3\right)=M_h\\ M_h=650\text{N}\cdot \text{m}\left(\text{anti-clockwise}\right)\end{array}$

Thus, for the beam (h), the equivalent force $\left(R_h\right)$ and couple $\left(M_h\right)$ are $\underset{\_}{500\text{N}(\downarrow )}$ and $\underset{\_}{650\text{N}\cdot \text{m}\left(\text{anti-clockwise}\right)}$ respectively.

(b)

To determine

The loading which are equivalent.

Answer to Problem 3.101P

The loading in all beams are equivalent.

Explanation of Solution

Refer part (a) loading calculation.

The loading condition in the case (a) and (e) are equivalent.